Answer
a. See sketch
b. $cos2\theta=-\frac{7}{25}$
c. $sin\theta= \frac{4}{5}$, $cos\theta =\frac{3}{5}$
d. $45^\circ \lt \theta \lt 90^\circ$
Work Step by Step
a. See sketch. The endpoint has coordinates $(-7,24)$ and the hypotenuse is $r=\sqrt {7^2+24^2}=25$
b. As $2\theta$ is in Quadrant II, we have $cos2\theta=-\frac{7}{25}$
c. Using the identities, we have
$sin\theta=\sqrt {\frac{1-(-7/25)}{2}}=\sqrt {\frac{25+7}{50}}=\frac{4}{5}$
and
$cos\theta=\sqrt {\frac{1+(-7/25)}{2}}=\sqrt {\frac{25-7}{50}}=\frac{3}{5}$
d. Since $90^\circ \lt 2\theta \lt 180^\circ$, we have $45^\circ \lt \theta \lt 90^\circ$ which is in Quadrant I. Thus both $sin\theta$ and $cos\theta$ are positive.
