Question : Given that A and B are second quadrant angles, $\sin A = \frac{1}{3}$ and $\sin B = \frac{1}{5}$, then find the value of $\cos(A-B)$.
Option 1: $\frac{4 \sqrt{3}+1}{15}$
Option 2: $\frac{8 \sqrt{3}-1}{15}$
Option 3: $\frac{8 \sqrt{3}+1}{15}$
Option 4: $\frac{4 \sqrt{3}-1}{15}$
Correct Answer: $\frac{8 \sqrt{3}+1}{15}$
Solution : $\sin A = \frac{1}{3}$ ⇒ $\cos A = \sqrt{1-\sin^2 A} = \sqrt{1-(\frac{1}{9})} = \frac{\sqrt8}{3}$ Also, $\sin B = \frac{1}{5}$ ⇒ $\cos B = \sqrt{1-\sin^2 B} = \sqrt{1-(\frac{1}{25})} = \frac{2\sqrt6}{5}$ Now, $\cos(A-B) = \cos A \cos B + \sin A \sin B$ $= \frac{\sqrt8}{3} × \frac{2\sqrt6}{5} + \frac{1}{3} × \frac{1}{5}$ $= \frac{8\sqrt3+1}{15}$ Hence, the correct answer is $\frac{8\sqrt3+1}{15}$.
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