Chapter 7, Problem 20CT

To determine

To calculate: The exact value of $\tan \left(2{\sin }^{-1}\frac{6}{11}\right)$ using sum, difference, product, or half-angle formulas.

Answer to Problem 20CT

Solution:

The exact value of $\tan \left(2{\sin }^{-1}\frac{6}{11}\right)$ is $\frac{12\sqrt{85}}{49}$.

Explanation of Solution

Given information:

The expression is $\tan \left(2{\sin }^{-1}\frac{6}{11}\right)$

Formula used:

Double angle formula of tan function: $\tan \left(2\theta \right)=\frac{2\tan \theta }{1-{\tan }^2\theta }$

Calculation:

Consider the expression $\tan \left(2{\sin }^{-1}\frac{6}{11}\right)$.

First find ${\sin }^{-1}\frac{6}{11}$.

Let $\theta ={\sin }^{-1}\frac{6}{11}$.

Then $\sin \theta =\frac{6}{11}$, 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 $\sin \theta =\frac{6}{11}=\frac{y}{r}$,

Let $y=6$ and $r=11$

The point on the circle of radius $r=11$ is $\left(x,6\right)$.

This implies that $x^2+6^2={11}^2$

$\Rightarrow x^2+36=121$

$\Rightarrow x^2=85$

$\Rightarrow x=\sqrt{85}\text{or}x=-\sqrt{85}$

Since $x>0$ in $I$ quadrant

$\Rightarrow x=\sqrt{85}$

Hence, $\tan \theta =\frac{y}{x}=\frac{6}{\sqrt{85}}=\frac{6\sqrt{85}}{85}$

Since ${\sin }^{-1}\frac{6}{11}=\theta$, it implies that $\tan \left(2{\sin }^{-1}\frac{6}{11}\right)=\tan \left(2\theta \right)$

Since $\tan \left(2\theta \right)=\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\tan \left(2\theta \right)=\frac{2\left(\frac{6\sqrt{85}}{85}\right)}{1-{\left(\frac{6\sqrt{85}}{85}\right)}^2}=\frac{\left(\frac{12\sqrt{85}}{85}\right)}{1-\frac{36\cdot 85}{{85}^2}}=\frac{\left(\frac{12\sqrt{85}}{85}\right)}{\frac{{85}^2-36\cdot 85}{{85}^2}}=\frac{\left(\frac{12\sqrt{85}}{85}\right)}{\frac{85\left(85-36\right)}{{85}^2}}=\frac{\left(\frac{12\sqrt{85}}{85}\right)}{\frac{49}{85}}=\frac{12\sqrt{85}}{85}\cdot \frac{85}{49}=\frac{12\sqrt{85}}{49}$

Therefore, the exact value of $\tan \left(2{\sin }^{-1}\frac{6}{11}\right)$ is $\frac{12\sqrt{85}}{49}$.