Question

sin inverse[sin 6pi/7]. I'm suppose to find the exact value.

Answer 1

that is just 6pi/7

Answer 2

that is true for all inverse functions
f inverse f(x)=x

Answer 3

so the same thing applies for this: cos inverse[cos pi/2]
is the the exact value pi/2 ?

Answer 4

yes

Answer 5

thanks!

Answer 6

this is what inverse functions do

Answer 7

what if its negative like in this case: tan inverse[tan(-pi/4)] ?

Answer 8

You need to be careful here, the inverse sine function has range (-pi/2,pi/2). So the answer can't be 6pi/7.

Answer 9

would you take the absolute value of "(-pi/4)" so it would be "pi/4"

Answer 10

watchmath, i'm confused.

Answer 11

"f inverse f(x)=x"

Answer 12

oh true, did not think of that

Answer 13

What I am trying to say that
\(\sin^{-1}(\sin x)=x\) only for \(x\in[-\pi/2,\pi/2]\)

Answer 14

So first you need to find an angle x so that sin(x)=sin(6pi/7).
And we know that sin(pi/7)=sin(6pi/7)
So
\(\sin^{-1}(\sin 6\pi/7)=\sin^{-1}(\sin (\pi/7))=\pi/7.\)

Answer 15

so the exact value is pi/7?

Answer 16

Yes
For the tangent, we have
\[\tan^{-1}(\tan x)=x\]only for \(x\in (-\pi/2,\pi/2)\)