use identities to find values of the sine and cosine functions of the function for the angle measure. 2x…

Explanation:

Step1: Determine the quadrant of x

Since $\tan x=- 3<0$ and $\cos x > 0$, x is in the fourth - quadrant.

Step2: Find $\sin x$ and $\cos x$ using $\tan x=\frac{\sin x}{\cos x}$ and $\sin^{2}x+\cos^{2}x = 1$

We know that $\tan x=\frac{\sin x}{\cos x}=-3$, so $\sin x=-3\cos x$. Substitute into $\sin^{2}x+\cos^{2}x = 1$: $(-3\cos x)^{2}+\cos^{2}x = 1$ $9\cos^{2}x+\cos^{2}x = 1$ $10\cos^{2}x = 1$ $\cos^{2}x=\frac{1}{10}$. Since x is in the fourth - quadrant, $\cos x=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}$, and $\sin x=-3\cos x=-\frac{3\sqrt{10}}{10}$.

Step3: Use double - angle formulas

The double - angle formula for $\cos2x$ is $\cos2x=\cos^{2}x-\sin^{2}x$. $\cos2x=\left(\frac{\sqrt{10}}{10}\right)^{2}-\left(-\frac{3\sqrt{10}}{10}\right)^{2}=\frac{10}{100}-\frac{90}{100}=-\frac{4}{5}$. The double - angle formula for $\sin2x$ is $\sin2x = 2\sin x\cos x$. $\sin2x=2\times\left(-\frac{3\sqrt{10}}{10}\right)\times\frac{\sqrt{10}}{10}=-\frac{3}{5}$.

Answer:

$\cos2x=-\frac{4}{5}$ $\sin2x=-\frac{3}{5}$