Question

If $\sin \alpha =\frac{15}{17}$, $\frac{\pi }{2}<\alpha <\pi$ and $\cos \beta =-\frac{4}{5}$, $\pi <\beta <\frac{3\pi }{2}$
Find $\sin (\alpha -\beta )$

• A. $\frac{88}{85}$
• B. $\frac{44}{85}$
• C. $\frac{84}{85}$
• D. $-\frac{84}{85}$

Answer 1

The correct option is D $-\frac{84}{85}$

$\sin \alpha =\frac{15}{17},\cos \alpha =\frac{-8}{17}$ $\frac{\pi }{2}<\alpha <\pi$

(REF. Image 1)

$\sqrt{{17}^2-{15}^2}=\sqrt{298-225}=8$

$\cos \beta =\frac{-4}{5}$, $\sin \beta =\frac{-3}{5}$ $\pi <\beta <\frac{3\pi }{2}$

(REF. Image 2)

$\sqrt{5^2-4^2}$

$=\sqrt{25-16}$

$=3$

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

$=\left(\frac{15}{17}\right)\left(\frac{-4}{5}\right)-\left(\frac{+8}{17}\right)\left(\frac{+3}{5}\right)$

$=\frac{-60-24}{85}=\frac{-84}{85}$.