Chapter 7, Problem 21CT

To determine

To calculate: The exact value of $\cos \left({\sin }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}\right)$ using sum, difference, product, or half-angle formulas.

Answer to Problem 21CT

Solution:

The exact value of $\cos \left({\sin }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}\right)$ is $\frac{2\sqrt{65}}{39}-\frac{6\sqrt{13}}{39}$.

Explanation of Solution

Given information:

The expression is $\cos \left({\sin }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}\right)$

Formula used:

Sum formula: $\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$

Calculation:

Consider the expression $\cos \left({\sin }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}\right)$.

By using the sum formula,

$\cos \left({\sin }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}\right)=\cos \left({\sin }^{-1}\frac{2}{3}\right)\cdot \cos \left({\tan }^{-1}\frac{3}{2}\right)-\sin \left({\sin }^{-1}\frac{2}{3}\right)\cdot \sin \left({\tan }^{-1}\frac{3}{2}\right)$. .. .. .. (*)

First find $\cos \left({\sin }^{-1}\frac{2}{3}\right)$.

Let $\theta ={\sin }^{-1}\frac{2}{3}$.

Then $\sin \theta =\frac{2}{3}$ where $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$.

Because $\sin \theta >0$, it follows that $0<\theta \le \frac{\pi }{2}$,

So $\theta$ lies in $I$ quadrant.

Since $\sin \theta =\frac{2}{3}=\frac{y}{r}$,

Let $y=2$ and $r=3$

The point on the circle of radius $r=3$ is $\left(x,2\right)$.

This implies that $x^2+2^2=3^2$

$\Rightarrow x^2+4=9$

$\Rightarrow x^2=5$

$\Rightarrow x=\sqrt{5}\text{or}x=-\sqrt{5}$

Since $x>0$,

$\Rightarrow x=\sqrt{5}$

Since $\cos \theta =\frac{x}{r}=\frac{\sqrt{5}}{3}$,

$\Rightarrow \cos \left({\sin }^{-1}\frac{2}{3}\right)=\cos \left(\theta \right)=\frac{\sqrt{5}}{3}$. .. .. . (1)

Now find $\cos \left({\tan }^{-1}\frac{3}{2}\right)$.

Let $\theta ={\tan }^{-1}\frac{3}{2}$.

Then $\tan \theta =\frac{3}{2}$, where $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$.

Because $\tan \theta >0$, it follows that $0<\theta <\frac{\pi }{2}$.

So $\theta$ lies in $I$ quadrant.

Since $\tan \theta =\frac{3}{2}=\frac{y}{x}$

Let $x=2$ and $y=3$

The point on the circle of radius $r$ is $\left(2,3\right)$.

This implies that $2^2+3^2=r^2$

$\Rightarrow 4+9=r^2$

$\Rightarrow r^2=13$

$\Rightarrow r=\sqrt{13}\text{or}x=-\sqrt{13}$

Since $r>0$

$\Rightarrow r=\sqrt{13}$

Since $\cos \theta =\frac{x}{r}=\frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}$,

$\Rightarrow \cos \left({\tan }^{-1}\frac{3}{2}\right)=\cos \left(\theta \right)=\frac{2\sqrt{13}}{13}$. .. .. .. . (2)

Now find $\sin \left({\sin }^{-1}\frac{2}{3}\right)$,

By using $\sin \left({\sin }^{-1}x\right)=x$ where $-1\le x\le 1$,

Since $-1\le x=\frac{2}{3}\le 1$,

$\sin \left({\sin }^{-1}\frac{2}{3}\right)=\frac{2}{3}$. .. .. .. (3)

Now find $\sin \left({\tan }^{-1}\frac{3}{2}\right)$,

Let $\theta ={\tan }^{-1}\frac{3}{2}$,

Then $\tan \theta =\frac{3}{2}$, where $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$,

Because $\tan \theta >0$, it follows that $0<\theta <\frac{\pi }{2}$,

So $\theta$ lies in $I$ quadrant.

Since $\tan \theta =\frac{3}{2}=\frac{y}{x}$,

Let $x=2$ and $y=3$

The point on the circle of radius $r$ is $\left(2,3\right)$.

This implies that $2^2+3^2=r^2$

$\Rightarrow 4+9=r^2$

$\Rightarrow r^2=13$

$\Rightarrow r=\sqrt{13}\text{or}x=-\sqrt{13}$

Since $r>0$

$\Rightarrow r=\sqrt{13}$

Since $\sin \theta =\frac{y}{r}=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}$,

$\Rightarrow \sin \left({\tan }^{-1}\frac{3}{2}\right)=\sin \left(\theta \right)=\frac{3\sqrt{13}}{13}$. .. .. .. . (4)

By using (1), (2), (3) and (4), in (*)

$\Rightarrow \cos \left({\sin }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}\right)=\frac{\sqrt{5}}{3}\cdot \frac{2\sqrt{13}}{13}-\frac{2}{3}\cdot \frac{3\sqrt{13}}{13}$

$\Rightarrow \cos \left({\sin }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}\right)=\frac{2\sqrt{65}}{39}-\frac{6\sqrt{13}}{39}$

Therefore, the exact value of $\cos \left({\sin }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}\right)$ is $\frac{2\sqrt{65}}{39}-\frac{6\sqrt{13}}{39}$.