Using the formula, sin(A – B) = sinA cosB – cosA sinB, find the value of sin 15º.

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$\frac{\sqrt{3 - 1}}{\sqrt{2}}$

$\frac{\sqrt{1}}{2 \sqrt{2}}$

$\frac{\sqrt{3}}{2 \sqrt{2}}$

$\frac{\sqrt{3} - 1}{2 \sqrt{2}}$

answer is B.

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Detailed Solution

Let A = 45º and B = 30º.

Then A – B = 15º.

Putting A = 45º and B = 30º in the given formula, we get:

sin(45º – 30º) = sin45º cos30º – cos45º sin30º

sin(45º – 30º) $= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

Hence, $\sin 15 ° = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

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