If $\alpha,\ \beta \in \left( 0,\ \frac{\pi}{2} \right),\sin{\alpha = \frac{4}{5}}$ and $\cos{(\alpha + \beta) = - \frac{12}{13}}$, then $\sin\beta$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
$\cos{(\alpha + \beta) = - \frac{12}{13}}$
Here, $0 < (\alpha + \beta) < \pi$
$	herefore\sin{(\alpha + \beta) = \sqrt{1 - \cos^{2}{(\alpha + \beta)}}}$
$= \sqrt{1 - \frac{144}{169}}$
$= \frac{5}{13}$
Now, $\sin{\beta =}\sin{\lbrack(\alpha + \beta) - \alphabrack}$
$= \sin{(\alpha + \beta)\cos{\alpha - \cos{(\alpha + \beta)\sin\alpha}}}$
$= \frac{5}{13}.\frac{3}{5} - \left( - \frac{12}{13} ight).\frac{4}{5}$
$= \frac{15}{65} + \frac{48}{65}$
$= \frac{63}{65}$

Q. If $\alpha,\ \beta \in \left( 0,\ \frac{\pi}{2} \right),\sin{\alpha = \frac{4}{5}}$ and $\cos{(\alpha + \beta) = - \frac{12}{13}}$, then $\sin\beta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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