Answer
$$\sin2x=\frac{15}{17}$$
$$\cos2x=-\frac{8}{17}$$
Work Step by Step
$$\tan x=\frac{5}{3} \hspace{2cm}\sin x\lt0$$
$$\sin 2x=?\hspace{2cm}\cos2x=?$$
1) First, we need to calculate $\sin x$ and $\cos x$ as they are essential to the calculations of $\sin 2x$ and $\cos 2x$.
We know that $\tan x=\frac{\sin x}{\cos x}$.
Thus, since $\tan x=\frac{5}{3}\gt0$ and $\sin x\lt0$, we deduce that $\cos x\lt0$.
According to Pythagorean Identities:
$$\tan^2x+1=\sec^2x$$
According to Reciprocal Identities:
$$\sec x=\frac{1}{\cos x}$$
Therefore, $$\tan^2x+1=\frac{1}{\cos^2x}$$
$$\cos^2x=\frac{1}{\tan^2x+1}=\frac{1}{\Big(\frac{5}{3}\Big)^2+1}=\frac{1}{\frac{25}{9}+1}=\frac{1}{\frac{34}{9}}=\frac{9}{34}$$
$$\cos x=-\frac{3}{\sqrt{34}}=-\frac{3\sqrt{34}}{34}\hspace{1cm}(\cos x\lt0)$$
Now we find $\sin x$, using Quotient Identities:
$$\tan x=\frac{\sin x}{\cos x}$$
$$\sin x=\tan x\cos x=\frac{5}{3}\times\Big(-\frac{3\sqrt{34}}{34}\Big)=-\frac{5\sqrt{34}}{34}$$
2) Now we can calculate $\sin2x$ and $\cos2x$ using Double-Angle Identities
$$\sin2x=2\sin x\cos x=2\times\Big(-\frac{5\sqrt{34}}{34}\Big)\times\Big(-\frac{3\sqrt{34}}{34}\Big)=\frac{2\times15\times34}{34^2}=\frac{15}{17}$$
$$\cos 2x=\cos^2x-\sin^2x=\Big(-\frac{5\sqrt{34}}{34}\Big)^2-\Big(-\frac{3\sqrt{34}}{34}\Big)^2=\frac{9}{34}-\frac{25}{34}=-\frac{8}{17}$$
