# Find $\sin x$ and $\cos x$ knowing $\tan x$

I know $$\tan x=-2\sqrt2$$. How to find $$\sin x$$ and $$\cos x$$ if $$x\in[-\frac{\pi}{2},0]$$? They probably would be $$-\frac{2\sqrt2}{3}$$ and $$\frac{1}{3}$$ respectively but I don't know how to prove it.

We know the signs.

We have $$\sin x = -2\sqrt2 \cos x$$ and

$$\sin^2 x+ \cos^2 x = 1$$

Substitute the first equation into the second, and you can solve for $$\cos x$$. Remember $$\cos x>0$$ in this quadrant. After that you should be able to recover $$\sin x$$ using the first equation.

Take $$\sin x=y$$ which means that $$\cos x= \pm\sqrt{1-y^2}$$ where the sign depends on the quadrant of interest. Now equate their ratio with $$\tan x$$ and solve. Again, you choose the appropriate value of $$y$$ depending on your coordinate of interest.

The given suggestions are good and simpler, as an alternative for a direct calculation recall that for $$\theta\in(-\pi/2,\pi/2)$$

$$y=\tan \theta \iff \theta=\arctan y$$

and therefore since in that case $$x\in(-\pi/2,0)$$ by $$y=-2\sqrt2$$ we can use that by composition formulas

• $$\sin x= \sin (\arctan y)=\frac{y}{\sqrt{1+y^2}}$$
• $$\cos x= \cos (\arctan y)=\frac{1}{\sqrt{1+y^2}}$$

Use the basic relations $$\cos^2x=\frac1{1+\tan^2x},\enspace\text{whence }\quad \sin^2x=\tan^2x\cos^2x=\frac{\tan^2x}{1+\tan^2x}.$$ So $$\sin x$$ is known up to its sign. On $$\bigl[-\frac\pi2,0\bigr]$$, it is negative or $$0$$.