Chapter 3.6, Problem 16E

  $\left(a\right)$

To determine

To prove: The inverse trigonometric relation ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ .

Answer to Problem 16E

The solution ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

Explanation of Solution

Given information:

The relation ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ .

Formula used:

Inverse sine trigonometric function is expressed as ${\sin }^{-1}\left(x\right)=y$ which is equivalent to $x=\sin y$ where $0\le y\le \pi$ .

Inverse cosine trigonometric function is expressed as ${\cos }^{-1}\left(x\right)=y$ which is equivalent to $x=\cos y$ where $0\le y\le \pi$ .

Calculation:

Consider the provided relation ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ .

Recall that the sine trigonometric function is expressed as ${\sin }^{-1}\left(x\right)=y$ which is equivalent to $x=\sin y$ where $0\le y\le \pi$ and inverse cosine trigonometric function is expressed as ${\cos }^{-1}\left(x\right)=y$ which is equivalent to $x=\cos y$ where $0\le y\le \pi$ .

Let ${\sin }^{-1}\left(x\right)=y$ . This implies that $x=\sin y$ .

Recall the relation $\sin x=\cos \left(\frac{\pi }{2}-x\right)$ . Apply it

  $\begin{array}{c}x=\sin y\\ =\cos \left(\frac{\pi }{2}-y\right)\end{array}$

Take $\arccos$ function on both the sides of above expression,

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

Since, ${\sin }^{-1}\left(x\right)=y$ , therefore,

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

Hence, it is proved that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ .

  $\left(b\right)$

To determine

To prove: The derivative of inverse cosine trigonometric function is $\frac{d}{dx}\left({\cos }^{-1}x\right)=\frac{-1}{\sqrt{1-x^2}}$ .

Answer to Problem 16E

function is $\frac{d}{dx}\left({\cos }^{-1}x\right)=\frac{-1}{\sqrt{1-x^2}},-1<x<1$ .

Explanation of Solution

Given information:

The expression ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ .

Formula used:

Inverse sine trigonometric function is expressed as ${\sin }^{-1}\left(x\right)=y$ which is equivalent to $x=\sin y$ where $0\le y\le \pi$ .

Inverse cosine trigonometric function is expressed as ${\cos }^{-1}\left(x\right)=y$ which is equivalent to $x=\cos y$ where $0\le y\le \pi$ .

Calculation:

Consider theexpression ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ .

Rewrite the above expression as, ${\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left({\cos }^{-1}x\right)=\frac{d}{dx}\left(\frac{\pi }{2}-{\sin }^{-1}x\right)\\ \frac{d}{dx}\left({\cos }^{-1}x\right)=\frac{d}{dx}\left(\frac{\pi }{2}\right)-\frac{d}{dx}\left({\sin }^{-1}x\right)\\ \frac{d}{dx}\left({\cos }^{-1}x\right)=0-\frac{d}{dx}\left({\sin }^{-1}x\right)\\ \frac{d}{dx}\left({\cos }^{-1}x\right)=-\frac{d}{dx}\left({\sin }^{-1}x\right)\end{array}$

Recall that the inverse sine trigonometric function is expressed as ${\sin }^{-1}\left(x\right)=y$ which is equivalent to $x=\sin y$ where $0\le y\le \pi$ .

Consider the function $x=\sin y$ .

Differentiate both sides with respect to x ,

  $\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(\sin y\right)$

Recall that chain rule for differentiation is if f is a function of g then $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Apply it. Also observe that y is a function of x,

  $\begin{array}{c}\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(\sin y\right)\\ 1=\cos yy\text{'}\\ y\text{'}=\frac{1}{\cos y}\end{array}$

Now, as value of $\sin y\ge 0$ therefore, $0\le y\le \pi$ .

Recall the Pythagorean identity ${\cos }^{2}x+{\sin }^{2}x=1$ , Apply it

  $\begin{array}{c}y\text{'}=\frac{1}{\cos y}\\ y\text{'}=\frac{1}{\sqrt{1-{\sin }^{2}y}}\end{array}$

Since, $x=\cos y$ , therefore,

  $\begin{array}{c}y\text{'}=\frac{1}{\cos y}\\ y\text{'}=\frac{1}{\sqrt{1-{\sin }^{2}y}}\\ y\text{'}=\frac{1}{\sqrt{1-x^2}}\\ \frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}\end{array}$

Since, $\frac{d}{dx}\left({\cos }^{-1}x\right)=-\frac{d}{dx}\left({\sin }^{-1}x\right)$ , therefore,

  $\frac{d}{dx}\left({\cos }^{-1}x\right)=\frac{-1}{\sqrt{1-x^2}},-1<x<1$

Hence, it is proved that derivative of inverse cosine trigonometric function is $\frac{d}{dx}\left({\cos }^{-1}x\right)=\frac{-1}{\sqrt{1-x^2}},-1<x<1$ .