# Evaluating an Inverse Trigonometric Function

## Homework Statement

Evaluate sin^-1(cos70°)

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

## Homework Statement

Evaluate sin^-1(cos70°)

## The Attempt at a Solution

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

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

Mark44
Mentor
Also, cos(70°) ≠ 1/2 and there is no real number θ for which sin(θ) = ##\pi/3## > 1.

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

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:
Mark44
Mentor
Ah a trig identity I see.
So cos(90°-70°)=sinθ
No, cos(90°-70°) = cos(20°) = sin(70°).
In general, cos(90° - θ) = sin(θ).
cos(20°)=sinθ
0.94=sinθ

sin^-1(sinθ)=θ
sin^-1(0.94)=θ
0.94=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.

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?

Mark44
Mentor
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.
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."
How is cos70° to be replaced by sinθ?
There's really no need for you to use θ, here, and I think that it is confusing you.
By the trig identities, sin(90°-x)=cosx, it is sin20° which is equivalent to cos70°.
Not equivalent to - equal to.
cos(70°) = cos(90° - 20°) = sin(20°)
Would my equation then be accurately represented by sin^-1(sin20°)=θ and then simply by using cancellation identities reduced to 20°=θ?
Make it simpler by getting rid of θ.

sin-1(sin(20°)) = ?

sin^-1(sin(20°))=20°=∏/9

Mark44
Mentor
sin^-1(sin(20°))=20°=∏/9
Yes.

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

Mark44
Mentor
I apologize for being so confused, independent study is not being kind to me. Thank you very much, your help is immensely appreciated.
Sure, you're welcome!

Thanks Mark for nicely guiding the OP.

Mark44
Mentor
That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.

That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.

and an excellent mentor 