Question

If $\cos \left(x\right)=-\frac{2}{3}$and $x$ is in quadrant III, find the exact value of $\sin \left(2x\right)$, $\cos \left(2x\right)$, and $\tan \left(2x\right)$ algebraically without solving for $x$.

Answer 1

Find the value of $\sin \left(2x\right)$, $\cos \left(2x\right)$, and $\tan \left(2x\right)$:

The double angle formula for $\sin \left(x\right)$ is $\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)$.

Since the value of $\cos \left(x\right)$ is given as $\cos \left(x\right)=-\frac{2}{3}$ , find the value of $\sin \left(x\right)$ using identity ${\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1$:

$\begin{array}{rcl}{\sin }^2\left(x\right)+{\cos }^2\left(x\right)& =& 1\\ & & \\ {\sin }^2\left(x\right)+{\left(-\frac{2}{3}\right)}^2& =& 1\\ \sin \left(x\right)& =& \sqrt{1-{\left(-\frac{2}{3}\right)}^2}\\ & =& \sqrt{1-\frac{4}{9}}\\ & =& \sqrt{\frac{5}{9}}\\ & =& \pm \frac{\sqrt{5}}{3}\end{array}$

Since the value of $\sin \left(x\right)$ is negative in quadrant III, taking $\sin \left(x\right)=-\frac{\sqrt{5}}{3}$.

Hence, the value of $\sin \left(2x\right)$ is given by as follows.

$\begin{array}{rcl}\sin \left(2x\right)& =& 2\sin \left(x\right)\cos \left(x\right)\\ & =& 2\left(-\frac{\sqrt{5}}{3}\right)\left(-\frac{2}{3}\right)\\ & =& \frac{4\sqrt{5}}{9}\end{array}$

Now, find $\cos \left(2x\right)$ by using the values $\sin \left(x\right)=-\frac{\sqrt{5}}{3}$and $\cos \left(x\right)=-\frac{2}{3}$ in the formula $\cos \left(2x\right)={\cos }^2\left(x\right)-{\sin }^2\left(x\right)$ as follows:

$\begin{array}{rcl}\cos \left(2x\right)& =& {\left(-\frac{2}{3}\right)}^2-{\left(-\frac{\sqrt{5}}{3}\right)}^2\\ & =& \frac{4}{9}-\frac{5}{9}\\ & =& -\frac{1}{9}\end{array}$

And now, find $\tan \left(2x\right)$using the $\sin \left(x\right)=-\frac{\sqrt{5}}{3}$and $\cos \left(x\right)=-\frac{2}{3}$ in the formula $\tan \left(2x\right)=\frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}$ where $\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}$.

$\begin{array}{rcl}\tan \left(x\right)& =& \frac{\sin \left(x\right)}{\cos \left(x\right)}\\ & =& \raisebox{1ex}{$-{\displaystyle \frac{\sqrt{5}}{3}}$}\!\left/ \!\raisebox{-1ex}{$-{\displaystyle \frac{2}{3}}$}\right.\\ & =& \frac{\sqrt{5}}{2}\end{array}$

Hence the value of $\tan \left(2x\right)$is given by:

$\begin{array}{rcl}\tan \left(2x\right)& =& \frac{2\left(\sqrt{5}/2\right)}{1-{\left(\sqrt{5}/2\right)}^2}\\ & =& \frac{\sqrt{5}}{1-(5/4)}\\ & =& \frac{\sqrt{5}}{-1/4}\\ & =& -4\sqrt{5}\end{array}$

Hence, the value of $\sin \left(2x\right)$, $\cos \left(2x\right)$, and $\tan \left(2x\right)$ is $\frac{4\sqrt{5}}{9}$, $-\frac{1}{9}$and $-4\sqrt{5}$respectively.