inverse trig func's

[main sedge]

how do i evaluate this i entered the value neg 1 then did the inverse tan of it on my calculator and pressed sine. wht am i doing wrong..apparently the answer is supposed to be - sqrt(2)/2 so im confused. or am i getting the representation in decimal form? i got.. = -0.707106781

[cedar star]

you shouldn't need to use a calculator to do this

using a unit circle, think about what arctan of -1 is

[main sedge]

is this value correct?

0.785398163

[main sedge]

oh

[cedar star]

this is arctan(1)

[main sedge]

ooh i figured out it now but..

how do i go from decimal to fraction..

if i can..

or do i gotta do it over?

the proper way

[cedar star]

dont use decimals to begin with

[main sedge]

oh ok

[cedar star]

no calculator, no decimals

[main sedge]

ok so now..

how do i figure out where arc tan is on the unit circle

[cedar star]

how are sin cos and tan related

[main sedge]

pythagorean theorem?

[cedar star]

sure

but more specifically

$ \tan x = \frac{\sin x}{\cos x} $

[twin bay]

Kaiser

[main sedge]

OHHH roght

right*

ok so lmk if im going in the right direction but..

would i take the inverse of sine and cosine to get the inverse of tangent?

or am i falling way far off the tracks..

[cedar star]

if youre asking whether arctan is equal to arcsin/arccos, then no

[main sedge]

ok then nvm...

[cedar star]

we can think of arctan(-1) as

$\tan \theta = -1$ for some $\theta$

[twin bay]

Kaiser

[main sedge]

uhh

so do i look in some certain quadrant where tan is negative?

[cedar star]

yea basically

[main sedge]

oh

[cedar star]

if $\tan \theta = -1$ then $\frac{\sin \theta}{\cos \theta} = -1$ also

[twin bay]

Kaiser

[cedar star]

so think about what angle $\theta$ on the unit circle will result in $\frac{\sin \theta}{\cos \theta} = -1$

[twin bay]

Kaiser

[cedar star]

since the ratio is -1, we can infer that $\sin \theta$ and $\cos \theta$ should be the same value but off by a negative sign

[twin bay]

Kaiser

[main sedge]

oh

[cedar star]

i.e. either $\sin \theta$ is positive and $\cos \theta$ is negative, or vice versa

[twin bay]

Kaiser

[main sedge]

kk

so by process of elimination do i look in the 3rd quadrant at all?

[cedar star]

what is true of the sign of sin and cos in each quadrant?

[main sedge]

well..

in the first quad sine and cos r pos

in the second

sine and sec are pos

in the third

tan and cot r

and the fourth cosine and sec

so im not sure..

[cedar star]

why are we talking about sec and cot

[main sedge]

idk😭

nvm tht

[cedar star]

do you understand how the unit circle works?

[main sedge]

in a sense

wht specially abt it tho

[main sedge]

idk wht im missing T-T

oh u askin me for terminal points?

[cedar star]

for any angle on the unit circle, lets call it $\theta$, the "x" value is given by $\cos \theta$ and the "y" value is given by $\sin \theta$

[twin bay]

Kaiser

[main sedge]

or neg, neg pos type thing

yup ik tht part

[cedar star]

so in the first quadrant, both x and y are always positive right?

[main sedge]

yes

[cedar star]

what about in the second quadrant

[main sedge]

x is neg y is pos

[cedar star]

right

so that would suggest that cos is negative and sin is positive in 2nd quadrant

[main sedge]

yup

[cedar star]

third quadrant, both are negative, and fourth quadrant, cos is positive but sin is negative

[main sedge]

yea

[cedar star]

now earlier we said we are looking for the angle $\theta$ where $\frac{\sin \theta}{\cos \theta} = -1$

[twin bay]

Kaiser

[main sedge]

mhm

[cedar star]

which quadrants can this happen in?

[main sedge]

ermm

wait am i focusing on the negative 1 part of sine and cos?

or*

[cedar star]

wdym

[main sedge]

OH WAIT

hold up-

does sine hve to be neg 1 or cos?

[cedar star]

when you divide sine of theta and cos of theta, that should be -1

in general if you have $\frac{a}{b}=-1$ what does that tell you about $a$ and $b$

[twin bay]

Kaiser

[main sedge]

one is neg and one is pos?

im confused even more..

[cedar star]

yea

which part of that is confusing

[main sedge]

oh ok

nvm i got tht part now

so wht happens now

[cedar star]

so we want to find $\theta$ where $\frac{\sin \theta}{\cos \theta} = -1$

[twin bay]

Kaiser

[main sedge]

hm so how do we know if its the 2nd or 4th quadrant?

[cedar star]

for now lets consider both

pick one, what is theta

[main sedge]

okay

quad 4?

[cedar star]

so lets look in quad 4

[main sedge]

alr

[cedar star]

what angle would make the x and y have different signs but also be neg 1 when you divide them

[main sedge]

in degress or radians?

[cedar star]

either

degrees

[main sedge]

ok

uhh

so is it btwn pi/6, pi/4, and pi/3?

or am i looking at pi/2 as well?

[cedar star]

it could be any angle in that quadrant

but hopefully there should be some intuition to point you to which one to look at first

[main sedge]

pi/2 since 0 is x and -1 is y?

idk..

i tried the others ones but idk if im doin it wrong but im not gettin -1 with the other 3

hmm hold on-

[cedar star]

walk me thru your calculation

[main sedge]

k

first pi/6

which is

sqrt(3)/2, 1/2

so i divided 1/2 by sqrt(3)/2

cancelled out the 2s

im left with 1/sqrt(3)

since u cant hve sqrt in bottom

i rationalized it

and made it

sqrt(3)/3

which didnt give me -1

so i crossed tht option out

[cedar star]

ok good

but one thing

[main sedge]

yes

[cedar star]

pi/6 (which is radians) is 30 deg

[main sedge]

ik

[cedar star]

but that is positive, so its in the 1st quadrant

[main sedge]

ohh

OHH RIGHT

mb mb

[cedar star]

to go to the fourth quadrant

you can think of it as -30 deg

or

360 - 30 = 330 deg

[main sedge]

i forgot we were doin - degrees

kk!

yea i should put -pi/6

[main sedge]

kk

[cedar star]

ok sure we can do -pi/6

[main sedge]

ooh wow ty!! thts actually a good trick i never thought of tht before haha. i always did it the long way

even for big numbers

ok -1/0 is undef

so thts not it either

[cedar star]

right

[main sedge]

WAIT

i think

i think i might have just found it

probably?

oh wait..

yes i think

is it

sqrt(2)/2, - sqrt(2)/2?

[cedar star]

yea thats right

[main sedge]

yayy:D

[cedar star]

if you divide those two you get -1

[main sedge]

yea

[cedar star]

what angle is that

[main sedge]

-pi/4, aka -45

[cedar star]

right

so now weve found the angle $\theta = -\frac{\pi}{4}$ such that $\frac{\sin \theta}{\cos \theta}=-1$

[twin bay]

Kaiser

[main sedge]

ohh

that makes so much sense now!!!

[cedar star]

awesome

[main sedge]

so would this same strategy work for other trig identites such as secant and cosecant and cotangent?

[cedar star]

yes

[main sedge]

ooh ok

[main sedge]

wait so wht was the sin for over here

was it as a guide to show tht tan and cos is somehow related to it?

or smthn..

[cedar star]

so lets consider any angle $\theta$ that is in the first or fourth quadrant, so the right half of the circle

[twin bay]

Kaiser

[cedar star]

then, $\arctan(\tan \theta) = \theta$

[twin bay]

Kaiser

[main sedge]

ok

[cedar star]

arctan is a way to undo the tan operation to get your original angle

[main sedge]

oh

[cedar star]

the problem asked for $\sin(\arctan -1)$

[twin bay]

Kaiser

[cedar star]

first, we want to calculate $\arctan (-1)$ and then take the sine of that to get the answer

[twin bay]

Kaiser

[main sedge]

oh ok

so we calculated it as -45 degrees

or do i hve to express this as a positive angle? or it doesn't matter?

or in radians?

[cedar star]

we were in the fourth quadrant, so it must be negative

otherwise its not in the fourth quadrant anymore

[main sedge]

alr

but wht if i said like 315 degrees?

still no?

[cedar star]

yea thats just another way to say -45 deg

they are the same thing

[main sedge]

okay

[cedar star]

in fact, sin (-45) = sin( 315)

[main sedge]

ooh i see

[cedar star]

this is also true for any multiple of 360, becauwse every 360, you do a full rotation on the circle

so sin(-45) = sin (315) = sin(315 + 360) = sin(315+ 360 + 360)

[main sedge]

mhm

that's pretty cool

[cedar star]

we calculated that arctan(-1) = -45 deg

[main sedge]

yea

and u said then we'd hve to take the sine of tht?

[cedar star]

thats what the original problem said right?

[main sedge]

yes but how would i do tht without a calculator and i mean when i inputed here earlier this showed up as correct so..

unless these two r equivalent in some way?

idk

[cedar star]

what is sine of -45 deg

[main sedge]

do i put tht into a calculator?

[cedar star]

look at the unit circle

[main sedge]

oh nvm

k

OHHH

ok

i see it now

tht makes sense

ohhh ok yea i get it now

i almost forgot abt the two values being x and y

[dawn ember]

Don’t mean to interrupt but when in school am I gonna have to do whatever this is

[main sedge]

wait so does tht mean u cant calcuate tangent for any other val other than cos and sine or would tht mean tht you'd have to use the trig conversions

[cedar star]

great

[cedar star]

wdym

[cedar star]

middle or high school or college depending on where youre at

[dawn ember]

Oh alright

[main sedge]

like say this said sec instead of sin. then wht?

[dawn ember]

Good luck guys

[main sedge]

ty

[cedar star]

$\sec x = \frac{1}{\cos x}$

[twin bay]

Kaiser

[main sedge]

oh ok

i figured

alr

soo..

[cedar star]

so if you wanna know what $\sec 30^\circ$ is then you would calculate $\cos 40^\circ$ first then do one dividded by that

[twin bay]

Kaiser

[cedar star]

*30 not 40

[main sedge]

wait why 30 degrees?

[cedar star]

just as an example

[main sedge]

ohh ok

kk

ok so in according this example would sec be 2 sqrt(2)/2

unless i did smthn wrong

[cedar star]

sec of -45deg?

[main sedge]

uhh yes

[cedar star]

yea

[main sedge]

so wht would i do with this now?

leave it like this?

[cedar star]

well what is the question

just sec -45 deg?

[main sedge]

this one but instead of sine sec

sec replaces the spot sin is in

so idk

[cedar star]

yea

leave it

[main sedge]

kk

[cedar star]

one thing i wanted to note earlier

when we were trying to find arctan(-1), we looked at where sin/cos was negative

[main sedge]

mhm

[cedar star]

and we said this would happen when sin is negative and cos is positive, or vice versa

[main sedge]

yea

[cedar star]

and we picked the second and fourth quadrants as places where that could happen

[main sedge]

we did

[cedar star]

and we first looked at the fourth quadrant which is how we got -45

[main sedge]

yea

oh i think ik where u r goin wit this

prolly

[cedar star]

but if you think about the second quadrant, there is also an angle that satisfies that condition of sin/cos = -1

[main sedge]

oh wait lemme guess tht angle doesnt satify this problem

bc its too big?

or small

[cedar star]

it has to do with how arctan or any of the inverse trig functions are defined

[main sedge]

does it have to be btwn -2 pi and -pi

[main sedge]

yea yea

[main sedge]

oops said this wrong i think-

is it -pi and pos pi?

or

-2 pi and pos 2 pi

[cedar star]

for arctan, -pi/2 to pi/2

[main sedge]

or neithe

ohh alr

[cedar star]

the distinction is

for $\tan \theta = -1$, both -45 deg in quad 4 as well as 135 deg in quad 2 are valid as values of $\theta$

[twin bay]

Kaiser

[main sedge]

is there a way to figure out the ranges for the other funcs by using this range or is it smthn tht im gonna hve to memorize like a formula?

[main sedge]

ok

[cedar star]

but for arctan, we have to restrict which values are valid, since for any function, you can't have it be two different values for the same input

[cedar star]

gotta memorize

its basically something mathematicians set

[main sedge]

right or it wont be a function anymore

[main sedge]

kk

[cedar star]

correct

[main sedge]

and if im also correct

im not exactlly sure which one but

i think-

sec/csc/cot r assymptotes and one of them doesnt hve a set range

so its like

(-infin, +infin)

smthn like tht

bc it differs from the sine and cosine func

and secant and csc graphs hve the kinda parabola shape

from [-1, inf]

etc etc...

[cedar star]

yea they all have repeating asymptotes

because sine cosine tangent are periodic, and cross through zero repeatedly

[main sedge]

ohhhh

[cedar star]

you can intuiteively think if you take the inverse of zero (1/0) that will shoot up to infinity or -infinity

which results in the asymptotic behvior

[main sedge]

mhm

oh and im sorry if i kept u here all day/night helping me understand these concepts

its officially the next day

12 am in the morning

and i have a final math exam at 8 am

and i need atleast a 65% to pass it so im doin as much practice as i can

my grades r pretty low thts why i gotta rlly take this serious and stuff so yea..

omg i yap too much

sorry..

[cedar star]

np

and good luck

its important to build intuition about these things instead of rote memorization

which is what ive been trying to encourage here

[main sedge]

Yeah I've noticed:)

and i rlly appreciate it

gosh i just hope my prof doesnt ask us to draw the inverse sine and cosine graphs too😞

but if he does i guess i should be fine

ill memorize the domain and ranges

[cedar star]

you should def be able to draw sin cos and tan

you can use these to help you draw the inverses

in general the graph of the inverse function is just a reflection about the line y=x of the original function

you just need to be careful about the domain and ranges (which as you said just memorize this part)

[main sedge]

ohh ok

Also I have like 3 more main units i want to go over(Angle measure, trig of right triangles, and trig identities in terms of verifying identities) i dont know if you'd like to stick around for tht or if you want to do other things but yeah..

and the there's one full chapter tht i want to cover but i dont want to bore u haha

[main sedge]

so ill try to do this one on my own

[cedar star]

feel free to ask and ill respond if im around

[main sedge]

alr

for the first one

angle measure

i just need help with these 2 questions

pretty short

idk how to specifically do this on my calculator but

how do i get this into regular radians

not pi radians

i remember some trick tht worked for pi radians

but idk abt regular radians

[cedar star]

not sure what u mean by regular vs pi radians

[main sedge]

wait nvm

wait this might be in decimal places i think

hmm lemme

[cedar star]

ig you mean whether you carry out the final multiplication with pi

[main sedge]

wht in pi radians is 20 degrees again

[cedar star]

to convert from deg -> rad, multiply by pi/180

[main sedge]

ohh

ok

[cedar star]

from rad -> deg, multiply by 180/pi

[main sedge]

yea ik tht but..

[main sedge]

where does tht 20 degrees go

before the pi/180?

where does it fit in the equation

[cedar star]

$20^\circ \times \frac{\pi}{180^\circ}$

[twin bay]

Kaiser

[main sedge]

ohh ok

gotcha

wht abt when its a "mixed fraction"

do i multiply the 3 by 2

wait does tht make sense?

[cedar star]

wdym

[main sedge]

oop nvm

i got it

last one for angle measure is..

ik how i would do it ish but

wht is the fastest way to find a coterminal angle withoutcounting every space on the unit circle

[main sedge]

over and over again

do i subtract by 2pi

ohh

[cedar star]

add any multiple of 2pi or 360

[main sedge]

i think i do

yea yea

lol silly me

perfect

and i got the same exact values from the problem

ok

now for

trigonometry of right triangles

do i hve to make this into an equilateral triangle?

ik wht all the angles r

but only 1 side

[cedar star]

use sine

[main sedge]

sine = x/11

do i do uh

or use

the 30 degree angle or smthn

and do the inverse or regular of tht and cross multiply?

i havent done this in a while

i forgot

[cedar star]

$\sin(30^\circ) = \frac{x}{11}$

[twin bay]

Kaiser

[main sedge]

ohhh

ok

alright umm

idk wht to do abt this cuz

it looks like i can use tan but i dont have the adjacent side

and i mean then i thought it could be cot but again dont hve adjacent side

and i cant do sine or cos either

cuz...

oh wait

wait nvm

yea im just stuck

[cedar star]

why is this any diff from the last problem

use the same logic

[main sedge]

oh rightt

uh

does adjacent or opposite go on top

x on top prolly

[cedar star]

soh-cah-toa

[main sedge]

ik but it seems weird for x to be on the bottom

it looks wrong

[cedar star]

why

not at all

[main sedge]

oh it isnt?

[cedar star]

there is no reason why x can't be on the bottom

[main sedge]

k

[cedar star]

i feel like teachers often freak out about things being on the bottom of the fraction

like radicals

but its actually all nonsense

none of it matters

[main sedge]

yea haha

[cedar star]

anything can be on the bottom

[main sedge]

kk

yea my math prof would agree

[cedar star]

(except 0)

[main sedge]

but i remember all my high school teachers being so strict abt it

[main sedge]

ofc

like id get points off just for not doin it tht way..even tho it was correct yk?

:/

[cedar star]

yea thats bs

[main sedge]

mhm

um wait when the x is on the bottom am i supposed to divide

i multiplied and got a completely different answer

wait hmm

i might hve done smthn wrong again actually :/

wtf in the..

ohh

it wanted me to take a different angle this time

thts stupid-ish

i had to take 59 degrees tan instead of 31 degrees tan

[cedar star]

for this problem?

[main sedge]

mhm

i had to take another angle

the missing angle for sum reason

[cedar star]

why

[main sedge]

it gave me a diff answer

idk

[cedar star]

$\tan(31^\circ) = \frac{11}{x} \ \implies x = \frac{11}{\tan(31^\circ)} = 18.30707$

[twin bay]

Kaiser

[main sedge]

hmm

can u try it with 59 to see wht u get

[cedar star]

well tan(31) is not tan(59) so the answer will be diff

[main sedge]

yea but why didnt i get 18.30707 for tan 31 like u

[cedar star]

try again

[main sedge]

i got sumthn else

strange i tried on both of my calculator

s*

[cedar star]

why are you multiplying

[cedar star]

look at this

[main sedge]

u told me to?

wdym

wait why r we dividing tho

[cedar star]

how do you solve a = b/x for x

i.e. if a=b\x then x= what

[main sedge]

divide both sides by uh

b?

[cedar star]

b is getting divided by x on the right side right?

[main sedge]

right

[cedar star]

we want to undo the division

make the x disappear fromt he right side

so that we can bring it to the left

how do we do that

b/x times x

becomes b

which is great because x has disappeared

but whatever we do on the right must also be done on the left

so we multiple left times x

so a * x = b

[main sedge]

oh

[cedar star]

is this familiar

[main sedge]

yes somewhat

[cedar star]

to be brutally honest

you prolly need to get a better grasp on this before worrying about the trig parts

[main sedge]

yea mb im just not used to doin it with 3 digits

but wait one question

[main sedge]

i think i remember being taught in high school one time tho tht u can cross multiply this? but then why when u multiply u still get the same result. and am i supposed to divide sin 30 by 11?

idk😭

im just mixing up a couple stuff a bit..my brain is goin all over the place-

[cedar star]

why can you not cross multiply

oh you said you can

sure

not sure what u mean tho

tell me how u would cross multiply that example

[main sedge]

which one

[cedar star]

either

[main sedge]

is this trick meant for sine as well

[cedar star]

there is no trick

whether you are using sin cos or tan

just set up the equation

[main sedge]

okay

[cedar star]

sin/cos/tan (angle) = some side ÷ another side

x will be some side or another side

then isolate the x by cross multiplying to so lve

[cedar star]

good

wb the tan one

[main sedge]

Sorry that it’s sideways

That’s all I could do for now

ok now onto the last section of trigonometry before i move on

Trig identities now

but more in-dept

now this one might be a long one...

my head is gonna break with this one

help-

asap

[cedar star]

which part is confusing you

[main sedge]

everything? uh start from sin(theta) above the first fill in the blank

and ik wht rewrite means but

just to make sure

wht does it mean by rewrite in this context?

[cedar star]

well our goal is to verify that the given identity is correct, using other facts we already know

[main sedge]

ahh

[cedar star]

and to do this, we will evolve the left hand side until it turns into the right hand side

[main sedge]

mm i see

firstly

how does sin(theta)/tan(theta) translate into sin(theta)/sin(theta)/cos(theta)

wth do they hve in common

[cedar star]

earlier we said that tan theta = sin theta / cos theta

this is something to remember

[main sedge]

ok

[cedar star]

so they simply plugged this in

i.e. they replaced the tan theta in the denominator with sin theta / cos theta

[main sedge]

kk

ok i see tht now

wht abt the cos(theta)/sin(theta) beside the sin(theta)

did they flip the num and denom?

[cedar star]

yep

[main sedge]

wait how come

wht does tht prove

[cedar star]

its just a necessary step to get the whole thing to simplify to cos

[main sedge]

oh

so do sin and sin cancel out or they just stay there

[cedar star]

yea they cancel

which leaves you with only cos

[main sedge]

kk

[main sedge]

throw me in a ditch..

not literally but-

i have more of these in these unit to go

like abt 7? or less

and some of them are longer

[main sedge]

also

how were these gotten

u there?

[cedar star]

tan^2 = sin^2/cos^2

and sec^2 = 1/cos^2

notice both have cos^2 in denominator meaning u can combine them into one single fraction wh ich becomes

(sin^2 x - 1)/(cos^2 x)

here we invoke the identity sin^2 x + cos^2 x = 1

rearrange the identity to get sin^2 x - 1 = -cos^2 x

plug this in to get

(-cos^2 x)/(cos^2 x) = -1

qed

similar for the next problem

convert csc into 1/sin x

combine 1/sin x - sin x into a single fraction by multiplying to get a common denominator

and continue simplifying til u get cot x

[main sedge]

im back