Question

2(sin6a+ cos6a) – 3(sin4a + cos40a) + 4(sin2a + cos2a)​

Answer 1

Step-by-step explanation:

We can prove the above by: 2(sin6a+cos6a) - 3(sin4a+cos4a)+1=0 =2[(sin2a)3+(cos2a)3] - 3[(sin2a)2+(cos2a)2] + 1 =2[(sin2a+cos2a)3 -3sin2a cos2a (sin2a+cos2a)] -3[(sin2a+cos2a)2 -2sin2acos2a] + 1 =2[1-3sin2acos2a] -3[1-2sin2a cos2a] + 1 =2 - 6sin2a cos2a - 3 + 6sin2a cos2a + 1 =0 Proved.

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Answer 2

Answer:

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Step-by-step explanation: