## anonymous 5 years ago sin inverse[sin 6pi/7]. I'm suppose to find the exact value.

1. anonymous

that is just 6pi/7

2. anonymous

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

3. anonymous

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

4. anonymous

yes

5. anonymous

thanks!

6. anonymous

this is what inverse functions do

7. anonymous

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

8. watchmath

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

9. anonymous

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

10. anonymous

watchmath, i'm confused.

11. anonymous

"f inverse f(x)=x"

12. anonymous

oh true, did not think of that

13. watchmath

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

14. watchmath

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 $$\sin^{-1}(\sin 6\pi/7)=\sin^{-1}(\sin (\pi/7))=\pi/7.$$

15. anonymous

so the exact value is pi/7?

16. watchmath

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