Question

Find the exact values of sin 2u, cos 2u, and tan 2u using the double-angle formulas. sec u = −2, π/2 < u < π

Answer 1

$\bf sec(\theta)=\cfrac{hypotenuse}{adjacent} \\\\\\ sec(u)=-2\iff sec(u)=-\cfrac{2}{1}\cfrac{\leftarrow hypotenuse}{\leftarrow adjacent}$

now... we have -2/1   but... which one is the negative?   well
we know the angle is " π/2 < u < π", that simply means, the 2nd quadrant
well, in the 2nd quadrant, the "x" or adjacent side is negative, so the adjacent side is then -1

now, let us use the pythagorean theorem, to get the opposite side

$\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{(2)^2-(-1)^2}=b \\\\\\ \pm\sqrt{3}=b$

now..which one though?  well, recall, we're in the 2nd quadrant, in the 2nd quadrant "y" or the opposite side, is positive, so is √(3) then

now... let's plug in those values in the double-angle identities then

$\bf \textit{Double Angle Identities} \\ \quad \\ sin(2\theta)=2sin(\theta)cos(\theta) \\ \quad \\\\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ \boxed{1-2sin^2(\theta)}\\ 2cos^2(\theta)-1 \end{cases} \\ \\\\ tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\ -----------------------------$

$\bf sin(2u)=2\left( \cfrac{\sqrt{3}}{2} \right)\left( \cfrac{-1}{2}\right) \\\\\\ cos(2u)=1-2\left( \cfrac{\sqrt{3}}{2} \right)^2 \\\\\\ tan(2u)=\cfrac{2\left( \frac{\sqrt{3}}{-1} \right)}{1-\left( \frac{\sqrt{3}}{-1} \right)^2}$

simplify away