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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.

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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.

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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.

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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

  • $\sin x= \sin (\arctan y)=\frac{y}{\sqrt{1+y^2}}$
  • $\cos x= \cos (\arctan y)=\frac{1}{\sqrt{1+y^2}}$
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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 $0$.

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