## Trigonometry (11th Edition) Clone

$$\sin x=-\frac{4}{5}$$ $$\tan x=-\frac{3}{4}$$ $$\cot(-x)=\frac{4}{3}$$
$$\cos x=\frac{3}{5}\hspace{2cm}\text{x in quadrant IV}$$ 1) Determining the signs of $\sin x$, $\tan x$ and $\cot x$ $x$ is in quadrant IV, where sines are negative but cosines are positive. As, $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$, this means both tangents and cotangents in quadrant are negative. Therefore, $\sin x\lt0$, $\tan x\lt0$ and $\cot x\lt0$. 2) Calculating $\sin x$ - Recall the Pythagorean Identities: $\sin^2x=1-\cos^2x$ Therefore, $$\sin^2x=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$$ $$\sin x=\pm\frac{4}{5}$$ But as $\sin x\lt0$, $$\sin x=-\frac{4}{5}$$ 3) Calculating $\tan x$ - Recall the Quotient Identities: $$\tan x=\frac{\sin x}{\cos x}=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$$ 4) Calculating $\cot(-x)$ - Recall the Quotient Identities: $$\cot x=\frac{\cos x}{\sin x}=\frac{-\frac{4}{5}}{\frac{3}{5}}=-\frac{4}{3}$$ - However, also recall the Negative-Angle Identities for cotangents: $$\cot(-x)=-\cot x=-\Big(-\frac{4}{3}\Big)=\frac{4}{3}$$