Chapter 4.7, Problem 130E

a.

To determine

To prove: $\arcsin \left(-x\right)=-\arcsin x$

Explanation of Solution

Let $y=-\arcsin x$

Then $-y=\arcsin x$

This implies that,

  $\sin \left(-y\right)=x$

But the sine function is an odd function and hence $\sin \left(-y\right)=-\sin y$ . This implies that,

  $-\sin \left(y\right)=x$

Then,

  $\sin y=-x$

And this gives that,

  $y=\arcsin \left(-x\right)$

Substituting $y=-\arcsin x$ in the above equation,

  $\arcsin \left(-x\right)=-\arcsin x$

Hence proved

b.

To determine

To prove: $\arctan \left(-x\right)=-\arctan x$

Explanation of Solution

Let $y=-\arctan x$

Then $-y=\arctan x$

This implies that,

  $\tan \left(-y\right)=x$

But the tan function is an odd function and hence $\tan \left(-y\right)=-\tan y$ . This implies that,

  $-\tan \left(y\right)=x$

Then,

  $\tan y=-x$

And this gives that,

  $y=\arctan \left(-x\right)$

Substituting $y=-\arctan x$ in the above equation,

  $\arctan \left(-x\right)=-\arctan x$

Hence proved

c.

To determine

To prove: $\arctan x+\arctan \frac{1}{x}=\frac{\pi }{2}$ , $x>0$

Explanation of Solution

Draw a right angled triangle with sides 1 unit and x unit as shown in figure below:

  

Here a and b are the acute angles of the triangle.

Since for a right triangle, the sum of acute angles is $\frac{\pi }{2}$ . This gives that,

  $a+b=\frac{\pi }{2}$

Now consider,

  $\begin{array}{c}\tan a=\frac{opp}{adj}\\ =\frac{x}{1}\\ =x\end{array}$

This gives that,

  $\arctan x=a$

Now consider,

  $\begin{array}{c}\tan b=\frac{opp}{adj}\\ =\frac{1}{x}\end{array}$

This gives that,

  $\arctan \frac{1}{x}=b$

Thus,

  $\begin{array}{c}\arctan x+\arctan \frac{1}{x}=a+b\\ =\frac{\pi }{2}\end{array}$

d.

To determine

To prove: $\arcsin x+\arccos x=\frac{\pi }{2}$

Explanation of Solution

Draw a right angled triangle with hypotenuse 1 unit and a side x unit as shown in figure below:

  

Here a and b are the acute angles of the triangle.

Since for a right triangle, the sum of acute angles is $\frac{\pi }{2}$ . This gives that,

  $a+b=\frac{\pi }{2}$

Now consider,

  $\begin{array}{c}\sin a=\frac{opp}{hyp}\\ =\frac{x}{1}\\ =x\end{array}$

This gives that,

  $\arcsin x=a$

Now consider,

  $\begin{array}{c}\cos b=\frac{adj}{hyp}\\ =\frac{x}{1}\\ =x\end{array}$

This gives that,

  $\arccos x=b$

Thus,

  $\begin{array}{c}\arcsin x+\arccos x=a+b\\ =\frac{\pi }{2}\end{array}$

e.

To determine

To prove: $\arcsin x=\arctan \frac{x}{\sqrt{1-x^2}}$

Explanation of Solution

Draw a right angled triangle with hypotenuse 1 unit and a side x unit as shown in figure below:

  

Here a and b are the acute angles of the triangle.

Then, the length of the remaining side can be found using the Pythagoras theorem as shown:

  $\begin{array}{c}BC=\sqrt{A{C}^2-A{b}^2}\\ =\sqrt{1^2-x^2}\\ =\sqrt{1-x^2}\end{array}$

Now consider,

  $\begin{array}{c}\sin a=\frac{opp}{hyp}\\ =\frac{x}{1}\\ =x\end{array}$

This gives that,

  $\arcsin x=a$ ….. (1)

Now consider,

  $\begin{array}{c}\tan a=\frac{opp}{adj}\\ =\frac{x}{\sqrt{1-x^2}}\end{array}$

This gives that,

  $\arctan \frac{x}{\sqrt{1-x^2}}=a$ ….. (2)

Thus, from (1) and (2),

  $\arcsin x=a=\arctan \frac{x}{\sqrt{1-x^2}}$

Hence,

  $\arcsin x=\arctan \frac{x}{\sqrt{1-x^2}}$