Chapter 5, Problem 25RE

To determine

To show: that the rate of change can also be written as $\cot x\sqrt{\sin x}$ .

Explanation of Solution

Given information:

The rate of change of the function $f\left(x\right)=2\sqrt{\sin x}$ is given by ${\sin }^{-1/2}x\cos x$ .

Calculation:

Consider the function

  $f\left(x\right)=2\sqrt{\sin x}$

The rate of change of the above function is given by the expression

  ${\sin }^{-1/2}x\cos x$

Rewrite the above expression as

  $\begin{array}{c}{\sin }^{-1/2}x\cos x=\left(\frac{1}{{\sin }^{\frac{1}{2}}x}\right)\cos x\\ =\left(\frac{1}{\sqrt{\sin x}}\right)\cos x\end{array}$

Multiply numerator and denominator by $\sqrt{\sin x}$ to get

  $\begin{array}{c}{\sin }^{-1/2}x\cos x=\left(\frac{1}{\sqrt{\sin x}}\right)\left(\cos x\right)\left(\frac{\sqrt{\sin x}}{\sqrt{\sin x}}\right)\\ =\left(\frac{\cos x}{\sin x}\right)\left(\sqrt{\sin x}\right)\mathrm{...........}(1)\end{array}$

Use the Quotient identity

  $\frac{\cos x}{\sin x}=\cot x$

Therefore, the expression in (1) reduces to

  $\begin{array}{c}{\sin }^{-1/2}x\cos x=\left(\frac{\cos x}{\sin x}\right)\left(\sqrt{\sin x}\right)\\ =\cot x\sqrt{\sin x}\end{array}$

Therefore, the expression ${\sin }^{-1/2}x\cos x$ is equal to $\cot x\sqrt{\sin x}$ .