Evaluating an Inverse Trigonometric Function

[thatguythere]

Homework Statement

Evaluate sin^-1(cos70°)

Homework Equations

The Attempt at a Solution

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

 

[Tanya Sharma]

Homework Statement

Evaluate sin^-1(cos70°)

Homework Equations

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]

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

 

[thatguythere]

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

 

[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

 

[Mark44]

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.

 

[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?

 

[Mark44]

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°)) = ?

 

[thatguythere]

sin^-1(sin(20°))=20°=∏/9

 

[Mark44]

sin^-1(sin(20°))=20°=∏/9

Yes.

 

[thatguythere]

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

 

[Mark44]

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!

 

[Tanya Sharma]

Thanks Mark for nicely guiding the OP.

 

[Mark44]

That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.

 

[Tanya Sharma]

That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.

and an excellent mentor