Question

Write the following trigonometric expression as an algebraic expression containing u and v. Give the restrictions required on u and v. sin ( cos^-1 u sin^-1 v)

Answer 1

Answer:

$-1 \le u, v \le 1$ --- restriction

$\sin(\cos^{-1}u + \sin^{-1}v) = \sqrt{(1 - u^2)(1 - v^2)} + uv$

Step-by-step explanation:

Given

$\sin(\cos^{-1}u + \sin^{-1}v)$

Required

Write in terms of u and v

Let:

$a =\cos^{-1}u$

Take arc cos of both sides

$\cos a = u$

Using the following trigonometry identity

$\sin^2a + \cos^2a = 1$

So:

$\sin^2a = 1 -\cos^2a$

$\sin^2a = 1 -u^2$

Take square roots of both sides

$\sin a = \sqrt{1 -u^2$

So, we have:

$\cos a = u$     $\sin a = \sqrt{1 -u^2$

Let:

$b =\cos^{-1}{\sqrt{1 - v^2}}$

Take arc cos of both sides

$\cos b = \sqrt{1 - v^2$

So:

$\sin^2b = 1 -\cos^2b$

$\sin^2b = 1 - (\sqrt{1 - v^2})^2$

$\sin^2b = 1 - (1 - v^2)$

$\sin^2b = 1 - 1 + v^2$

$\sin^2b = v^2$

Take square roots

$\sin b = v$

So, we have:

$\cos b = \sqrt{1 - v^2$     $\sin b = v$

So, we have:

$\sin(\cos^{-1}u + \sin^{-1}v)$

$\sin(\cos^{-1}u + \sin^{-1}v) = \sin(a + b)$

Apply sine rule

$\sin(\cos^{-1}u + \sin^{-1}v) = \sin a\ cos b + \sin b \cos a$

$\sin(\cos^{-1}u + \sin^{-1}v) = \sqrt{1 - u^2} * \sqrt{1 - v^2} + v *u$

$\sin(\cos^{-1}u + \sin^{-1}v) = \sqrt{(1 - u^2)*(1 - v^2)} + v *u$

Expand

$\sin(\cos^{-1}u + \sin^{-1}v) = \sqrt{(1 - u^2)(1 - v^2)} + uv$

The restriction on u and v is:

$-1 \le u, v \le 1$