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Question
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
Solution
Sin4θ – cos4θ = 1 – 2cos2θ
LHS = Sin4θ – cos4θ
LHS = (Sin2θ)2 – (cos2θ)2
LHS = (Sin2θ + cos2θ)(Sin2θ - cos2θ) ...[a2 – b2 = (a + b)(a – b)]
LHS = (Sin2θ – cos2θ).(1) ...(Sin2θ + cos2θ = 1)
LHS = 1 – cos2θ – cos2θ ...(1 – Sin2θ = cos2θ)
LHS = 1 – 2cos2θ
RHS = 1 – 2cos2θ
LHS = RHS
Hence proved.
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