Question

arccos(sin(pi/6)) Help?

Answer 1

https://www.desmos.com/calculator

Answer 2

hmm ohhh of ... sine... ok

Answer 3

@jdoe0001 I think I'm supposed to try without a calculator?

Answer 4

http://math.info/image/325/trig_cofunction.gif <--- notice the sine cofunction

so we can say that we need a value that subtracted from 1/2 gives us 1/6

that is \(\bf \cfrac{1}{2}-\theta=\cfrac{1}{6}\implies \cfrac{1}{2}-\cfrac{1}{6}=\theta\implies \cfrac{3-1}{6}=\theta\implies \cfrac{1}{3}=\theta\qquad thus\\ \quad \\ \quad \\

sin\left(\frac{\pi}{6}\right)\implies sin\left(\frac{\pi}{2}-{\color{blue}{ \frac{\pi}{3}}}\right)\implies cos\left({\color{blue}{ \frac{\pi}{3}}}\right)\qquad thus\\ \quad \\
cos^{-1}\left[sin\left(\frac{\pi}{6}\right)\right]\implies
cos^{-1}\left[sin\left(\frac{\pi}{2}-{\color{blue}{ \frac{\pi}{3}}}\right)\right]\\ \quad \\
\implies cos^{-1}\left[cos\left({\color{blue}{ \frac{\pi}{3}}}\right)\right]\)

Answer 5

see what it is?

Answer 6

Not.. At.. All..

Answer 7

well.. did you see the cofunction identities?

Answer 8

Where did the 1/2 come from? And why theta?

Answer 9

Yes..

Answer 10

well, I use \(\theta\) I could have used any variable really
just needed a value that when subtracted from 1/2, would give 1/6

Answer 11

notice the sine cofunction there \(\Large \bf sin\left(\frac{\pi}{2}-{\color{blue}{ \square }}\right)=cos({\color{blue}{ \square }})\)

Answer 12

any part you find confusing?

Answer 13

All of it. So first I find 1/3. Then I.. Then it can all be equal to cos(1/3 because of those identities you showed... So then I do cos^-1(cos(pi/3))?

Answer 14

BRB

Answer 15

well... see, lemme use the angles themselves

\( \bf \cfrac{\pi}{2}-\cfrac{\pi}{3}=\cfrac{3\pi-2\pi}{6}\implies \cfrac{\pi}{6}\qquad \implies sin\left(\frac{\pi}{2}-{\color{blue}{ \frac{\pi}{3}}}\right) \iff sin\left(\frac{\pi}{6}\right)\)

anything confusing there?

Answer 16

Okay, got you so far.

Answer 17

The question asks this: Find the exact value of arccos(sin(pi/6)). For full credit, explain your reasoning.

Answer 18

right

Answer 19

so you see why we bother doing the \(\bf \cfrac{1}{2}-\theta=\cfrac{1}{6}\implies \cfrac{1}{2}-\cfrac{1}{6}=\theta\implies \cfrac{3-1}{6}=\theta\implies \cfrac{1}{3}=\theta\) part?

Answer 20

Yes. Got it. But what do you do after you get the cos(pi/3)?

Answer 21

say what's the \(\bf cos(\pi)\quad ?\)

Answer 22

Huh?

Answer 23

heheh, just a quick example :)

Answer 24

I didn't get it.. :c

Answer 25

as an example.... just as example... if you check your Unit Circle, what would you get for the \(\bf cos(\pi)\) ?

Answer 26

Not very good at the unit circle.. 180?

Answer 27

as an example.... just as example... if you check your Unit Circle, what would you get for the \(\bf cos(180^o)\) ?

Answer 28

-1?

Answer 29

right.... so
what if take arccos() of that, what would you get from \(\bf cos^{-1}({\color{blue}{ -1}})\quad ?\)

Answer 30

Oh, whoa. It's pi... cool.

Answer 31

So how do I connect that to what we did?

Answer 32

pi = 180 degrees

well, what does that mean? it means \(\bf {\color{blue}{ cos(180^o)}}=-1\qquad cos^{{\color{blue}{ -1}}}(-1)=180^o\\ \quad \\\qquad \implies cos^{-1}[{\color{blue}{ cos(180^o)}}]=180^o\\ \quad \\ \quad \\
thus\implies cos^{-1}[cos(\square )]=\square \)

Answer 33

\(\bf sin\left(\frac{\pi}{6}\right)\implies sin\left(\frac{\pi}{2}-{\color{blue}{ \frac{\pi}{3}}}\right)\implies cos\left({\color{blue}{ \frac{\pi}{3}}}\right)\qquad thus\\ \quad \\
cos^{-1}\left[sin\left(\frac{\pi}{6}\right)\right]\implies
cos^{-1}\left[sin\left(\frac{\pi}{2}-{\color{blue}{ \frac{\pi}{3}}}\right)\right]\\ \quad \\
\implies cos^{-1}\left[cos\left({\color{blue}{ \frac{\pi}{3}}}\right)\right]\implies \huge ?\)

Answer 34

So, afer we get 1/3, then it becomes cos^-1(cos(pi/3)). And because of the unit circle, the answer is pi/3?

Answer 35

yeap

Answer 36

\(\large \bf cos^{-1}\left[cos\left({\color{blue}{ \frac{\pi}{3}}}\right)\right]\implies {\color{blue}{ \frac{\pi}{3}}}\)

Answer 37

I think I understand now. Thanks! :)

Answer 38

yw