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Mathematics

If A = 30°; show that :

4 cos A cos (60° - A). cos (60° + A) = cos 3A

Trigonometric Identities

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Answer

4 cos A cos (60° - A). cos (60° + A) = cos 3A

L.H.S.=4 cos A cos (60° - A). cos (60° + A)=4 cos 30°. cos (60° - 30°). cos (60° + 30°)=4 cos 30°. cos 30°. cos 90°=4×32×32×0=0\text{L.H.S.} = \text{4 cos A cos (60° - A). cos (60° + A)}\\[1em] = \text{4 cos 30°. cos (60° - 30°). cos (60° + 30°)}\\[1em] = \text{4 cos 30°. cos 30°. cos 90°}\\[1em] = 4 \times \dfrac{\sqrt3}{2} \times \dfrac{\sqrt3}{2} \times 0\\[1em] = 0

R.H.S.=cos 3A=cos (3 x 30°)=cos 90°=0\text{R.H.S.} = \text{cos 3A}\\[1em] = \text{cos (3 x 30°)}\\[1em] = \text{cos 90°}\\[1em] = 0

∴ L.H.S. = R.H.S.

Hence, 4 cos A cos (60° - A). cos (60° + A) = cos 3A.

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