Question : If $\sin(A+B)=\sin A\cos B+\cos A \sin B$, then the value of $\sin75°$ is:
Option 1: $\frac{\sqrt{3}+1}{\sqrt{2}}$
Option 2:
$\frac{\sqrt{2}+1}{2\sqrt{2}}$
Option 3:
$\frac{\sqrt{3}+1}{2\sqrt{2}}$
Option 4:
$\frac{\sqrt{3}+1}{2}$
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Correct Answer:
Solution : Given: $\sin(A+B)=\sin A\cos B+\cos A\sin B$ So, $\sin75°=\sin(45°+30°)$ = $\sin 45°\cos30°+\cos45°\sin30°$ = $\frac{1}{\sqrt{2}}×\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}×\frac{1}{2}$ = $\frac{\sqrt{3}+1}{2\sqrt{2}}$ Hence, the correct answer is $\frac{\sqrt{3}+1}{2\sqrt{2}}$.
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