Convert the following sums or differences into products sin 12 A - sin 4 A?

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Solution

$\sin 12 A - \sin 4 A = 2 \cos (\frac{12 A + 4 A}{2}) . \sin (\frac{12 A - 4 A}{2}) [As, \sin C - \sin D = 2 \cos (\frac{C + D}{2}) . \sin (\frac{C - D}{2})] = 2 \cos (\frac{16 A}{2}) . \sin (\frac{8 A}{2}) = 2 \cos 8 A . \sin 4 A$

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Convert the following sums or differences into products sin 12 A - sin 4 A?

sin 12 A - sin 4A= 2 cos12A + 4A2 .sin12A - 4A2 As, sin C - sin D = 2 cosC + D2 . sin C - D2=2 cos16A2 . sin8A2=2 cos 8A . sin 4A

$\sin 12 A - \sin 4 A = 2 \cos (\frac{12 A + 4 A}{2}) . \sin (\frac{12 A - 4 A}{2}) [As, \sin C - \sin D = 2 \cos (\frac{C + D}{2}) . \sin (\frac{C - D}{2})] = 2 \cos (\frac{16 A}{2}) . \sin (\frac{8 A}{2}) = 2 \cos 8 A . \sin 4 A$