Can someone walk me through the following problem:
Find sin2(x) , cos2(x), and tan2(x) if cos(x) = -3/[(13)^(1/2)] and x terminates in quadrant II
Thank You.
Hi super,
Since $\displaystyle \cos X = \frac{-3}{\sqrt{13}}$, this means that $\displaystyle x = -3$ and $\displaystyle r=\sqrt{13}$.
Using $\displaystyle r^2=x^2+y^2$, you can determine that $\displaystyle y=2$.
This would make $\displaystyle \sin X = \frac{2}{\sqrt{13}}$
Note that $\displaystyle \sin 2X = 2\sin X \cos X$
Substituting, we arrive at $\displaystyle \sin 2X=2\left(\frac{2}{\sqrt{13}}\right)\left(\frac{-3}{\sqrt{13}}\right)=-\frac{12}{13}$.
Use the double angle identities for $\displaystyle \cos 2X$ and $\displaystyle \tan 2X $ to complete your assignment.