# Evaluating an Inverse Trigonometric Function

1. Oct 14, 2013

### thatguythere

1. The problem statement, all variables and given/known data
Evaluate sin^-1(cos70°)

2. Relevant equations

3. The attempt at a solution

sin^-1(cos70°)=θ
sinθ=cos70°
sinθ=1/2
sinθ=∏/3

2. Oct 14, 2013

### Tanya Sharma

First use cos(90° - θ ) = sinθ and then sin-1(sinθ) = θ

3. Oct 15, 2013

### Staff: Mentor

Also, cos(70°) ≠ 1/2 and there is no real number θ for which sin(θ) = $\pi/3$ > 1.

4. Oct 15, 2013

### thatguythere

Could you possibly explain the cos(90°-θ)=sinθ?

5. Oct 15, 2013

### thatguythere

Ah a trig identity I see.
So cos(90°-70°)=sinθ
cos(20°)=sinθ
0.94=sinθ

sin^-1(sinθ)=θ
sin^-1(0.94)=θ
0.94=θ

Last edited: Oct 15, 2013
6. Oct 15, 2013

### Staff: Mentor

No, cos(90°-70°) = cos(20°) = sin(70°).
In general, cos(90° - θ) = sin(θ).
Most of the above makes no sense, with each line having little or no relation to the preceding line. Your value for θ is incorrect, and you could easily check that it is incorrect by using a calculator.

7. Oct 15, 2013

### thatguythere

Alright. So if cos(90°-x)=sinx, which in my case is 70°, I still do not understand why Tanya then encourages me to use sin^-1(sinθ)=θ. How is cos70° to be replaced by sinθ? By the trig identities, sin(90°-x)=cosx, it is sin20° which is equivalent to cos70°. Would my equation then be accurately represented by sin^-1(sin20°)=θ and then simply by using cancellation identities reduced to 20°=θ?

8. Oct 15, 2013

### Staff: Mentor

Because f-1(f(x)) = x, as long as x is in the domain of f. That's a basic property of a function and its inverse. It's also true that f(f-1(y)) = y, when y is in the domain of f-1. In short, the composition of a function and its inverse "cancels."
There's really no need for you to use θ, here, and I think that it is confusing you.
Not equivalent to - equal to.
cos(70°) = cos(90° - 20°) = sin(20°)
Make it simpler by getting rid of θ.

sin-1(sin(20°)) = ?

9. Oct 15, 2013

### thatguythere

sin^-1(sin(20°))=20°=∏/9

10. Oct 15, 2013

### Staff: Mentor

Yes.

11. Oct 15, 2013

### thatguythere

I apologize for being so confused, independent study is not being kind to me. Thank you very much, your help is immensely appreciated.

12. Oct 15, 2013

### Staff: Mentor

Sure, you're welcome!

13. Oct 16, 2013

### Tanya Sharma

Thanks Mark for nicely guiding the OP.

14. Oct 16, 2013

### Staff: Mentor

That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.

15. Oct 17, 2013

### Tanya Sharma

and an excellent mentor