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:
$\frac{\sqrt{3}+1}{2\sqrt{2}}$
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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