geometry-and-trigonometry

[dire jackal]

yeah

at your level if you see any trig ratio squared try to see if using any of the square identites work

since thats the only thing you can do with them

you havent been taught double angle formulas right, like cos(2x)?

[simple vigil]

no what is that

BTW my sister passed 12TH i saw her sheet of trigonometry identity i was shocked

around 20-25 or more identity

[dire jackal]

dont worry about it now(unless ur curious)

yeah its crazy

[simple vigil]

but most of them was of inverse trigo

[dire jackal]

there are a bunch of normal trigo too

[simple vigil]

in 10th its only 3 identity

that is making this chapter one of the hardest in boards

what about 11th and 12th

https://tenor.com/view/the-office-i-am-dead-inside-depressed-sad-apathetic-gif-15213089479568496581

[dire jackal]

these are just two pages from my formula book

(im in 12th)

[dire jackal]

again my point is just try those 3 identites and they should work out

[simple vigil]

looks very hard to remember

[worthy eagle]

||with undergraduate role||

[worthy eagle]

well if you are in cbse board, if a question comes to find cos60 cos45 cos75, will you use the formula from here or calculate it long way? I asked cuz many formulas aren't in book so I wonder if it will be ok to use them in exam

[dire jackal]

I'm pretty sure you gotta calc it the long way if the formula doesn't exist in the syllabus

[worthy eagle]

[edgy phoenix]

when I get sinx = 0 how do I find principle angle using unit circle?

[spiral lodge]

Well, you should know that sin(0) and sin(π) are 0

Then, all the other solutions are those values with a multiple of 2π added

Therefore, sin(x) = 0 has the following solutions:

x = 0 + 2kπ V x = π + 2kπ, where k in Z

[worthy eagle]

I think by principal angle, you mean the smallest angle in [0, 2π] that gives the given value of sin.

So which angles in [0,2π] give sinx=0? 0 and π. Since 0<π, principal angle is 0.

For using unit circle, sin x gives the y-coordinate of a point on circle. So you have to ask, which point on circle has y-coordinate 0? It's (1,0) and (-1,0). Which one comes first starting from (1,0) and moving counter clockwise? It's (1,0). So whatever angle (1,0) makes with +x-axis should be principal angle which is 0 rad.

[junior iris]

Struggling with refraction, watched countless videos. Is this correct? I don't think it is.

[ruby latch]

Those lines should be perpendicular to the side of the prism, you've drawn them very strangely

[junior iris]

The normal lines?

[ruby latch]

yes

[junior iris]

Don't they have to be 90?

[ruby latch]

Yes, 90 deg = perpendicular
and on your image they are clearly not perpendicular

[junior iris]

Is this correct?

[ruby latch]

Yep, this is much better
Here's what I've drawn

[junior iris]

Thank you so much!!, exactly what I needed.

[night cape]

hi, im confused, what do the trig functions yeild?

what answer do they give

is it the angle?

[maiden brook]

No, a real number

[spiral lodge]

Rather, their input is an angle (and if this is measured in radians, it is a real number) @night cape

[night cape]

ohhh and what do i do with this number?

[spiral lodge]

Wdym?

[night cape]

what do i do with the number i dont know how else to ask

[spiral lodge]

You do what you need to do

It's like asking: what do we do with the word "beautiful"?
Answer: you use it when it's useful/required

I really don't understand your question

[ripe bison]

It depends so heavily on what exact trig function you're dealing with. Super contextual, which is why the others are giving you a "vague" response. Each trig function has a different geometric interpretation

As for what to do with it, that just depends on the specific question that needs it, like Alberto said

[night cape]

ohhh interesting i see

then i will give the example Im workign with

[knotty quiver]

ratios

[night cape]

how do i use that number for a rotation of a vector

[ripe bison]

That, above all else, yes

[knotty quiver]

by using a rotation matrix?

[night cape]

yes

Now i know, from seeing, what the rotation matrix is

So what from what you all told me thus far

the resulting rotation from the functions are added to the starting vector to result a rotation

[knotty quiver]

resulting rotation from the functions?

[night cape]

yes

im so sorry if im not using the wright terms

i started getting into linear algerab and trig this week

[spiral lodge]

Have you not done trig in precalculus?

https://youtu.be/Ta8cKqltPfU?feature=shared

[spiral lodge]

By the way, are you probably referring to this?

[obsidian harness]

a key concept of linear algebra is the concept of basis vectors

so in the 2D plane, the standard basis vectors are $\hat{e_1} = (1, 0)$ and $\hat{e_2} = (0, 1)$

[somber coyote]

south

[obsidian harness]

if you multiply a vector by the rotation matrix, the columns tell you that $e_1$ goes to $(\cos \theta, \sin \theta)$ and $e_2$ goes to $(-\sin \theta, \cos \theta)$

[somber coyote]

south

[obsidian harness]

it's an exercise to show that $-\sin \theta = \cos(\theta + \pi/2)$ and $\cos \theta = \sin(\theta + \pi/2)$

[somber coyote]

south

[obsidian harness]

and then every vector $(x, y)$ is a linear combination of $e_1, e_2$, cause $(x, y) = x \cdot (1, 0) + y \cdot (0, 1)$

[somber coyote]

south

[obsidian harness]

so if you track where the basis vectors go after the linear transformation, you can just do x * (new basis 1) + y * (new basis 2) to find out where any vector goes

[night cape]

I found that video thank you :D

[night cape]

and yes i was :D

[night cape]

I have never done precalculus until last week :c

[fluid night]

Im teaching school maths next year, i went through a past exam paper and I swear to god I have no clue how to solve this question neatly. I came up with an answer but I think my working is using methods above this level (UK GCSE). Is there a trick to finding easy formula for the single final transformation that students would do at age 16?

like i feel ridiculous ive completed a degree in maths but i just have no clue how students at that low level are expected to get to an answer for this, like it has to be something quite simple for 3 marks but i only came up with complicated ways

[fluid night]

well i found one of biggest issues was at one step i got signs mixed up on x and y for rotation

[grave pond]

Here's what I would do: Draw triangle B in red, and then triangle C in blue -- the right-angled "handles" connecting the apex to (1,2) help eyeball the rotation.
Now, I know that since the directiions of A and C are not the same, the transformation from one to the other is either a rotation or a reflection, but it's clearly not a reflection, since (a) I notice that both of the two transformations A-B and B-C are orientation-preserving (but that might be slightly too highbrow for the target group), (b) both triangles have the same orientation: if I stand on the right angle looking toward the hypotenuse, the short leg is to my left.
So it's a rotation. If it's not visually clear what the center of the rotation is, then: Each point in triangle A moves to its corresponding point in triangle C along some circular arc. The straight line between the original and corresponding point is a chord in the circle, and therefore the center of rotation is somewhere along the perpendicular bisector of the chord. I've drawn two sets of chord and perpendicular bisector in different green nuances, sneakily chosen to make everything of interest happen at grid intersections. The two perpendicular bisectors intersect at (-5,2) (-4,1), so that must be the rotation center.

Except I botched the initial translation. 😭 Moment, new diagram coming up.

This is better:

[fluid night]

i would have to spend a while understanding why that works lol, i found the answer analytically using clockwise rotation transform equation, putting the start and end points in and doing coordinate change relative to the unknown centre of rotation, then solving for that centre of rotation using the start and end points

but i dont think thats how students would do it due to level of exam

so yours is probably the right one

[also as in, i did the analytic work after finding the first translation doing everything between A and the blue one]

[grave pond]

Hmm, my description was perhaps more convoluted than it really ought to be. What's going on is really just

• the rotation center has the same distance to P as to P'
• the rotation center has the same distance to Q as to Q'
  The two perpendicular bisectors are simply the loci of those two conditions.

[fluid night]

Yh makes sense cause ik you need it to be same distance from all points

i am not geometry brained i found it by DEFINITELY NOT this level of learning analytical method

[grave pond]

And the best thing is there are no calculations to get signs mixed up in, just drawing on the helpfully provided graph paper.

[fluid night]

yh signs i got mixed up was did CCW rotation instead of CW rotation when i did working out first time lol

i really need to revise geometry stuff i was never good at remembering stuff from that

i like swimming in algebra soup

[last palm]

hello!

can anyone help me understand reflections on graphs?

[hazy tree]

What are your questions?

[last palm]

how do i relfect y=x

i dont have any actual questions, im just trying to grasp the understanding of itr

[hazy tree]

With respect to which axis?

y=x is a straight line

a diagonal

if you reflect it with respect to the vertical axis

you'd get another straight line

[last palm]

ohh

[hazy tree]

that instead of going up from left to right, goes down from left to right

[last palm]

what about y=-x?

[hazy tree]

That's the reflection of x=y, with respect to the vertical axis

[last palm]

not understanding

[hazy tree]

let me draw it

[last palm]

ok!

[hazy tree]

just a second

[last palm]

so how would i reflect it?

like how would i work it out?

[hazy tree]

if you want a reflexion of y=f(x), with respect to the vertical axis, you have to plot y=f(-x)

like this

[last palm]

ohh

so we inverse it?

[hazy tree]

nooo

the inverse would be f^-1(y)=x

because f^-1(y)=f^-1(f(x))=x

this is the reflection of that curve with respect to the horizontal axis

y=f(x) is the original one, the reflection is y=-f(x)

with respect to the horizontal axis

[last palm]

ohh

but i have a question

[hazy tree]

say it

[last palm]

isnt the equation of the p(x) just the inveresed equation of h(x)

[hazy tree]

no

you just fliped the sign of the argument of the function

the inverse of a function is another concept

[last palm]

ohh okay

so thats all i have to do for those two relfection lines?

flip the sign?

[hazy tree]

with respect tho the horizontal and vertical axis

you could also reflect with respect to an axis that's just shifted up or down

left or right

or even in a diagonal

[last palm]

can i have an example

[hazy tree]

but you always reflect with respect to a straight line

[last palm]

oki

[hazy tree]

sure

[last palm]

ty

[hazy tree]

I've reflected the red curve with respect to the vertical orange line

like a mirror

[last palm]

is x=-10 the reflection line?

im very confused

[hazy tree]

imagine that there's a mirror at x=-10

[last palm]

oki

[hazy tree]

look that at x=-2, f(x)=0

and that's 8 units of lenght away from the mirror

to the right

you have to have the reflection at that distance from the mirror, but to the left

got it?

[last palm]

yes

so whats ever done from the orange line to the right has to be done from the orange line to the left?

but in a mirrored way?

[hazy tree]

yes

[last palm]

okay!

[hazy tree]

at the same distance from the mirror

[last palm]

alrighty

thanks a lot!

[hazy tree]

you're welcome

[cinder ledge]

Hello

I have a quick question

Can sin(x) be represented as a polynomial?

[lone panther]

it can

[jagged sonnet]

Do trig identities appear in calc AB

Or only unit circle type trig

[cinder ledge]

Is it Taylor series?

Or something

[cunning lion]

not unless you count infinite sums of polynomials as polynomials (which they aren't really)

[cinder ledge]

Oh ok!

So it would have an infinite amount of terms?

[cunning lion]

yes, and power series ("polynomials with infinite terms") can behave quite differently from polynomials

[cinder ledge]

Ah alright!

Thanks for the help you all!

[real socket]

Yeah I think so

Idk what I’m talking about but

[knotty quiver]

,w define Taylor series

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by usrcard

[cinder ledge]

Oh thanks!

[chrome glacier]

sinx doesn't show the properties to call itself a polynomial though

[grave pond]

In particular, it has infinitely many zeros, and the only polynomial that has that is the zero polynomial (which clearly isn't the sine function).

[simple vigil]

should i call tan 90 infinity or not defined from exam pov

[grave pond]

If you’re setting the exam, strive to ask questions in a way where the difference won't matter.
(Unless your point is to enforce making the right terminology choice, but then you wouldn't be asking...)

[strange reef]

Is the convex hull of two line segments in R^3 a (possibly degenerate) parallelepiped?

[grave pond]

In general it's an irregular tetrahedron.

Same as the convex hull of four points; the interior of the line segments don't add anything to the convex hull anyway.

[strange reef]

Ah, of course. Thanks

[granite umbra]

is this the math2/3 chat?

[maiden brook]

But how would u know for example sin(x) defined on [0, 2pi] can’t be expressed as a polynomial

[maiden brook]

oh nvm

[real socket]

😮

[grave pond]

Yeah, that requires a more intricate argument. For example that all the higher derivatives of sine exist and are nonzero, whereas if you differentiate a polynomial sufficiently many times you get an everywhere zero function.

[maiden brook]

👍

[grave pond]

On the other hand, thanks to the Weierstass approximation theorem there are polynomials whose values are very close to the sine over the entire interval [0,2pi].

[cinder ledge]

So...?

[maiden brook]

Would that not be an “obvious” fact? Can’t u get a polynomial to approximate very well basically any smooth function over an interval

[grave pond]

Not equal, but approximate arbitrarily well.

"Obvious" is in the eye of the beholder. It doesn’t feel obvious to me that one can do that, even though it's true.

[maiden brook]

I feel like it seems easier to approximate with polynomials than with other functions eg trig ones

[stiff spoke]

Anyone know héros formula

[cinder ledge]

Herons?

[stiff spoke]

Yes

[cinder ledge]

Oh

How to do latex

Oh

Uh

Mb

What

[somber coyote]

ILoveCalculus

[cinder ledge]

There we go

,,Where~s = \frac {a+b+c}{2}

[somber coyote]

ILoveCalculus

[cinder ledge]

FINALLY

[turbid vortex]

Called Hero’s too :)

[cinder ledge]

Oh

[jagged sonnet]

How did you guys become more fluent/ quick with your trig before calculus?

[shadow yacht]

how resolve?

[shadow yacht]

I just practiice exercises every single day, I spend lik 30 minutes to 2 hours

[jagged sonnet]

Yeahhh I should probably do that

Do you use chat gpt to generate problems?

[shadow yacht]

i prefer deepseek or gemini, I use prompts to make specific difficult levels of exercise

[jagged sonnet]

hmm can you give me an example of a prompt you'd enter?

I'm new to using AI to study I def wanna start though

[shadow yacht]

but I see a guy in discussions said the gemini can be switched, so

[shadow yacht]

yeah, i´ll get, one sec

"Use clear, direct language and avoid complex terminology. Aim for a Flesch reading score of 80 or higher. Use the active voice. Avoid adverbs. Avoid buzzwords and instead use plain Germany. Use jargon where relevant. Avoid being salesy or overly enthusiastic and instead express calm confidence."

I use it to german exercises, but u can make some mods to math!

[dark sparrow]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[jagged sonnet]

I see I see

Thank you

oh

okay so I will not be using AI to study

[shadow yacht]

i dont use GPT, but I think the others AI are more good, i dont use it to generate answers too

[dark sparrow]

all AIs suffer from the same flaws.

[shadow yacht]

I see

sorry for something about it, I used it so many times-

[hoary totem]

do smth that involves a lot of trig

[old swift]

what does that even do

[dark sparrow]

prints the factoid you see just below my msg

[old swift]

oh i thought-

gpt

chat gpt

nevermind

[warm shuttle]

I generally don't trust humans either

[old swift]

fair enough

[granite umbra]

more like
trigwall```

[whole imp]

trigonometry is just memorizing stuff for the most part

[delicate chasm]

fr

[delicate chasm]

are u in high school?

[visual flume]

help pls

aaaaaaaaaaaaaa ;-;

[whole imp]

yes

[whole imp]

and no

I dropped out :p

You will encounter trigonometry for the rest of your life... some hate it, some love it

it's a necessary evil nonetheless

[worthy eagle]

have you tried anything?

[whole imp]

@visual flume use #1021175428326633542

I can help you out...

tho

[visual flume]

ok what the sigma

i got 25.5√2+25.5

definetely wrong

[whole imp]

I just imagined a triangle from the dot's position

[visual flume]

triangle?

i did this

most likely wrong

…

[zealous pike]

The small circle has radius R₁ and the bigger circle radius R₂, both tangent at point Q. Line s and r are tangent in both circles. Find the distance between P and Q

I tried taking similar triangles but failed miserably

The hypotenuse of the smaller triangle would be x + 2R₁ and that of the bigger one, x + 2R₁ + R₂

I dunno y I failed to find x

[worthy eagle]

PT1C1 is similar to PT2C2
T1C1/T2C2 = PC1/PC2

C1, C2 are centres, T1, T2 are points where tangents touch the circles. I think you can find x in terms of R1, R2 from here.

[marble oriole]

First day on geometry

Yo this is so sick guys

Bro screw learning algebra first

[zealous pike]

basically, $\frac{R_{1}}{R_{2}} = \frac{x + R_{1}}{x + 2R_{1} + R_{2}}$

y

[marble oriole]

Cuz no

[worthy eagle]

yes

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

I tried but failed smh

[worthy eagle]

what x u get from here?

,w y/z = (x+y)/(x+2y+z), find x

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by shubham1029

[zealous pike]

Yea I got that one

[worthy eagle]

x = 2R1²/(R2-R1)

PQ = x+2R1

Is it not the answer

[zealous pike]

$\frac{2R_{1}R_{2}}{R_{2} - R_{1}}$

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

that's the answer

thx

[whole imp]

It is

wrong

This actually matches a common application for the quadratic formula since you are given two possible scenarios for R

just get the distance of (R_1, R_2) to (24, 27)

$\text{Using the distance formula, we get } \sqrt{(R_1 - 24) + (R_2 - 27)} = R$

Since R is just the distance of (R_1, R_2) to (24, 27), it works out well

now just rewrite that expression as a polynomial with a degree of 2

[zealous pike]

I have another exercise that I didn't understand at all

[whole imp]

Something like R^2 - 102R + 1305 = 0

then apply the quadratic

[zealous pike]

Its translation is as follows:

Given perpendicular lines r and s that intersect each other in the plane at point (3, 3), and r intersects the x axis at (2, 0), find the area of the triangle bounded by the two lines and the x axis.

Is it possible to solve it with this information?

[somber coyote]

Andy

[worthy eagle]

yes

you need to know at which point s intersects x axis

[zealous pike]

how do I find it?

[worthy eagle]

try to find equation of line s with the given information in y=mx+c form then put y=0 and you get x=-c/m

[zealous pike]

but to find the slope wouldn't I need a second "x" for s? I only have one pair of points for s

[worthy eagle]

hints: y-y1=(y2-y1)/(x2-x1), m1 m2 = -1

[worthy eagle]

||you can start by finding equation of line r||

[warm shuttle]

pls plot

becomes much easier if you can see the picture

[whole imp]

You need a tangent slope?

[zealous pike]

wait a min

I have that the equation for line r is 3x - 6

[warm shuttle]

That is correct

[zealous pike]

The line perpendicular to r would be the decreasing version of r, wouldn't it?

I wonder if x is perpendicular to -x? for example

[warm shuttle]

y = x is indeed perpendicular to y=-x

but you can't just swap the sign in general

[zealous pike]

yea

but to be a decreasing function, a needs to necessarily be < 0

[whole imp]

I get it now

[zealous pike]

At this point I'm sure I skipped something really important on linear functions on stewart's precalculus book

[warm shuttle]

yep, if i remember correctly for this line representation, you flip the sign, and then find the multiplicative inverse for a

[zealous pike]

-x/3 + 6?

[whole imp]

Is s = -r?

[warm shuttle]

Well + c

-x/3 + c

c moves it around

You also need to satisfy (3,3) intersection

[zealous pike]

OH OHOHOH

I UNDERSTANDIT

-x/3 + 4

[warm shuttle]

Yeah

[zealous pike]

then 0 = -x/3 + 4
x = 12

now comes the sick part that is calculating the distances

[warm shuttle]

Also, this is something that's bothering me personally, but i would've liked if you wrote either
y = mx + c
or
ax + by + c = 0
to represent lines

[whole imp]

Man, this got confusing real quick

I was just thinking of getting the first tangent

[warm shuttle]

You don't actually need to do that

Not through pythagorean formula at least

You just need to find the base and height of your triangle

[zealous pike]

the base and height aren't the distances between (0, 2), (3, 3) and (12, 0), (3, 3) for r and s, respectively?

[zealous pike]

answer is 15

[hoary totem]

its fun

[gray osprey]

Can anyone help with trig question

It’s quite basic

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[hoary totem]

just ask

[tribal plank]

geometry dash

[simple vigil]

how to solve this question

if 1+sin^2A = 3sinAcosA find cotA

[dire jackal]

convert sin^2(A) to 1-cos^2(A) and divide both sides by sin^2(A)

[summer cradle]

divide both sides by cos²(A)

to make in terms of tan(A)

[dire jackal]

yeah that works too

[summer cradle]

convert sec²(A) in the LHS to 1+tan²(A)

solve for tan(A) then cot(A)=1/tan(A)

[simple vigil]

can someone show me in paper

[simple vigil]

it will give sec^2A and tan ^2A

[dire jackal]

@simple vigil

[simple vigil]

ok thanks

but i have doubt how to think like this in exam

where time pressure is there

[dire jackal]

well since you needed to find cotA, i looked for ways in which i could change the terms in the given equation to tan and sec^2 (since sec^2x=1+tan^2x.) you see that dividing by cos^2x gives you that

[simple vigil]

so first use identity if any works

if not then divide by a ratio to convert

[dire jackal]

ye

youll see what you need to do in equation with practice

[simple vigil]

i practise around 10 a day

long

[dire jackal]

you understood what i did to 3sinAcosA right

[dire jackal]

nice

[simple vigil]

and 20-25 mcqs

[dire jackal]

oki

[dire jackal]

since you asked then deleted

[simple vigil]

ya but i understood we divided whole eq by cosA

not only one side

[dire jackal]

yess

[simple vigil]

let me go i have 20 more question in my assignment

that i have to submit tomorrow

[dire jackal]

alr see ya

[short lynx]

what does "A vertical stretch by a factor of 4" mean?

[tepid jolt]

@short lynx "A vertical stretch by a factor of 4" means that every y-value (height) on the graph of the function is multiplied by 4, making the wave four times taller compared to the original.

For example, the original sine function
𝑦

sin
⁡
(
𝑥
)
y=sin(x) oscillates between -1 and 1. After a vertical stretch by a factor of 4, the new function becomes:

𝑦

4
sin
⁡
(
𝑥
)
y=4sin(x)
Now, instead of oscillating between -1 and 1, it oscillates between -4 and 4. The peaks are four times higher, and the troughs are four times lower.

In the graph you shared, this means the blue curve's amplitude (the height from the centerline to a peak) is four times the amplitude of the black sine curve.

If you combine that with the vertical shift up by 3 units, the function becomes:

𝑦

4
sin
⁡
(
𝑥
)
+
3
y=4sin(x)+3
This means the whole graph is also shifted upward by 3 units on the y-axis, so the midline moves from
𝑦

0
y=0 to
𝑦

3
y=3.

[grave pond]

Do not attempt to answer questions by copy-pasting from an AI bot response.

[short lynx]

ChatGPT work

wild

[grave pond]

"Vertical stretch by a factor of 4" would mean to multiply all of the y-coordinates by 4. Every vertical distance in the diagram becomes 4 times as large, whereas the horizontal distances stay the same.

In this case I think the problem author has expected the stretch to leave the x-axis in place, even though you have already moved the curve away from the x-axis. So after the stretch, the marked red and blue points on the curve end up with y=12 instead of the y=3 they had after the shifting.

[short lynx]

ohh alright so then the equation would be like y=4sin(x)+3?

right?

[grave pond]

No, y=4(sin(x)+3)

[short lynx]

isnt that the same thing

here, y=-2?, is 3 like the vertical shift up?

oh wait the amplitude

(max-min)/2=amplitude?

[simple vigil]

which class Q is this?

[grave pond]

No -- for example for x=0, we have 4sin(x)+3 = 4·0 + 3 = 3 but 4(sin(x)+3) = 4·(0+3) = 4·3 = 12.

[short lynx]

Ohh

[short lynx]

precalc

[zealous pike]

Find the values of theta such that: 14sencos + 7(sen² - cos²) = 0, for theta in [0, π]

Any hints to solve this?

[spiral lodge]

Do you recognise the double angle formulas for sine and cosine?

Hint: 14 = 7•2

[zealous pike]

No

[spiral lodge]

At least, do you know them?

[zealous pike]

Oh

No but just looked uo

7×2sincos +7sin² - 7cos²

7×sin(2Θ) + 7(sin² - cos²)

7sin(2Θ) + 7(-(cos² - sin²))

sin(2Θ) - cos(2Θ) = 0

[spiral lodge]

Awesome

So you have sin(2θ) = cos(2θ)

Now, I would use the fact that cos(a) = sin(π/2 - a)

[ocean isle]

Yo I'm a new to trigonometry highschooler, I'm tryna advanced learning ahead of others in my batch. Is there any advice you would give me for learning this subject

[spiral lodge]

So that you have:
sin(2θ) = sin(π/2 - 2θ)

Other ways could be writing the equation as R•sin(...) = 0 or dividing by cos to get tangent

[knotty quiver]

me, i would put the cosine to the RHS

[spiral lodge]

For the latter beware the fact that you can't divide by 0, so ...

[spiral lodge]

Yeah sure

[zealous pike]

In this case wouldn't it be π - 2Θ?

[spiral lodge]

Nope

[knotty quiver]

oh my, my vision is so bad i thought the = sign was -

[spiral lodge]

Because I used this with a = 2θ @zealous pike

[zealous pike]

Ohh

Ix

Ic

Soo

How do I

Find the values that satisfy this?

[spiral lodge]

Of course 😅

That's what solving an equation means

[zealous pike]

It was given that Θ is in the interval [0, π]

sin(0) = sin(π)

No

[knotty quiver]

0≠1

[knotty quiver]

this is your answer?

[zealous pike]

No

Hey

[knotty quiver]

hey

[knotty quiver]

you were here

[knotty quiver]

(or here)

[zealous pike]

Is there any value that satisfies this?

[spiral lodge]

Sure, there are infinite values

sinA = sinB
iff
B = A + 2kπ V B = π - A + 2kπ

Then you have to choose the ones belonging to your interval

[zealous pike]

What's k?

[knotty quiver]

i would simplify it first to find theta

[knotty quiver]

practice deriving the trig identities

[zealous pike]

π/8 is one of the values

[ocean isle]

I'm still stuck on the law of sines after learning Pythagoras 💔

this you mean?

[knotty quiver]

yep

ngl i still don't know how the law of sine is derived

but i know for law of cosine

[spiral lodge]

An Integer...
Haven't you been taught how to solve trig equations? 🤔

[zealous pike]

No, I'm self studying

[knotty quiver]

it looks a lot but it's simpler than you thought. they're just derived from one another, for most of them

,w plot 14sinxcosx+7(sin²x-cos²x) from 0 to π

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by usrcard

[simple vigil]

how you are guys are so good at maths

what should i do such that trigo becomes easy for me

[maiden brook]

Area of a triangle in three ways

[simple vigil]

i know 2

1/2 x b x h

s = a+b+c/2
area = underroot s(s-a)(s-b)(s-c)

[maiden brook]

No I mean there’s a particular formula u can use in three ways that quickly gets u the law of sines

[zealous pike]

-2Θ + π/2 = 2Θ + 2kπ

-2Θ = -π/2 + 2Θ - 2kπ
-4Θ = -π/2 - 2kπ
Θ = (π/2 + 2kπ)/4

?

For B = π/2 - 2Θ and A = 2Θ

[upper karma]

Stuck at this , anyone please help

[ocean isle]

Oh thank you, Im just not familiar so I guess it's natural to feel alot

[zealous pike]

π/8 + 5π/8 = 6π/8 = 3π/4

Yes THATS THE ANSWER

[zealous pike]

thx BRO

[ocean isle]

what does derives even mean tbh 🥀

[knotty quiver]

like to transform from one form to another

[knotty quiver]

for example, you see the Pythagorean identities? any idea why they're called that?

[knotty quiver]

does substituting 3π/4 to the equation gives 0?

[zealous pike]

No

The values that satisfy the equation are π/8 and 5π/8

Only

Meow

[warm shuttle]

Honestly the coolest proof i know is using radius of the circumcircle

See https://en.wikipedia.org/wiki/Law_of_sines section Relation to the Circumcircle, Proof

[dark sparrow]

gonna share this thing i made here if anyone ever wants to get a good visual for trig values

this is a recreation of my own thought process when calculating values like these

[dark sparrow]

<@&268886789983436800> may i ask for this to get pinned

[dark sparrow]

yay!!!

[weary drift]

trig sheet 2: electrig boogaloo

[whole imp]

Is linear algebra overrated

[whole imp]

I just memorized a reference value: pi/4 rad = 45 deg

[vocal verge]

No it isn't

[whole imp]

Nic

e

[fresh anchor]

Yo

[marble oriole]

Yo guys I got a question

So I started geometry yesterday right

Couldn’t understand what the book was telling me when it was trying to explain how to create a 90 degree angle

So I did it myself

And then learned that you can keep splitting angles right

Did I do good

[zealous pike]

The book said that an identity is an expression that'd be true for all values of x (in this case, x = theta). While reading this chapter I've come up with a thought: is this really an identity? It may indeed work for all values of theta, but if theta = pi/2, won't this be undefined? Does it "work" even if the function is undefined at some value?

[zealous pike]

hey, awesome work

[tough lark]

the correct way of saying it may be that it works for all values of theta in the domain of the function

pi/2 is not in secant's domain, nor in the other expression's domain

so it doesnt matter

[zealous pike]

I see. Sorry for the dumb question and thank you for the clarification

[tough lark]

basically all computational or early math textbooks are very handwavy with domains of functions so your confusion makes sense

not a dumb question at all

[willow kiln]

nope, it's used just about everywhere in "university" math and in algorithms such as those for computer graphics, machine learning, digital audio processing (FFT)

in case you wanted a serious answer lol

[ocean isle]

Cause they're used in trying to find something that they all have relation to with each other?

Idk how to put into words

[obsidian harness]

it's cause this literally comes from the Pythagorean theorem

okay, so for this part I'm assuming you're familiar with the definitions of csc, sec, and cot

starting from sin^2 t + cos^2 t = 1

what happens if you divide everything by cos^2 t?

[dark sparrow]

that's a conversion from degrees to radians, not really directly related to what i wrote here

[zealous mountain]

I have sucessfully remember the value of the angles of sin cos tan cosec sec cot, anyone have exercise for me to do?

[dim jay]

Is anyone is their who knows integration

Class 12 questions

[simple vigil]

yes i like to do is ask your friend to ask you value at any point

for faster recalling

[noble swan]

Does anyone know how to solve a not really a midpoint of a line segment?

[simple vigil]

means you want to prove that a line segment has 1 mid point only?

[radiant sparrow]

if its not the midpoint, its just some other point along the line. use the section formula with the ratio they give u

[upper karma]

The angles are in a pattern, i only remember sinx° and tanx°, and get every else by the pattern

The pattern is for 0°,30°,45°,60° and 90

But am pre sure theres a pattern for ur table too

[visual flume]

isnt this like the golden ratio or smt

like 1+√5/2

[orchid storm]

the standard way of constructing a right angle is something like this

[marble oriole]

Yeah I know that now

🤣🤣🤣

[marble oriole]

Okay what about getting a 90 degree angle from any point like this

[orchid storm]

yeah thats how you do it

[marble oriole]

Oh I see

I’m trying to get a parallel line from the line

Not trying to get 90 degrees rn actually

But don’t tell me

I’m trying to figure this out on my own

This is incredibly difficult

Nvm

don’t look parralel to me tho 🤣🤣

We out here son

Definitely not the conventional way to do it

Dude what am I

[queen shadow]

Yup

I think I do

[marble oriole]

Ah so this is how Descartes made his coordinate plane

Makes sense

Ah my mind

[zealous pike]

Are u using a compass?

[queen shadow]

No I think it'll be 2/1

[visual flume]

huh

wait howd u get that

[marble oriole]

Yes

He actually did it a different way I came to find out

He started from algebra

[zealous pike]

I was just wondering how your circles were so perfect

[marble oriole]

Look at this thing

What a beast of a compass

[zealous pike]

indeed

[visual flume]

ok what did i do wrong here

obviously 3√5 is a valid answer but if u think about it aops wouldnt require u to solve for a radical

sooooo…

paefup0a39urp49t8gsof;dk'pfs[[[eiruhgdfjo;;;rsusiw3yapdfiojglke

老天保佑金山银山全都有

[marble oriole]

What is even the question

[upper echo]

2sqrt(7) is not valid, bc it is less than 6.
So, you got extra roots when squaring the first line.
LHS must be positive.

[queen shadow]

Well I do have a method to do it

A little different from this

I used similarity of triangle for that

[queen shadow]

Well using properties of circle I proved that MN is equal to CN and NQ is half of CN

[final terrace]

hy everybody

[visual flume]

help

(again)

[upper echo]

draw trapezoid, use its midline and properties of intersecting chords

[quick tangle]

Guys, I'm taking Geometry for my freshman year of highschool. Any tips?

[quick tangle]

Ping me with any tips if you have any.

[obsidian harness]

1. Always draw a diagram if one is not provided
2. Know the formal names of geometrical objects and theorems: in maths you always need a reason for why something is true
3. Understand why the non-obvious theorems are true: don't just memorise them!
4. Draw an extra line or something if you can't figure out how to start with a problem
5. You got this! Good luck!

[dark sparrow]

@quick tangle im gonna expand a bit on point (2) from south -- there's quite a lot of important vocab:

• nouns (straight line, segment, ray, angle, triangle, bisector, trapezoid, square, circle, ...)
• adjectives (parallel, isosceles, right, acute, obtuse, supplementary, interior, inscribed, ...)
• verbs (to construct, to drop [a perpendicular], to draw [a line through something], ...) -- these are smaller in number but it'll still help you a lot if you can explain what you do properly.

[spark yacht]

Old school

[ionic zodiac]

Also get a good compass if the course is heavy on construction

[simple vigil]

what makes a compass good BTW

this is the one i use

[obsidian harness]

all of them are equally good

as long as it doesn't break or something

[simple vigil]

that depends on your frustration while doing maths

[obsidian harness]

some main theorems (this is not an exhaustive list):

• logical reasons: given, substitution property of equality, transitive property, addition/subtraction/multiplication/division property
• triangle congruence theorems: SAS, SSS, ASA, AAS
• correspondence theorems: corresponding parts of congruent triangles are congruent (CPCTC), corresponding parts of similar triangles are proportional
• angle congruence theorems: in parallel lines, alternate angles, corresponding angles are equal, co-interior angles sum to 180 degrees; vertically opposite angles are equal
• circle theorems: angle in the centre is twice the angle at the circumference, angle between a tangent and radius is 90 degrees, angles in the same segment are equal, intersecting chord theorems...

[nocturne sigil]

hi,im new,and i have a question,cuz im not like american and i just saw ur message about theorems and if u are american,in what grade do u guys learn this stuff?

[obsidian harness]

there's a big difference between American school geometry and Olympiad geometry
Americans usually learn this in 9th grade, but 8th grade is also very common for those who have the ability

[nocturne sigil]

9th grade?????

i learned that in 6th

like the congruences and about the circle

[obsidian harness]

well yes, unless they do integrated maths, they're going to be learning geometry for the entire year

which is weird by global standards

so there's also some stuff about right triangle trigonometry, Pythagoras in 3D

honours geometry might also have conic sections: the circle, ellipse, hyperbola, and parabola

[obsidian harness]

also you probably didn't need to prove any theorems, like give proper reasons step-by-step

[nocturne sigil]

nooo,we had to prove

we were doing complex problems

not only learn them yk?

[obsidian harness]

most American students don't go to special STEM schools

there's always a choice to (they're called magnet schools)

and this is changing but maybe 10, 20 years ago, American students didn't need tutoring for 9th grade

I'm sure the average American student writes better essays not just in their native language, but in the social sciences, than students in your country

cause there's less reliance on a memorisation-based curriculum for these subjects

[nocturne sigil]

yeah that is right tho

we havr very diffrent ways of teaching

[cold cargo]

I never had that, idk that thing's name in English

[cold cargo]

Teach me

Basic

Formulas

Pls

[simple vigil]

you never used a compass??

how you draw circles then?

[cold cargo]

I dont

Teachers were and are like

"Well, then just watch"

[simple vigil]

means you dont draw in notebook

then how draw in exam

[marble oriole]

you have never in ur life

actually learned what got us to our current advancement

in maf

[queen shadow]

Of what ??

[quick tangle]

Good point

[knotty quiver]

-# is #advanced-algebra technically trig

[fervent dirge]

going into geometry as of tomorrow, is it necessarily harder than algebra 1?

[fast widget]

It really depends on your own personal strength honestly, but if you’re personally better at logic and visualizing shapes, geometry might be easier. It really depends on you

[marble oriole]

Is yo grandpa technically yo momma

Sorry

[pliant hawk]

i was js asking, never looked at anyone of the material

[pale sentinel]

Assuming this as a /gen, not at all

It's more about algebras over a field, which are annoyingly also called... algebras

tf happened to the embed

https://en.wikipedia.org/wiki/Algebra_over_a_field

[proven zealot]

Anyone have tips to get better at geo?

Im like a former geo main back in 7th grade but I’m kinda bad at it now

Like coordbash is my only saving grace lol

[fast widget]

I’d skim over Geometry Revisited by Harold Scott, however if you know your basics, I would look at the AoPS introduction to geometry , I think it’s personally really helpful.

[proven zealot]

already finished both intro to geo and the other books

[proven zealot]

thx for info anyway

[maiden brook]

coxeter is good but a bit difficult as well. i think it is beter to start with aops and then do coxeter that book goes beyond aops

[proven zealot]

ok yeah

nice

wait so another thing is like is it better to grind theorems or to acc just do problems?

like beyond the well known ones like stewarts potlemy, etc.

[keen plover]

is this place appropriate for olympiad geometry?

[polar leaf]

????

[proven zealot]

wait

oops

[proven zealot]

perhaps not

oops

where is the place if u dont mind me asking

[keen plover]

i mean this channel

[maiden brook]

this channel works i believe, so does competition math. if its difficult, competition math or a help channel might be better

[upper karma]

Sup chat

Can someone help me wit this

I got like a hindred solutions for it

(TanA/1-cotA)+(cotA/1-tanA)=1 + secA×cosecA

Kts just using the simple inedetities

But uff

[dire jackal]

do you have to find A or prove this?

[obsidian harness]

,w solve tanA/(1-cotA)+cotA/(1-tanA)=1 + secA * cosecA

[obsidian harness]

isn't this an identity

should be true for all A where both sides are defined

[upper karma]

you have to prove it

simplify it and shi

[upper karma]

prove*

we have to do RHS = LHS

[worthy eagle]

did you try anything?

[upper karma]

yup

tried simplifying it in alot of diff ways

cant get it same on both sides

[worthy eagle]

try converting to sin cos

[upper karma]

we have to simplify both sides to get a common value on both sides

[marble oriole]

Lemme see it

I got this

[upper karma]

[marble oriole]

Watch magic

[upper karma]

watchin

...

[marble oriole]

Let’s see if chat gpt can do it 🤣🤣

[dire jackal]

yoo i think i got it

wait

yess

convert everything to sin and cos on the LHS

then simplify

[marble oriole]

Did it do it

[upper karma]

am doin it

kinda stuck rn ig

[worthy eagle]

you can send a pic to show us where you are stuck

[dire jackal]

ig this is correct yeah

[warm shuttle]

I don't see any mistakes

[marble oriole]

GPT Olympiad strikes again

[upper karma]

didnt thought cubic equation to come

anyways thnx

||to chat gpt||

[marble oriole]

🪬

I can’t believe it

[warm shuttle]

Trig identities are wild

[upper karma]

ffr

[worthy eagle]

i know this question, actually if you convert everything to sin, cos, you get
s²/c(s-c) + c²/s(c-s)
the trick is to see that (c-s)=-(s-c) so it becomes
s²/c(s-c) - c²/s(s-c)
and after this it's ez

[upper karma]

its quite lenghty by that way tbh, i tried to do it, just doin cot to 1/tan etc. is simpler

[obsidian harness]

if you're completely stuck it's usually a good idea to convert it in terms of s, c only however

[cold cargo]

Then they'd give me that circle making thingy

Compass? Isn't this compass 🧭

[cold cargo]

Of circle, square, long Square, triangle, etc. Js basics

[pale sentinel]

the forbidden Long Square

[haughty root]

🤔🤔🤔.

[tender lark]

I started trig tdy, pls dm me if ur good at the subject with any tips that will help me

[foggy swallow]

can someone help me do a sum of trigo i dont understand it

sin + cos -1 / sin + cos - 1 = sec + tan

[rain scaffold]

? Use latex

[dark sparrow]

nobody knows what you mean by this

write it on paper

[foggy swallow]

this

[dark sparrow]

(sin(t) - cos(t) + 1)/(sin(t) + cos(t) + 1) = sec(t) + tan(t)

this is how you write it

when you write fractions in plain text, you need brackets

also just gonna fix your awful crop...

so now that this is out of the way, can you tell us what's troubling you here @foggy swallow

[foggy swallow]

ok the question was easy but i found my self struggling in finding out commons

tan(t) + sec(t) - (sec + tan)(sec - tan)

all are theta

making commons trouble me

pls help me Ann of maths

[dark sparrow]

i have no idea what "making commons" is supposed to mean.

[foggy swallow]

T-T

[dark sparrow]

also i'm confused if you're finding the question easy or can't do it at all bc clearly both cannot be true at once

[foggy swallow]

well the process of question is easy but i found a part in it hard which isnt related to trigo

and its like taking out common terms

and then cutting stuff to get answer

[dark sparrow]

can you show your work maybe

[foggy swallow]

um actually i got over that question and i understood thanks to my friends help

but

now i have come across another question and am stuck

[dark sparrow]

ok sure let's see that one + tell us exactly where you're stuck

[foggy swallow]

(1+cotθ−cscθ)(1+tanθ+secθ)=2

now my answer came out to be 2cot (t) = 2

which seems wrong

problem is i dont have an answer key

so i dont even know if i did right or wrong

but obviously its wrong

i cant take picture of my work

as im using laptop

[dark sparrow]

is this meant to be an identity or an equation?

[foggy swallow]

its a equation

a to prove question

[dark sparrow]

if it's "to prove" then it should be an identity.

[dark sparrow]

no picture of work, no diagnosis.

[foggy swallow]

im in class 10th and i dont have such identities so idk

if u want i can send a inverted image

[dark sparrow]

identity == equation that's true for all values of x (or theta or whatever) where both sides make sense.

[foggy swallow]

which will be hard af to read

[dark sparrow]

inverted meaning?

[foggy swallow]

ambulence uses inverted texts

those kind of text

[dark sparrow]

no idea what that means.

ok but maybe send your image anyway.

[foggy swallow]

alr wait i have to bring my phone if u wanna take a gaignosis

[dark sparrow]

if "inverted" turns out to mean "rotated 180 degrees" then that isnt an issue

[foggy swallow]

See

Ann help me find what i did wrong

come onn

[dark sparrow]

sorry, i had some shit come up IRL.

line 4 looks sus, how did it happen?

[foggy swallow]

well 1 - 1 = 0 and 1 + 1 = 2

[dark sparrow]

no, not line 5

[foggy swallow]

oh

i converted

them into tan and cot

[dark sparrow]

how did you go from this to this

[foggy swallow]

ok so cos/sin = cot

and sin/cos = tan

[dark sparrow]

(cos(t)-1)/sin(t) is not cot(t)-1

[foggy swallow]

aaa

should i conver 1 into some cos and sin prop?

[dark sparrow]

also these were cot-csc and tan+sec just a minute ago, and you mysteriously turned them into cot-1 and tan+1. red flag

[foggy swallow]

hehe

[dark sparrow]

i would probably add the 1's into the fractions in each bracket

[foggy swallow]

how do you know?

what trick is behind knowing what to do

[dark sparrow]

tons and tons and tons of practice i guess

[foggy swallow]

so hard work is the only trick

so do i just multiply it after adding a sin and cos?

[dark sparrow]

i would not say hard work

overwork exists

abbreviating sin(t) and cos(t) to just s and c respectively, you will get:

(s+c-1)/s * (c+s+1)/c

those brackets on the numerator can be read as a difference of squares

[foggy swallow]

abbreviating meaning?

[dark sparrow]

shorten

[foggy swallow]

oh

[dark sparrow]

im on my phone and want to save time/effort while still writing the thing clearly

[foggy swallow]

DONEEEE

i did it

thanks alottt

sensei?teach?Ann?senpai? what should i use ?

your clearly way older than me so calling you Ann would be arrogant especially when your taking out your time to help me in maths

[dark sparrow]

you can just call me by my name without any honorifics, it's fine.

i am 26.

[foggy swallow]

im 15 T-T

if its possible could you accept my friend request as your way of teaching was really good

most of the people just ignore and spoon feed me and i end up getting worse and worse in maths

but i want to , no i need to understand how maths work

ok i wont let the motivation in me burn up i can do 5 more trigo questions

[foggy swallow]

uh Ann

if your still there i have a doubt

can i cancel

(cosA + CosA.SinA)/SinA

like can i cut Sin A

and make it cos A + Cos A

wait nvm dont even asnwer that

[dark sparrow]

the answer would have been no

[foggy swallow]

T-T its ok tho i solved a question own my own

i did not take help i will take that as a small achievement and i will treat myself 1 cookie

but damn cookies are so full of calories like i treated my self 1 cookie per 3 trigo questions i ate 3 of them and will eat another now so 4 its like 450 calories

[dark sparrow]

calorie counting is the road to mental hell

[foggy swallow]

well then welcome to my hell

are you free? if yes im sending a question could you solve it for me pleaseee

i dont know how to do it

its totally fine to decline as you probably have alot of work yourself

ok so

the question is

if cos (t) + sin (t) = root 2 cos (t)

prove

i mean show that

cos (t) - sin (t) = root2 sin (t)

i cant do this at all

could you please send me a pic of this solved

pleaseeee

[knotty quiver]

if $\cos(t)+\sin(t)=\sqrt2\cos(t)$, show that $\cos(t)-\sin(t)=\sqrt2\sin(t)$.

[somber coyote]

glass

[knotty quiver]

@foggy swallow this?

[foggy swallow]

yess

can u solve it for me

[viral plover]

how do i do this??? idek what type of math this qualifies as?? algebra?? geometry??

[knotty quiver]

sorry, I don't understand it

[dark sparrow]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[knotty quiver]

i actually tried solving it and, idk how are they equal

unless it assumes t to be some value

[foggy swallow]

so the question itself is wrong?

[knotty quiver]

like π/4

[dark sparrow]

i said nothing about the question itself

look what the bot wrote

[knotty quiver]

π/8

[foggy swallow]

ok

[knotty quiver]

looks like about functions

[foggy swallow]

the answer comes 0

it doesnt match

[knotty quiver]

what's 0

[foggy swallow]

zero

[knotty quiver]

why would it come to 0

[foggy swallow]

did your answer match?

did your answer match

OHHH NOO

i didnt put a value when i was solving

mb

maybe the question is correct

hehe mb

[foggy swallow]

wow did i get reported on my first day

i hope i dont get banned

[dark sparrow]

lol this is not a report

[foggy swallow]

damn i have litral truama of reports i once said i was 11yrs old for fun and someone actually reported me even tho was 14 i got banned from dc and had to do so many shinanigans to get my acc back

[dark sparrow]

im sorry that happened to you

[knotty quiver]

-# i once said I'm 10 or 11 (i forgot) as a joke and i was detained and made to join vc with a mod to prove it. i didn't talk. In the end, i was let go

[foggy swallow]

I appreciate it, but it’s ancient history now.

[upper karma]

Dayum

My Book have the exact same ques

[foggy swallow]

what book?

my teacher gave me a worksheet

[upper karma]

It was a maths exampler

[foggy swallow]

ohh

[upper karma]

The diff is i was given this eq, and was told to prove that tanA+cotA=2 from that equation

[foggy swallow]

could you explain

how you did it

or let me try one last thing at my disposle

[upper karma]

Firstly square on both sides

[foggy swallow]

let me try one last time

bro

i just have one problem

im stuck on

2 cossine = 2cos sq - 1

what do i do after that

ahuhauh im stuck on this question

i just dont get it

i even asked a friend for help but he seemed to have used the to prove part

but thats not how your supposed to solve it right

@upper karma

[dire jackal]

i think i got it

[dire jackal]

this is the question right

[foggy swallow]

yessss

im hard stuck on it

[dire jackal]

so what i did was i transfered the root 2 to LHS

and split so i get cos(pi/4)cost+sint(pi/4)sint=cost

[foggy swallow]

it would be nice if u could show work

bro pi????

why is class 10th maths so hard

[dire jackal]

1/root2 is sin(pi/4)

and 1/root2 is cos(pi/4) too

wait ill send the work

[foggy swallow]

emu

still there?

[dire jackal]

uhh im not sure i have it lmao

i thought i did but not quite

[foggy swallow]

its ok either it feels like your way might not work for me

[dire jackal]

close though wait a sec

[foggy swallow]

as its too high of grade

could you check this photo

my teacher did this

but i dont understand the

[dire jackal]

ok

[foggy swallow]

second half

[knotty quiver]

$2\cos^2x-1$ is an identity

[somber coyote]

glass

[dark sparrow]

it's half of an identity

by itself it is simply an expression

[knotty quiver]

oh i see

[dark sparrow]

the identity you're looking for is cos(2x) = 2cos^2(x) - 1

[foggy swallow]

damn so cool

[knotty quiver]

yea i knew it sounded wrong to call it an identity itself

[dire jackal]

ok so simplifying eqn 1 here we get sin2t=cos2t right

that should give us something

though it might not be in ara's 10th syllabus

[knotty quiver]

exercise: prove it using the identity $\cos(a+b)=\cos a\cos b-\sin a\sin b$.

[dire jackal]

we didnt have double angle formula in 10th

[somber coyote]

glass

[dark sparrow]

are we still on the question "if cos(t) + sin(t) = sqrt(2) cos(t) then prove cos(t) - sin(t) = sqrt(2) sin(t)" btw?\

[foggy swallow]

ye...

[dark sparrow]

i believe there is an easier way

in both of these equations you can solve for tan(t)

with some rearrangement and [at one point] a division of both sides by cos(t)

[dire jackal]

i mean yeah we get tan2t=1 right

hence 2t can be pi/4

t is pi/8

but this doesnt seem to be the correct approach

[foggy swallow]

t means theta

nvm

[foggy swallow]

um but we can only use 1 equation as the other one we have to prove

[dark sparrow]

you can still do it to the goal equation as rough work

and then reproduce the same-ish work backwards

like "if the equation cos(t) - sin(t) = sqrt(2) sin(t) were true then what would tan(t) be"

just on your own piece of paper

[foggy swallow]

FINALLLYYYYYY

its donee

AND THE BEST PART

i was really going to just ask for someone to give me direct answer

but

thanks to your strict nature of not providing answer

I DID IT BY MYSELFFF

and it wasnt even tough

no need to use pie and all

i might have not used your tan (t) advice as it flew past my head

what i did was

sub cos

and got sin as

root 2 cos - cos

and i took cos as comman

so

sin = (root 2 - 1)cos

and then i just placed its value on the to prove

both side