Question

Inverse Trig Function help.

Answer 1

\[\cot ^{-1}\left( \tan \frac{ \Pi }{ 3 } \right)\]

Answer 2

tangent of pi/3 is sqrt3

Answer 3

i dont know where to go after im left with cot^-1(sqrt3)

Answer 4

\[\cot ^{-1}\left( \tan D \right)=\frac{1}{\tan ^{-1}\left( \tan D\right)}=\frac{1}{D}\]

Answer 5

umm where did that variable come from?

Answer 6

I am just using an angle of D as an example

Answer 7

What happened to D would be strictly the same with \(\pi\)/3

Answer 8

hmmm.

Answer 9

If you have an "x" there instead you would also have that:

\[\color{blue}{\cot ^{-1}\left( \tan x \right)=\frac{1}{\tan ^{-1}\left( \tan x\right)}=\frac{1}{x}}\]

Answer 10

so for my answer i get pi/6

Answer 11

No, how did you get this?

Answer 12

I am guessing that what I posted doesn't make full sense for you

Answer 13

Lets take it peace by peace, if you will...

Answer 14

sure

Answer 15

You know that:
\(\large \displaystyle \color{blue}{\cot(x)=\frac{1}{\tan(x)}}\)

right?

Answer 16

yeah

Answer 17

yes, and if you are dealing with inverse-trigonometric functions, then they would also satisfy this property.

This way,

\(\large \displaystyle \color{blue}{\cot^{-1}(x)=\frac{1}{\tan^{-1}(x)}}\)

Answer 18

yes

Answer 19

Knowing that, we are also able to conclude that:

\(\large \displaystyle \color{blue}{\cot^{-1}\left(~\tan(x)~\right)=\frac{1}{\tan^{-1}(~\tan(x)~)}}\)

(Except that this time you are taking an inverse tangent, not of x, but of a bigger quantity, and that is tan(x) )

Answer 20

yes

Answer 21

is my last blue equation making sense so far?

Answer 22

oh, glitched was just about to ask...
anyway...

Answer 23

\(\large \displaystyle \color{blue}{\tan^{-1}(~\tan(x)~)=x}\)

Answer 24

do we know that \(\Uparrow\) ?

Answer 25

yeah

Answer 26

And this way:

\(\large \displaystyle \color{blue}{\cot^{-1}\left(~\tan(x)~\right)=\frac{1}{\tan^{-1}(~\tan(x)~)}=\frac{1}{x}}\)

Answer 27

Right?

Answer 28

right. so basically it was just an entire reciprocal ?

Answer 29

Well, what happens really is that you ended up taking an inverse trigonometric function of its trigonometic function.

After that all trig peaces dissapear.

Such as,

\(\large \displaystyle \color{black}{\tan^{-1}(~\tan(x)~)=x}\)

Answer 30

\(\large \displaystyle \color{blue}{\cot^{-1}\left(~\tan(x)~\right)=\frac{1}{\tan^{-1}(~\tan(x)~)}=\frac{1}{x}}\)

and this is thus true for any x....
you can do the same with \(\pi\)/3
that is,

\(\large \displaystyle \color{blue}{\cot^{-1}\left(~\tan(\pi/3)~\right)=\frac{1}{\tan^{-1}(~\tan(\pi/3)~)}=\frac{1}{\pi/3}}\)

Answer 31

All right so this is what i did in the beginning. the equation is cot^-1(tanpi/3). So i dissected it and worked out tan of pi/3 and i got sqrt of 3. so now im left with cot^-1(sqrt3). SO i take the reciprocal of that, and get tang^-1(1/sqrt3). My teavher doesn't like sqrt on denominator so i change it to sqrt3/3. And in this lesson I've been answering in radians. so in my head (where confusion starts) i ask myself what is tang^-1(sqrt3/3) and for some reason i get pi/6.

Answer 32

Yeah, whenever you see an inverse trig function, OF a [regular] trig function, then there is a way to rewrite the inverse trig function in terms of that same [regular] trig function.

Answer 33

and then

\(\large \displaystyle \color{blue}{\cot^{-1}\left(~\tan(\pi/3)~\right)=\frac{1}{\tan^{-1}(~\tan(\pi/3)~)}=\frac{1}{\pi/3}=\frac{3}{\pi}}\)

Answer 34

oh ok. one more? \[\sin^{-1}(\frac{ \sqrt{2} }{ 2 })\]

Answer 35

i got pi/4

Answer 36

yes

Answer 37

I also thought of a 45-45-90 triangle :)

Answer 38

sin(x)=√2/2
x=45° (or \(\pi /4\))

Answer 39

\[\sin^{-1} \left( \cos \frac{ 5\Pi }{ 6 } \right)\]

Answer 40

-pi/3??

Answer 41

Please write sin(x) in terms of cos(x).

Answer 42

thats what the problem says on my paper.

Answer 43

I know, and that is the first thing you have to know how to do to easily obtain the answer.

Answer 44

1=sin²x+cos²x

right?

Answer 45

ummm that looks kinda complicated. Teacher never went over that yet

Answer 46

Well that is just a pythegorean identity.

Answer 47

You can look up proofs online for this theorem, but for now accept it as it is, I guess, if you want to do your problem.

Answer 48

1 = sin\(^2\)x+cos\(^2\)x

is true for any x. (This is an identity)

When rearranged, you get:

1 = sin\(^2\)x+cos\(^2\)x

1-cos\(^2\)x = sin\(^2\)x

\(\sqrt{1-\cos^2x}\) = sin x

Answer 49

(I am on purpose leaving ±, because it should be positive)

Answer 50

Do you see how I get that:

\(\sin(x)=\sqrt{1-\cos^2(x)}\) ?

Answer 51

yeah

Answer 52

Ok, and same thing would be true by inverse trigonometric functions.

That is,

\(\large\color{black}{ \displaystyle \sin^{-1}(x)=\sqrt{1-\left(\cos^{-1}x\right)^2} }\)

Answer 53

if any confusion, ask...
if not then I will go on

Answer 54

No i get it. i was just over complicating these problems. I appreciate your help. I have to go now, but thanks for your time.

Answer 55

Anytime, but we aren't done yet...

Answer 56

i know im going somewhere so i cant continue

Answer 57

Well, we kind of are... if you think about it all of these problems when you are taking inverse trig function of a trig function, -- are "snap-able" , but we need some work still

Answer 58

I will not be on for a while, so I will post it...

Answer 59

So we clarified that:

\(\large\color{black}{ \displaystyle \sin(x)=\sqrt{1-(\cos x)^2} }\)

and concluded that inverse trig functions would also satisfy the same property.
This way,

\(\large\color{black}{ \displaystyle \sin^{-1}(x)=\sqrt{1-\left(\cos^{-1}x\right)^2} }\)

And since we know that for a quantity x it is true:

\(\large\color{black}{ \displaystyle \sin^{-1}(\color{red}{x} )=\sqrt{1-\left(\cos^{-1}\color{red}{x}\right)^2} }\)

then,

\(\large\color{black}{ \displaystyle \sin^{-1}(\color{red}{\cos(x)} )=\sqrt{1-\left(\cos^{-1}\left(\color{red}{\cos(x)}\right)~\right)^2} }\)

and you know that inverse function of same regular function would leave you just x. Same thing happens now:

\(\large\color{black}{ \displaystyle \sin^{-1}(\color{red}{\cos(x)} )=\sqrt{1-\left(x\right)^2} }\)

Answer 60

And this way you know that:

\(\large\color{black}{ \displaystyle \sin^{-1}(\color{red}{\cos(x)} )=\sqrt{1-x^2} }\)

(you can actually use this one as an identity)

Answer 61

So if I had, for example:

\(\large\color{black}{ \displaystyle \sin^{-1}(\cos\frac{\sqrt{\pi}}{2} )=\sqrt{1-\left(\frac{\sqrt{\pi}}{2}\right)^2}= \sqrt{1-\frac{\pi }{4}} }\)

\(\large\color{black}{ \displaystyle \sqrt{1-\frac{\pi }{4}}=\sqrt{\frac{4}{4}-\frac{\pi }{4}}=\sqrt{\frac{4-\pi }{4}}=\frac{ \sqrt{4-\pi }}{ \sqrt{4}}= \frac{ \sqrt{4-\pi }}{ 2}}\)

Answer 62

good luck