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Miscellaneous Exercise 2 · Q90

Q.Prove that sin⁡6A+cos⁡6A=1−3sin⁡2A+3sin⁡4A\sin^6 A + \cos^6 A = 1 − 3\sin^2 A + 3\sin^4 A.

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Step 1. From the a³+b³ identity, sin⁡6A+cos⁡6A=1−3sin⁡2Acos⁡2A\sin^6A+\cos^6A = 1-3\sin^2A\cos^2A.

Step 2. Substitute cos⁡2A=1−sin⁡2A\cos^2A=1-\sin^2A: 3sin⁡2Acos⁡2A=3sin⁡2A(1−sin⁡2A)=3sin⁡2A−3sin⁡4A3\sin^2A\cos^2A = 3\sin^2A(1-\sin^2A) = 3\sin^2A-3\sin^4A. …

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