Question

cos^-1(cos 7pi/6) = ?

Answer 1

@pooja195 @rational @Miracrown @Zale101 @KingGeorge @sleepyjess @Nnesha

Answer 2

Are you assuming that \(\cos^{-1}\) is the inverse cosine function?

Answer 3

Yeah!

Answer 4

In that case, the only thing you have to be careful about, is the range of \(\cos^{-1}\). Usually with inverse functions you have that \(f^{-1}(f(x))=x\). But with trig functions it's a bit trickier. So the first step, is to figure out what \(\cos(7\pi/6)\) is. So what is it? How well do you remember your unit circle?

Answer 5

sec im glitched

Answer 6

its not letting me post :/

Answer 7

Not very well, I remember it well for the first quadrant and sort of okayish on the 2nd quadrant. Probably why these problems are harder.

But I do know [0, pi] is the principle value branch for cos and that cos^-1(cos 7pi/6) would give 7pi/6?

Answer 8

hey
\[\cos 7\pi/6 = \cos(\pi + \pi/6)\]

Answer 9

proceed further

Answer 10

Why take cos 7pi/6 as cos(pi + pi/6)?

Answer 11

remember??
\[\cos^{-1} (\cos \theta) = \cos \theta~for~all~~ \theta \epsilon[0 , \pi ]\]

Answer 12

Yeah I do! :)

Answer 13

But cos^-1 and cos cancel each other and we get theta only?

Answer 14

What rishavraj said. The problem is that \(7\pi/6\) isn't in that interval. So we can't just cancel like you want to.

Answer 15

here its 7pi/6
\[7\pi/6 > \pi\]

Answer 16

Oh so we get 7pi/6 into a form which belongs to the principle value branch of cos^-1

Answer 17

yeah now just apply cos (pi + x) =????

Answer 18

cos( pi + x) = cos x well actually 2pi + x

Answer 19

\[\cos (\pi + x) = -\cos x\]

Answer 20

https://www.adelaide.edu.au/mathslearning/handouts/useful-trig-identities.pdf

Answer 21

sec writing that down, my bad

Answer 22

Oh my god. Lovely pdf, I missed so many! I'll get that printed and copy it down, learn them, thank you O_O

Answer 23

While I do like that approach, I think it may not be the best approach as it requires the memorization of a bunch of extra identities that aren't really necessary...

Answer 24

yeah i know only few r important .....its kinda rubbish......lol

Answer 25

Just when I thought I'm finally done with this chapter and its collection of identities, more come flying at me xD

Answer 26

If you want to try (what I think is) an easier approach, we need to know what \(\cos(7\pi/6)\) is.

Answer 27

@KingGeorge what u think cos7pi/6 =
i m blank.....

Answer 28

The back of the book says the answer should be 5pi/6

Answer 29

And that's the answer since \(\cos(7\pi/6)=\cos(5\pi/6)\). So in theory if you can figure out what \(\cos(7\pi/6)\) is all you need to do is match it up with an \(x\in[0,\pi]\) such that \(\cos(x)=\cos(7\pi/6)\).

Answer 30

@kennybm its correct........

Answer 31

ugh this is frustrating me so much.

2pi - 7pi/6?