help-27

[thick knoll]

The identities stop there

This is a bit different

[nimble mirage]

There is one more I guess

[thick knoll]

Go on

[nimble mirage]

Sry my device got off

Yes

[thick knoll]

.

[nimble mirage]

tan( 90° -∅ ) = cot ∅ same with cot also

[thick knoll]

.

Different but you could consider them to be similar

.

Can you work with them

[nimble mirage]

Do u want more,?

[thick knoll]

No

Like if I give you (1-cosx)(1+cosx)

What will that be

*give me a min brb

[nimble mirage]

1 - cos²0 = sin²0

Simple ig

[nimble mirage]

K

[thick knoll]

Simple, sure

But very useful nonetheless

[fierce viper]

@nimble mirage okay great so you know a lot more than i expected.
Here's some more things to probe at your knowledge

1. What are the addition/subtraction formulae for sin, tan, and cos?
2. What are the double-angle formulae?
3. What is the R-formula?
4. What are the period, amplitude, and range of sin, cos, and tan?
5. How do you convert from degrees to radians?
6. State the period, amplitude, and range of asin(b x) + c.

all 9th grade trigo btw

[thick knoll]

Can you do these @nimble mirage

[nimble mirage]

1.sin(A ± B) = sinA·cosB ± cosA·sinB
cos(A ± B) = cosA·cosB ∓ sinA·sinB
tan(A ± B) = (tanA ± tanB) / (1 ∓ tanA·tanB)

Is it right?

Doing 2nd one wait

[thick knoll]

Yes

[fierce viper]

yes

[nimble mirage]

5 is easy

[thick knoll]

Go on

[nimble mirage]

Radians = Degrees * π ÷ 180

[thick knoll]

Yes

So what will 35 deg be in rad

Or 37.5 deg

[nimble mirage]

35 * π ÷ 180 = 7 π ÷ 36 rad

?

[thick knoll]

Yes

[nimble mirage]

2. Is sin2A = 2sinA·cosA
   cos2A = cos²A − sin²A

[thick knoll]

Yes

[fierce viper]

there are two more forms for cos2A

[nimble mirage]

Then the answer I think
tan2A = (2tanA) / (1 − tan²A)

[nimble mirage]

Yeah

[thick knoll]

What bout the rest

[thick knoll]

Go on

[nimble mirage]

I don't know:(

I don't know domain

[thick knoll]

One will be in only cos and the other in sin

[fierce viper]

hint: use the pythagorean identities to guide you

[nimble mirage]

It is of higher class

[thick knoll]

Which one

[nimble mirage]

Not in mine ig

[thick knoll]

Domain

[nimble mirage]

3 and 4

[thick knoll]

What about this

[nimble mirage]

Sry 🙁

I just know basics

[thick knoll]

No problem

[thick knoll]

.

[nimble mirage]

Ok

[thick knoll]

Now cos2a=cos^2a-sin^a

Right

[nimble mirage]

Basic

[thick knoll]

Do you have just these

[nimble mirage]

No there is more

[thick knoll]

What else

The double angles

And then

[nimble mirage]

This

I just know the top 5

[thick knoll]

Ok

[nimble mirage]

Can you make me understand onwards 5

[thick knoll]

We will work on the rest

[nimble mirage]

Thank you 🙂

[thick knoll]

6 is like 4 and 5

In the tan(a+b) formula

Put a=x and b=2x
Do this

[nimble mirage]

Can you help in all I just studied them not their proofed also pls😅

[thick knoll]

Or vice versa

[nimble mirage]

🤔

[thick knoll]

Do it

And look at the result

[nimble mirage]

Ok

[thick knoll]

Did you do it

[nimble mirage]

Yes

But I don't understand Why Sin ( A + B ) = sinAcosB + cosAsinB

Can you explain

[thick knoll]

You want to prove them

[nimble mirage]

Yea

[thick knoll]

It would be better to just verify and run with them imo

[nimble mirage]

Can you tell 🙁

You r Pro pls

[thick knoll]

I can

If want me to

[nimble mirage]

Yes with diagram

[thick knoll]

K

Can you want a bit

*wait

[nimble mirage]

Yes ofc 🙂 thank you

Have You Done

[thick knoll]

We have the values of all four used right

[nimble mirage]

Wow your handwriting is nice

[thick knoll]

Not that much

[nimble mirage]

But I don't understand why 😔

[thick knoll]

We'll get to it now

This is just getting the values

[nimble mirage]

This?

[thick knoll]

Here, angle B looks like a right angle

[nimble mirage]

?

[thick knoll]

In mine

[nimble mirage]

Oh

[thick knoll]

Don't worry about this

[nimble mirage]

Ok

[thick knoll]

If the line AC was higher, the angle B will be more

But I didn't see that while drawing it

Anyway

We will do it the easier way

Just take sinA, cosC

Multiply them

And do the same with cosA, sinC

What will you get

[nimble mirage]

I got the diagram can you explain now pls friend

[nimble mirage]

@thick knoll

[thick knoll]

Ok

There are two angles right

A and B

[nimble mirage]

Yes

[thick knoll]

And two triangles

Both of them being right angled

[nimble mirage]

Yes

But how can be sin b be side of triangle?

[thick knoll]

It is bsinB right

Remember that in sin or cos is the angle

[nimble mirage]

Yea but sinb is side but how it is the study of angle you said?

[nimble mirage]

Oh

so sinB is side but how in dia

[thick knoll]

SinB is the opp to B/hyp

Right

Right?

[nimble mirage]

Yes Sin0 = opp / hyp

Yeah

[thick knoll]

And what is the length of said hyp

Hello?

[nimble mirage]

Hello

[thick knoll]

.

It's 1 unit right

[nimble mirage]

?

Yes

[thick knoll]

So now sinB=(opp to B)/1

[nimble mirage]

Ok

[thick knoll]

So side opp to B=sinB

Similarly side adj to B = cosB

[nimble mirage]

Ok

[thick knoll]

Now in the triangle with A

What is sinA

[thick knoll]

.

[nimble mirage]

This

[thick knoll]

Yes

Now you get it

[nimble mirage]

Ohhh thanks

But one more thing

[thick knoll]

Go on

[nimble mirage]

How we get the sides of both triangle?

sinA and sinB?

[thick knoll]

You don't need their values

[nimble mirage]

Yes but why sides are sinA cosA like that..

[thick knoll]

We assume that we have the values of A and B

[nimble mirage]

No I know that..

[thick knoll]

If we have them then sinA and the like will just be constants right

[nimble mirage]

I understood till here

[thick knoll]

Leave that sin(A+B) out for now

[nimble mirage]

Ok

[thick knoll]

And the side a

Take it as 1 unit

[nimble mirage]

Why?

[thick knoll]

Or it will just get messy

[nimble mirage]

Ok

Side A = 1

[thick knoll]

Yes

[nimble mirage]

And Side x and w be Sin0 and cos0

[thick knoll]

Now sinB is x/1

[thick knoll]

Yes

[nimble mirage]

Because sin²0 + cos²0 = 1² pythagoras theorem

[thick knoll]

Not like that

[nimble mirage]

SIDE Y = sinAcosA

[nimble mirage]

Then how

[thick knoll]

But by definitions of sinA and cosA

[thick knoll]

Correct

[nimble mirage]

Side z we don't need to find out

[thick knoll]

Yes

[nimble mirage]

Because we have to just derive sin ( A + B )

[thick knoll]

Yes

[nimble mirage]

So we found all sides now what ?

We got this see

[thick knoll]

In this figure, there is another triangle

[thick knoll]

Not that one

[nimble mirage]

9h

Oh

[thick knoll]

But the big one

With A+B as an angle

[nimble mirage]

y2 + sinAcosA = sin ( A + B )

Oh

Ok

But still we will have to find y2

See

[thick knoll]

Yes

[nimble mirage]

Because Sin0 = opp ÷ hyp

[thick knoll]

Yes we do need to find it

[nimble mirage]

Yes now how to find y2

[thick knoll]

Look at the figure

Do you know why the angle is A at the top

[nimble mirage]

?

[thick knoll]

Near the arrow

Do you know why is it A

[nimble mirage]

Alternate ig?

[thick knoll]

Yea

Now in that triangle, find cosA

[nimble mirage]

Alternate angles

[thick knoll]

Yes

[thick knoll]

.

[nimble mirage]

Ok

[thick knoll]

Not sinAcosA but sinAcosB

[nimble mirage]

But where is cos A I found but where is it 🙁

[thick knoll]

This adj is your y2 isn't it

[nimble mirage]

Yes

y2 = cosAsinB

[thick knoll]

Now, put that in this

[nimble mirage]

But where is cosA can you help

[nimble mirage]

See

Ohh

I understood sry

[thick knoll]

We took cosA because it gives a relationship between the y2 and some known value

[nimble mirage]

cos A forming at the triangle is sides division yes

[nimble mirage]

Yeah

[thick knoll]

Yes

[thick knoll]

.

[nimble mirage]

Ok now y2 + sinAcosA = ?

[thick knoll]

But it is y2+sinAcosB

You made a small mistake

[nimble mirage]

Oh sry

[thick knoll]

.

[nimble mirage]

Sry

[thick knoll]

The hyp is cosB

Not cosA

[nimble mirage]

Yes

Sry friend

[thick knoll]

Np

[nimble mirage]

Ohhh thanks so muchhhh

[thick knoll]

Now put y2 and the other correction here

[nimble mirage]

I understood 😭😭😭😭😭😭

[thick knoll]

Good

Now onto the next one

Cos(A+B)

[nimble mirage]

☠️

[thick knoll]

🤣

[nimble mirage]

😭

[thick knoll]

As I said before

[nimble mirage]

Math Is Hard

[thick knoll]

You don't need to do these

Somebody did all this and gave you the formulae

[nimble mirage]

That Somebody was very dangerous ☠️💀

😭

[thick knoll]

So the least you could do is verify them

And use them

[thick knoll]

I agree

[nimble mirage]

Ok I ll try

[thick knoll]

Now for the verification

[nimble mirage]

I m pushing ig

[thick knoll]

Just put A=30 and B=60

And check the result

[thick knoll]

Well pushing is something we need to do

[nimble mirage]

Here

[thick knoll]

Shall we continue

[nimble mirage]

Ok

[nimble mirage]

Yes ofcc

[thick knoll]

For 7

Isn't that just 1

With A replaced by A/2

And 8 is 2

[nimble mirage]

Hi

[thick knoll]

.

Right?

[nimble mirage]

What is this?

[nimble mirage]

But what is this bro

[thick knoll]

The formula number 7

Is just number 1

With the angle changed to A/2

Instead of A

[nimble mirage]

Oh

[thick knoll]

Right?

Do you see it

[thick knoll]

The same with 8 and 2

[nimble mirage]

Ohh

Isee

I seee yes

[thick knoll]

Now

Do you remember we talked about cos2a

And that it has 2 other forms

[nimble mirage]

I understood

See this pls

[thick knoll]

Other than cos2a=cos^a-sin^a

[thick knoll]

Go on

[nimble mirage]

Ohhh

[thick knoll]

Yes

That is correct

[nimble mirage]

Now the following part also

[thick knoll]

Go on

[nimble mirage]

I will write

[thick knoll]

K

[nimble mirage]

You tell ok

[thick knoll]

K

[nimble mirage]

?

Is this right

@thick knoll

@thick knoll

[thick knoll]

Yes

[nimble mirage]

1. Identity done ✅

[thick knoll]

And for cos2A

[nimble mirage]

2. Let's do

Cos2A

Wait

Let me get diagram

[thick knoll]

Go on

[devout snow]

@nimble mirage Has your question been resolved?

Channel closed

Closed by @nimble mirage

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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[nimble mirage]

Anyone online

[carmine veldt]

post the question

[untold ravine]

no

[nimble mirage]

Can you explain this

[midnight echo]

What about it don't you understand? We can't read your mind from a screenshot

[nimble mirage]

@midnight echo dot product

?

[carmine veldt]

its an operation

hmm, do you want like a physical example of what it is or smn?

[nimble mirage]

Yes like how we use it

[midnight echo]

https://www.mathsisfun.com/algebra/vectors-dot-product.html

[nimble mirage]

Thanks bud

[midnight echo]

It's a measure of how "parallel" two vectors are, scaled by their magnitude. Perfectly parallel vectors have a larger dot product. Perpendicular vectors have a 0 dot product.

Beyond that, it's hard to figure out what else to explain until you tell us exactly what about the dot product you don't understand

[nimble mirage]

@carmine veldt can you tell why sin theta and cos theta used

why*

[carmine veldt]

its a projection of a vector along another vector

like you know how we write 2d vectors are v = xi + yj

basically you project 1 vector into the span of another, Idrk what else to say

[nimble mirage]

I understood dot product thanks

@midnight echo

[nimble mirage]

Yes

[midnight echo]

You use cos because it scales inversely to angle. cos(0) (parallel) = 1, and cos(90) (perpendicular) = 0. It makes the dot product of "more parallel" vectors greater. That's why cos is in the dot product (and sin does the opposite in the cross product),

!solved

Factoids not working but you can close with .close when youre done

[nimble mirage]

And sin?

[midnight echo]

Well whats sin(0) and sin(90)?

[nimble mirage]

Sin0 = 0

Sin90 = 1

[midnight echo]

So sin measures how perpendicular two vectors are, not how parallel

[nimble mirage]

Cos0 = 1
Cos90 = 0

[midnight echo]

Thats why sin is used in the definition of the cross product, which measures how perpendicular two vectors are

[nimble mirage]

Ok

[midnight echo]

While cos measures how parallel two vectors are, in the dot product

[nimble mirage]

So cos measures how perpendicular they are ok

Thank you

[midnight echo]

parallel, not perpendicular

[nimble mirage]

Oh my bad

Parallel

[midnight echo]

cos -> parallel (dot product)
sin -> perpendicular (cross product)

[nimble mirage]

Ok so dot product= how parallel they are?

[nimble mirage]

Can you explain these formulas also pls

@midnight echo

And what is i j and k also

@carmine veldt

[carmine veldt]

directions

[nimble mirage]

Ok

[carmine veldt]

or rather unit vectors we use to represent them

[nimble mirage]

Can you show a diagram representing i j and k pls

[carmine veldt]

x corresponds to i, y corresponds to j, z corresponds to k

ofc the exact directions depend on your frame of reference

[nimble mirage]

Can we change the variables also?

[carmine veldt]

yeah sure

[nimble mirage]

Ok ty

Can you help me in understanding of these formulas also

Just few only not all 🙂

@midnight echo

@carmine veldt

[carmine veldt]

sure

which ones do you have doubts in

[nimble mirage]

All 😭😭😭😭

I m new that's why

[carmine veldt]

alr

[nimble mirage]

What is 1 formula

First

a vector • b vector =

abcostheta

[carmine veldt]

in Dot product, formula 1 is just the definition

you get it by considering 2 vectors a and b like I've drawn

and computing the projection of b along a

or a along b

and you multiply those

[nimble mirage]

Can you explain like I am 10 yrs 🙁

a vector is first • second vector

ab cos theta

[carmine veldt]

that's the dot product

[nimble mirage]

Yea

[carmine veldt]

a vector is something with size and direction

[nimble mirage]

K

[carmine veldt]

hmm so here we're looking at a and b from the side right?

[nimble mirage]

Yes I understood

Like something with magnitude and direction both

[carmine veldt]

yeah, I'm tryna explain what projection is now

[nimble mirage]

If you walk 5km north

5 km is magnitude and north is its direction

Like that?????

[carmine veldt]

yup precisely

so lets talk abt projections now

suppose you look at the image from the direction I've marked with the red arrow

it'd look something like this right? can you tell me why b appears "shorter"?

[nimble mirage]

Shadow vector it is?

Because the projection only shows the part ofA that lies along B, its length becomes which is smaller than A unless the two vectors are in the same direction

@carmine veldt

@carmine veldt

[carmine veldt]

yup

so in dot product, all we're doing is multiplying those lengths together

and that's where the formula comes from

for (2)

apply formula number 1 and see what you get

[nimble mirage]

So 2 vectors are lines?

With magnitude and direction?

[carmine veldt]

terminated arrows

[nimble mirage]

Yep I understood

[nimble mirage]

Ok

[nimble mirage]

These formulas are very hard to understand 😭😭😭😭

Can you help

[carmine veldt]

ify 😭

did you get what a vector is and what a dot product is?

once that's done, we can wrap up the dot product part pretty fast

[nimble mirage]

Yes 🙂

[carmine veldt]

awesome

[nimble mirage]

But not dot product 100 percent

I understood vector 100 percent

[carmine veldt]

what didn't you understand

[nimble mirage]

Cos0 😭

Cos theta

Sin theta

[carmine veldt]

forget about sin for now

that comes in cross product

[nimble mirage]

Ok

[carmine veldt]

in this diagram

[nimble mirage]

Yes

[carmine veldt]

OB' times OA is a dot product

with me so far?

[nimble mirage]

Yep

[carmine veldt]

the values are that known to you, are the lengths OA, OB and the angle theta

[nimble mirage]

1 vector OB multipled by second vector OA

[carmine veldt]

so how will you find the length of OB' using these known quantities?

[nimble mirage]

Is Dot Product

[nimble mirage]

Yes

Multiplication of 2 vectors

I get it

[carmine veldt]

yea

but like

think geometric

how will you find length OB'

[nimble mirage]

Ok

Can we make right angle triangle?

[carmine veldt]

yes, OBB' is a right triangle

[nimble mirage]

Oh

Ok

OBB' is right angle as BB' is perpendicular to side OB'

Yes

[carmine veldt]

right

[nimble mirage]

Now?

[carmine veldt]

use trigonometry

[nimble mirage]

Ok let me

[carmine veldt]

OB is the hypotenuse

[nimble mirage]

Wwait

Cos theta= adjacent ÷ hypotenuse

[carmine veldt]

right

[nimble mirage]

hypotenuse*

[carmine veldt]

so whats the adjacent side to $\theta$ and what's the hypotenus

[nimble mirage]

Adjacent = OB'
Hypotenuse= OB
Opposite= BB'

[woven radish]

Donkey

[carmine veldt]

yes

[nimble mirage]

cos theta= OB' ÷ OB

But now what?

[carmine veldt]

multiply OB on both sides

so $OBcos\theta = OB'$

[woven radish]

Donkey

[nimble mirage]

cos Theta • OB = OB'

?

[carmine veldt]

yeah

so done right

that's where cos theta comes from

cuz the value you want is (OB')(OA)

that's the same as $(OBcos\theta)(OA)$

[woven radish]

Donkey

[carmine veldt]

aka $(OB)(OA)cos\theta$

[woven radish]

Donkey

[nimble mirage]

Cos∅ • Hypotenuse= Adjacent

[carmine veldt]

yes

[nimble mirage]

And?

[carmine veldt]

I mean that's it

you found OB'

you wanted (OB')(OA)

[nimble mirage]

OB' is adjacent

[carmine veldt]

yes

[nimble mirage]

Okk

[carmine veldt]

did you understand where cos comes from in dot product now?

[nimble mirage]

Ohhhhhh thank you so muchhh

[nimble mirage]

Yes

[carmine veldt]

awesome

so when we do stuff like i•i

that's the same as $|i|^2cos\theta$

[woven radish]

Donkey

[carmine veldt]

where |i| is the length of the unit vector i

[nimble mirage]

Yes I know that

[carmine veldt]

and theta is the angle between i and i

so can you tell me the value of theta here?

[nimble mirage]

Let me try ok

0° ?

@carmine veldt

[carmine veldt]

exactly

and cos(0°) is?

[nimble mirage]

1°

cos90 = 0°

[carmine veldt]

only 1, degree isn't required there

and you're right

[nimble mirage]

Can you tell me this also

[carmine veldt]

so $|i|^2cos\theta = |i|^2$ right?

[woven radish]

Donkey

[carmine veldt]

?

[nimble mirage]

2 formula also

[carmine veldt]

that's what I'm explaining rn

[nimble mirage]

Oh

[carmine veldt]

the LHS of this equation is the dot product eqn for i.i

[nimble mirage]

Ok

[carmine veldt]

and $|i|^2 = 1$ cuz the magnitude of a unit vector is just 1

[woven radish]

Donkey

[carmine veldt]

same logic applies for j.j and k.k

so that's formula (2) done

for (3), (4)

[nimble mirage]

Oh I understood

[carmine veldt]

we know angle between the unit vectors is 90 deg

so cosine becomes 0

so dot product becomes 0

[nimble mirage]

use formula
A vector • B vector= A magnitude • B magnitude cos ∅

[carmine veldt]

yes

[nimble mirage]

A vector= i
B vector= i

[carmine veldt]

okay

[nimble mirage]

cos 0 = 1

1³ = 1

Solved

Same with j and k

[carmine veldt]

exactly

[carmine veldt]

try the same formula for formulae (3) and (4)

[nimble mirage]

Okkk

[carmine veldt]

but you'll see that angle between each vector is 90 degree

[nimble mirage]

Yes because of cos90

[carmine veldt]

yea

and cos90 is?

[nimble mirage]

Cos90 is there because they are in different axises

[carmine veldt]

correct

[nimble mirage]

😎

[carmine veldt]

🔥

[nimble mirage]

so
I • J = I magnitude × J magnitude × cos 90

[carmine veldt]

correct

[nimble mirage]

Now 5th?

Same with I , K and J , K

[carmine veldt]

yes

you got it

[nimble mirage]

Let's do 5th ig

[carmine veldt]

oh 5th is multiplication

yk how (x+y+z)(a+b+c) = xa + xb+ xc + ya + yb + yc + za + zb + zc?

do the same with a vector and b vector

note that wherever you have perpendicular directions

like i.j, j.k and all

it collapses to 0

so the only terms that survive are ones that face the same direction

so i.i, j.j and k.k

and their coefficients get multiplied as the magnitude

[nimble mirage]

Yes

[carmine veldt]

you got it?

[unborn osprey]

ohhh

that makes sense

[carmine veldt]

thanks

[nimble mirage]

I understood yes

[nimble mirage]

Yes

[carmine veldt]

awesome

now @unborn osprey will explain cross product

[nimble mirage]

Bro one more left this

[unborn osprey]

the same thing but you take the other side of the right angle triangle

the side perpendicular to the other vector

[nimble mirage]

Gundu

R u Indian

[unborn osprey]

in dot product u took the component along the other vector

[nimble mirage]

Your name is Indian ig?

[unborn osprey]

in cross product you take the component perpendicular to the other vector

thats why its sin instead of cos

[unborn osprey]

im not indian

[nimble mirage]

Oh

Ok continue

Wait can you explain cos∅ formula

[nimble mirage]

6 th formula

@carmine veldt

[unborn osprey]

you know that a.b is abcos theta right

[nimble mirage]

@unborn osprey

[unborn osprey]

?

[nimble mirage]

Yes

[unborn osprey]

then what is (abcos theta)/ab

[nimble mirage]

Cos Theta = adjacent ÷ hypotenuse

[unborn osprey]

yeah but that is using right angle triangle

this is using the dot product formula

[nimble mirage]

a vector • b vector / magnitude a • magnitude b

[unborn osprey]

yeah so if you simplify it what happens

the ab ab gets cancelled right

and you're left with cos theta

[nimble mirage]

yes

[unborn osprey]

yeah so thats the formula

[nimble mirage]

Can you make an equation?

[unborn osprey]

wdym

[nimble mirage]

Like how they cancelled out

@unborn osprey

[unborn osprey]

a . b=|a||b|cos theta
a.b/|a||b|= (|a| |b| cos theta)/(|a| |b|)
(|a| |b|/|a| |b|)=1
so a.b/|a| |b|=cos theta
plug in a.b value through the formula for dot product

breh

[nimble mirage]

Oh I see ty

How did we find the formula of magnitude a

@unborn osprey

|a|

How

?

@unborn osprey

[unborn osprey]

by the distance formula

from the origin to the end of a

[nimble mirage]

Equation formed?

Can you show the equation

[unborn osprey]

distance between 0,0,0 and a1,a2,a3= sqrt(a1^2+a2^2+a3^3)

[nimble mirage]

Yes

I know

[unborn osprey]

then

what were u asking

[nimble mirage]

|a| = √((a1)²+(a2)² + (a3)² )

I know

But can you make an equation to prove it pls @unborn osprey

And diagram also

[unborn osprey]

i think this would be a good thing for you to try and prove on your own

think about how the pythagoras theorem can be applied

[nimble mirage]

Yes

But there are 3 sides

[unborn osprey]

i agree

you can use pythagoras twice

[nimble mirage]

Can you please tell how

By formula , using the technique

[unborn osprey]

at least try bro

itll improve your thinking skills

apply yourself

[devout snow]

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[devout snow]

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[quaint ether]

please help me understand the intuition behind this

[nimble mirage]

Hi

[devout snow]

@quaint ether Has your question been resolved?

[devout snow]

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[burnt cradle]

How do i do #8 ?

Starting with a ig, idk what formula to do

Or like how to get an equaion

[misty crest]

$L(x) = f(a) + f'(a)(x - a)$

[woven radish]

knief

[burnt cradle]

Ah i dont think i ever learned that, what rule is that?

[misty crest]

point slope

[burnt cradle]

Y = mx +b?

[misty crest]

this is just y - y_1 = m(x - x_1)

[burnt cradle]

Oh

The y-y1 ah

[misty crest]

where y_1 = f(a), m = f'(a), x_1 = a

[burnt cradle]

.close

[devout snow]

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[nimble mirage]

Anyone online

?

[midnight echo]

#❓how-to-get-help

[nimble mirage]

Hi bro

[wintry flare]

nm

[nimble mirage]

Hi

Anyone can explain formula 2

[wintry flare]

of

oh

[nimble mirage]

Do you know it?

[wintry flare]

double angle forumla

yah, i dont use it often

what do you mean explain it

[midnight echo]

@nimble mirage claim your own channel. this one will close soon & you wont be able to send more messages

[nimble mirage]

cos ( A + A ) = cosAcosA - sinAsinA

[nimble mirage]

Ok

[nimble mirage]

The Graph I want

[wintry flare]

you want the graph of it??

[nimble mirage]

Sry my bad

Diagram

Like this

[wintry flare]

where did you find this one

[devout snow]

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[nimble mirage]

Ok

[devout snow]

Please don't occupy multiple help channels.

[faint gorge]

why did you open a second channel?

[nimble mirage]

?

Hey bro

Can you help

[faint gorge]

you are in #help-44｜stanley-🌲-v2-dans already

[nimble mirage]

Yes I want this group

[faint gorge]

ok, then close one, and post your math problem/question

[nimble mirage]

You can remove me from that if you want ..

Ok

2 formula whole proof

[wintry flare]

https://www.mathdoubts.com/cos-double-angle-identity-proof/

read this

[nimble mirage]

Ty

[devout snow]

@nimble mirage Has your question been resolved?

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[plucky portal]

.close

[devout snow]

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[devout snow]

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[tall sand]

Consider the function y defined by the series (image1)
Show that there exists constants a and b such that the differential equation (image2) is satisfied.
When I put the function and its second derivative side by side, it seems to me that a = 1 and b = 0.
However I'm not sure what is the actual way of finding the answer?

[jaunty mantle]

Could a be negative?

[tall sand]

the question only mentions a is a constant

[jaunty mantle]

Well maybe think about if a is negative is there a good answer

[tall sand]

hm

if I try a = -2 I would get y/(x^-2) = 1/(x^-2) + x/6 +... which would still have terms matching powers with y'' except 1/(x^2)

but not matching coefficients

idk

wait I have an idea

[jaunty mantle]

Not quite -2

[tall sand]

Okay

I'm not sure what could work

[jaunty mantle]

Look at the first term

[tall sand]

I thought I was cooking 💔

[jaunty mantle]

y’’ has an x

y has a 1

How can you turn x into 1

[tall sand]

I think we’re rather trying to turn 1 to x

[tall sand]

power of 0

[jaunty mantle]

Oh

Oops I misread it

y’’ = x^a y + b

Let’s ignore the b for now

[tall sand]

Okay

[jaunty mantle]

First term says x = x^a * 1

What is a

[tall sand]

first term of y'' ?

[jaunty mantle]

And of y

[tall sand]

first term of y is 1
first term of y'' is x^1 * 1

[jaunty mantle]

Okay look at the next time

See if a = 1 works

[tall sand]

it looks like we multiplied each term of y by x^1 to get to y''

I just don't know if "it seems like we multiply by x^1 each term and add nothing" is acceptable to say a = 1 and b = 0

[devout snow]

@tall sand Has your question been resolved?

[devout snow]

@tall sand Has your question been resolved?

[devout snow]

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[orchid wasp]

for a i got y/2<=x<=sqrt(pi)/2 and 0<=y<=sqrt(pi)

how i do b and c

[orchid wasp]

this is right btw?

[craggy dagger]

looks rght

[craggy dagger]

have you sketched and shaded R

you should be able to tell me what shape R has

[devout snow]

@orchid wasp Has your question been resolved?

[devout snow]

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[orchid wasp]

I sketch it like unit circle?

[craggy dagger]

the shape of R is not a circle

[orchid wasp]

Then how am I suppose to sketch the pi

Idk how to graph it

[craggy dagger]

$\pi$ is just a number

[woven radish]

Element118

[craggy dagger]

between 3 and 4

[devout snow]

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[devout snow]

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[gray prawn]

We can convert a triangie into parallelogram, a parallelopiped into cuboid
We can see that visually
But is that how we formally show that any object can be converted into other by re-arrengement of it's pieces?

[devout snow]

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[fierce rune]

quick question hopefully, how would you compute the operation we define III with? like, would it be evaluated like $N_u\cdot N_u(du)^2+N_v\cdot N_v(dv)^2$ like a dot product or $N_u\cdot N_u(du)^2+2N_uN_vdudv +N_v\cdot N_v(dv)^2$?

[woven radish]

Syrenate

[nimble mirage]

Hi

Everyone

[fierce rune]

hello

[nimble mirage]

Hi

What topic are you studying?

[fierce rune]

differential geometry

are you looking for general chat

because this is a help channel

[nimble mirage]

No I also want that's whyy

[fierce rune]

ah

#❓how-to-get-help

check this out

[nimble mirage]

Yes I saw

Can you help?

[fierce rune]

you can open your own thread by talking in one of the help chats above this one

one without a username

like #help-12

[nimble mirage]

Ok

[fierce rune]

oh is this the wedge product?

[devout snow]

@fierce rune Has your question been resolved?

[devout snow]

@fierce rune Has your question been resolved?

[devout snow]

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[devout snow]

@lunar eagle Has your question been resolved?

[fathom vault]

e=lim((1+(x/n))^n)

so use that ig

not sure if it can do it

[lunar eagle]

?

[fathom vault]

wait let me just try

[lunar eagle]

i tried using L'h i got

1/4

but answers given pi/8

[fathom vault]

wait what are you using L'h for

like im actually curious

[gloomy aurora]

how are you using L'h here?

[fathom vault]

by the way L'h is not super applicable

its just substitution

[lunar eagle]

Int / 1/n^2

Int limits are from 0 to 0

Denominator is also 0

[lunar eagle]

Wait

Ik its wrong

I wnana know why

[fathom vault]

the differentiation of the integral of the numerator is way more complex i think

[lunar eagle]

Why

Iant it just newton lebinitz

Isnt*

oh wait i cant use it if n is also a part of function can i?

[fathom vault]

yes

[lunar eagle]

ye

mb

how do we solve it then

[fathom vault]

oh

so e^x-1 approximates to x when x converge to 0

so that basically cleans out the e

[lunar eagle]

ik like x is small but there isnt a limit on x

so can we use it

[fathom vault]

the limit is taken at n -> infinity

so x basically converge to 0

cuz 1/n converge to 0

[lunar eagle]

Ye fair

[fathom vault]

next

[lunar eagle]

[fathom vault]

we let u=nx

so when x=0 then u=0, x=1/n u=1

substitute in there

x=u/n

uhh hello?

[fathom vault]

tbh this is the key idea

so itll be done easily after

[devout snow]

@lunar eagle Has your question been resolved?

[lunar eagle]

oh ye

makes so much sense

[fathom vault]

so did you do it

[devout snow]

@lunar eagle Has your question been resolved?

[devout snow]

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[devout snow]

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[lofty verge]

I wrote this program on my calc

it gives 33 triangles

but the answer is 8

i check that they are distinct with the not in part

so i'm confused as to why im getting so many

and maybe there is a way to do this without a program?

and also looking at the results, i see no duplicates

[vital edge]

You can't have a triangle with all possible sets of side lengths that add up to 20

Are you familiar with the triangle inequality

[lofty verge]

oh yeah

a+b<c

right?

[vital edge]

Not just that

[carmine veldt]

greater

[lofty verge]

a+b>c

[vital edge]

Sum of any two sides is greater than the third

So you need three inequalities

[vital edge]

I mean there is, but it's manyal work

[lofty verge]

ah, not worth it then

and yeah, adding that to my program

it results in 8 triangles

thank you, can't believe i overlooked that lol

.close

[devout snow]

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[vital edge]

Lolol no worries

[devout snow]

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[ionic terrace]

hello , worked out the integral of sec^4x. tan^5x by using a trick but im struggling to work out the intergral with the powers switched

[nimble mirage]

How R You

@carmine veldt

Can you help

Cross product

[ionic terrace]

sorry im new here the help bot just said send ur question. im doing good and yourself?

[nimble mirage]

@carmine veldt

@ionic terrace

Can you help

Hi

[ionic terrace]

?? im so confused rn

[nimble mirage]

Hi

[ionic terrace]

hello

[nimble mirage]

Tell

[wind marten]

make ur own channel bro

[nimble mirage]

Ok

[carmine veldt]

hm

so integral of sec^5 tan^4?

[pseudo basin]

!noping

[devout snow]

Please do not ping individual helpers unprompted.

[pseudo basin]

also actually yeah this is butchie's channel

[carmine veldt]

Ig reduction formula works well there

rewrite the eqn in terms of $sec^nx$

[woven radish]

Donkey

[carmine veldt]

apply IBP to that general eqn to get a formula

and work it out

$\int sec^nx \dd{x} = \frac{sec^{n-2}xtanx}{(n-1)} + \frac{n-2}{n-1}\int sec^{n-2}x \dd{x}$ if I recall correctly, check once tho

[woven radish]

Donkey

[carmine veldt]

@ionic terrace have a blast mate 😭

[ionic terrace]

thanks bro, dont think its an easy one

[carmine veldt]

nah dw, js apply that reduction formula again and again

it'll go down within 6 terms

try deriving that one tho, its fun

[ionic terrace]

ill give it a go cheers, how do i close chat>

[carmine veldt]

.close

[nimble mirage]

@woven radish tan²x + 1 = ?

[ionic terrace]

.close

[devout snow]

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[nimble mirage]

Q. ( A vector × B vector ) • ( B vector × C vector)

[devout snow]

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[ionic jasper]

when i got 2 points in an 3d coordinate system how do i know the function of the linear graph that goes through the points

[carmine veldt]

$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$

[woven radish]

Donkey

[hollow wasp]

They asked for points in 3 dimension

To start, note that we only need a direction vector and a point on that line

[carmine veldt]

oh mb

[hollow wasp]

From there we can describe every point on that line

[ionic jasper]

what

wym plane

[marble rapids]

nvm sorry

[jaunty mantle]

What is the general form of a line in 3D

[ionic jasper]

1sec

its in german but there is it

point C and D is given

and now i wanna make an function like the g and h above

[hollow wasp]

Yes. What are the two vectors in the functions?

(The first is on their own, the second is multiplied by a constant r/s)

As in what they mean

[ionic jasper]

idk

[hollow wasp]

Do you mind working through a 2D problem?

[ionic jasper]

idm

[hollow wasp]

Like finding the function through the points say (2,3) and (5,6)

How would you do that

[ionic jasper]

idk

[hollow wasp]

A good method would be to start at a point that is on the line, and then walk in a direction parallel to the line

That would describe every point on that line yes?

[ionic jasper]

ye