Question

Angle α lies in quadrant II , and tan α = rac{12}{5}" alt="-\frac{12}{5}" align="absmiddle" class="latex-formula"> . Angle β lies in quadrant IV , and cosβ=3/5 .
What is the exact value of sin(α+β) ?

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sin(α+β) =

Answer 1

Since $\alpha$ lies in quadrant II and $\beta$ lies in quadrant IV, we expect $\sin\alpha>0$, $\cos\alpha$, and $\sin\beta$.

Recall the Pythagorean identities,

$\sin^2x+\cos^2x=1\iff1+\cot^2x=\csc^2x\iff\tan^2x+1=\sec^2x$

It follows that

$\sec\alpha=\dfrac1{\cos\alpha}=-\sqrt{\tan^2\alpha+1}=-\dfrac{13}5\implies\cos\alpha=-\dfrac5{13}$

$\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{12}{13}$

$\sin\beta=-\sqrt{1-\cos^2\beta}=-\dfrac45$

Recall the angle sum identity for sine:

$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$

So we have

$\sin(\alpha+\beta)=\dfrac{12}{13}\dfrac35+\left(-\dfrac45\right)\left(-\dfrac5{13}\right)=\boxed{\dfrac{56}{65}}$