Question

Angles θ and φ are angles in standard position such that:

Answer 1

When $\theta$ terminates in quadrant III, both $\cos\theta$ and $\sin\theta$ are negative, and

$\sin^2\theta+\cos^2\theta=1\implies\cos\theta=-\sqrt{1-\sin^2\theta}=-\dfrac{12}{13}$

When $\varphi$ terminates in quadrant II, $\cos\varphi$ is negative and $\sin\varphi$ is positive, so

$1+\tan^2\varphi=\sec^2\varphi\implies\sec\varphi=-\dfrac{17}{15}$

which gives

$\cos\varphi=\dfrac1{-\frac{17}{15}}=-\dfrac{15}{17}$

$\tan\varphi=\dfrac{\sin\varphi}{\cos\varphi}=-\dfrac8{15}\implies\sin\varphi=\dfrac8{17}$

Now,

$\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi=-\dfrac{21}{221}$