sin^6+cos^6+3sin²cos²
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Using simple formula a³+b³=(a+b)(a²-ab+b²) and a²+b²=(a+b)²-2ab
And identity sin²x+cos²x=1
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bhaveshjain281:
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Solution :
Now, LHS = sin⁶θ + cos⁶θ
= (sin²θ)³ + (cos²θ)³
= (sin²θ + cos²θ) {(sin²θ)² + (cos²θ)² - sin²θ cos²θ}
= 1 * {(sin²θ + cos²θ)² - 2sin²θ cos²θ - sin²θ cos²θ}
= 1 - 3 sin²θ cos²θ
= RHS (Proved)
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