Evaluating an Inverse Trigonometric Function

  • #1

Homework Statement


Evaluate sin^-1(cos70°)


Homework Equations





The Attempt at a Solution



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

Answers and Replies

  • #2
1,540
135

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
 
  • #3
34,876
6,615
Also, cos(70°) ≠ 1/2 and there is no real number θ for which sin(θ) = ##\pi/3## > 1.
 
  • #4
Could you possibly explain the cos(90°-θ)=sinθ?
 
  • #5
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:
  • #6
34,876
6,615
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.
 
  • #7
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
34,876
6,615
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°)) = ?
 
  • #11
I apologize for being so confused, independent study is not being kind to me. Thank you very much, your help is immensely appreciated.
 
  • #12
34,876
6,615
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!
 
  • #13
1,540
135
Thanks Mark for nicely guiding the OP.
 
  • #14
34,876
6,615
That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.
 
  • #15
1,540
135
That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.

and an excellent mentor :smile:
 

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