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 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<-fracpi2,0igr>\$, that is an unfavorable or \$0\$.

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