First slide

Trignometry

Question

Find the value of $2 \sin 15 ° \cos 45 °$

Moderate

A

1
B

$(\frac{\sqrt{2} - 1}{2})$
C

0
D

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

Solution

Apply
$\sin C - \sin D = 2 \sin (\frac{C - D}{2}) \cos (\frac{C + D}{2})$
$C = 60 °$
and
$D = 30 °$
$∴ 2 \sin 15 ° \cos 45 ° = \sin 60 ° - \sin 30 °$
$= (\frac{\sqrt{3} - 1}{2})$

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