Chapter 7.5, Problem 87AYU

In Problems 87-92, write each trigonometric expression as an algebraic expression containing $U\text{ and }V$. Give the restrictions required on $U\text{ and }V$.

$\sin \left({\tan }^{-1}u-{\text{ sin}}^{-1}v\right)$

To determine

To find: The trigonometric expression $\sin \left({\tan }^{-1}u-{\sin }^{-1}v\right)$ into algebraic expression containing $u$ and $v$ and to give the restrictions on $u$ and $v$.

Answer to Problem 87AYU

Solution:

The trigonometric expression $\sin \left({\tan }^{-1}u-{\sin }^{-1}v\right)$ into algebraic expression containing $u$ and $v$ $=u\cdot \sqrt{1-v^2}-v$ and the restrictions on $u$ and $v$ $=-\alpha <u<\infty \text{ and }-1\le v\le 1$.

Explanation of Solution

Given:

The trigonometric expression $\sin \left({\tan }^{-1}u-{\sin }^{-1}v\right)$.

Calculation:

Let $\alpha ={\tan }^{-1}u$, $\text{and }\beta ={\sin }^{-1}v$.

$u=\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{u}{1}$, $\frac{-\pi }{2}<\alpha <\frac{\pi }{2}$, $-\infty <u<\infty \text{ and }v=\sin \beta$, $\frac{-\pi }{2}\le \beta \le \frac{\pi }{2}$, $-1\le v\le 1$

$\cos \alpha =1$, $\sin \alpha =u$, $\cos \beta =\sqrt{1-v^2}$, $\sin \beta =v$

$\frac{-\pi }{2}<\alpha <\frac{\pi }{2}$, $\cos \alpha \ge 0 \text{and }0\le \beta \le \pi$, $\cos \beta \ge 0$

$\sin \left({\tan }^{-1}u-{\sin }^{-1}v\right)=\sin \left(\alpha -\beta \right)$   -----1

By difference formula of $\text{sine}$ .

$\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$   -----2

From equation 1 and 2.

$\sin \left({\tan }^{-1}u-{\sin }^{-1}v\right)=u\cdot \sqrt{1-v^2}-1\cdot v$