Prove that- cos3 theta

We have, LHS= 4 cos θ cos ( π/3+θ ) cos ( π/3 θ ) =2cosθ [2cos ( π/3+θ ) cos ( π/3 θ ) ] =2cosθ [ 2cos ( π/3+θ + π/3 θ ) +cos ( π/3+θ π/3+θ ) ] =2cosθ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

Prove that:= ...

Question

Prove that:

$4 c o s θ (\frac{π}{3} + θ) c o s (\frac{π}{3} - θ) = c o s 3 θ$

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Solution

We have,

LHS $= 4 c o s θ c o s (\frac{π}{3} + θ) c o s (\frac{π}{3} - θ) = 2 c o s θ [2 c o s (\frac{π}{3} + θ) c o s (\frac{π}{3} - θ)] = 2 c o s θ [2 c o s (\frac{π}{3} + θ + \frac{π}{3} - θ) + c o s (\frac{π}{3} + θ - \frac{π}{3} + θ)] = 2 c o s θ [c o s \frac{2 π}{3} + c o s 2 θ] = 2 c o s θ [c o s (\frac{π}{2} + \frac{π}{6}) + c o s 2 θ] = 2 c o s θ [- s i n \frac{π}{6} + c o s 2 θ] = c o s θ [- \frac{1}{2} + c o s 2 θ] = - 2 c o s θ \frac{1}{2} + 2 c o s θ c o s 2 θ = - c o s θ + [c o s (θ + 2 θ) + c o s (2 θ - θ)] = - c o s θ + c o s 3 θ + c o s θ = c o s 3 θ = R H S$
$∴$ LHS=RHS.

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Prove that: 4cosθ(π3+θ) cos (π3−θ)=cos3θ

Prove that:

$4 c o s θ (\frac{π}{3} + θ) c o s (\frac{π}{3} - θ) = c o s 3 θ$