I recognize $ an x=-2sqrt2$. Just how to uncover $sin x$ and also $cos x$ if $xin<-fracpi2,0>$? They most likely would be $-frac2sqrt23$ and also $frac13$ respectively yet I don"t know exactly how to prove it.

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We understand the signs.

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

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

Substitute the an initial equation into the second, and you deserve to solve for $cos x$. Remember $cos x>0$ in this quadrant. After that you should have the ability to recover $sin x$ making use of the first equation.


Take $sin x=y$ which means that $cos x= pmsqrt1-y^2$ wherein the sign depends on the quadrant the interest. Currently equate their ratio with $ an x$ and also solve. Again, you select the proper value of $y$ depending upon your name: coordinates of interest.


The given suggestions are great and simpler, together an alternate for a direct calculation recall the for $ hetain(-pi/2,pi/2)$

$$y= an heta iff heta=arctan y$$

and therefore since in that instance $xin(-pi/2,0)$ through $y=-2sqrt2$ we can use that by composition formulas

$sin x= sin (arctan y)=fracysqrt1+y^2$$cos x= cos (arctan y)=frac1sqrt1+y^2$

Use the simple relations$$cos^2x=frac11+ an^2x,enspace extwhence quad sin^2x= an^2xcos^2x=frac an^2x1+ an^2x.$$So $sin x$ is recognized up to its sign. ~ above $igl<-fracpi2,0igr>$, that is an unfavorable or $0$.

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