Answer:
[tex]-1 \le u, v \le 1[/tex] --- restriction
[tex]\sin(\cos^{-1}u + \sin^{-1}v) = \sqrt{(1 - u^2)(1 - v^2)} + uv[/tex]
Step-by-step explanation:
Given
[tex]\sin(\cos^{-1}u + \sin^{-1}v)[/tex]
Required
Write in terms of u and v
Let:
[tex]a =\cos^{-1}u[/tex]
Take arc cos of both sides
[tex]\cos a = u[/tex]
Using the following trigonometry identity
[tex]\sin^2a + \cos^2a = 1[/tex]
So:
[tex]\sin^2a = 1 -\cos^2a[/tex]
[tex]\sin^2a = 1 -u^2[/tex]
Take square roots of both sides
[tex]\sin a = \sqrt{1 -u^2[/tex]
So, we have:
[tex]\cos a = u[/tex] [tex]\sin a = \sqrt{1 -u^2[/tex]
Let:
[tex]b =\cos^{-1}{\sqrt{1 - v^2}}[/tex]
Take arc cos of both sides
[tex]\cos b = \sqrt{1 - v^2[/tex]
So:
[tex]\sin^2b = 1 -\cos^2b[/tex]
[tex]\sin^2b = 1 - (\sqrt{1 - v^2})^2[/tex]
[tex]\sin^2b = 1 - (1 - v^2)[/tex]
[tex]\sin^2b = 1 - 1 + v^2[/tex]
[tex]\sin^2b = v^2[/tex]
Take square roots
[tex]\sin b = v[/tex]
So, we have:
[tex]\cos b = \sqrt{1 - v^2[/tex] [tex]\sin b = v[/tex]
So, we have:
[tex]\sin(\cos^{-1}u + \sin^{-1}v)[/tex]
[tex]\sin(\cos^{-1}u + \sin^{-1}v) = \sin(a + b)[/tex]
Apply sine rule
[tex]\sin(\cos^{-1}u + \sin^{-1}v) = \sin a\ cos b + \sin b \cos a[/tex]
[tex]\sin(\cos^{-1}u + \sin^{-1}v) = \sqrt{1 - u^2} * \sqrt{1 - v^2} + v *u[/tex]
[tex]\sin(\cos^{-1}u + \sin^{-1}v) = \sqrt{(1 - u^2)*(1 - v^2)} + v *u[/tex]
Expand
[tex]\sin(\cos^{-1}u + \sin^{-1}v) = \sqrt{(1 - u^2)(1 - v^2)} + uv[/tex]
The restriction on u and v is:
[tex]-1 \le u, v \le 1[/tex]
