2(sin6a+cos6a)-3(sin4a+cos4a)+1
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hope it helps you........
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crowny:
But the answer should be equal to 0 not 1
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2(sin6a+cos6a)
-3(sin4a+cos4a)+1=0
=2[(sin2a)3+(cos2a)3]
-3[(sin2a)2+(cos2a)2]+1
=2[(sin2a+cos2a)3
-3sin2acos2a(sin2a+cos2a)]
-3[(sin2a+cos2a)2
-2sin2acos2a]+1
=2[1-3sin2acos2a]
-3[1-2sin2acos2a]+1
=2-6sin2acos2a
-3+6sin2acos2a+1=0
-3(sin4a+cos4a)+1=0
=2[(sin2a)3+(cos2a)3]
-3[(sin2a)2+(cos2a)2]+1
=2[(sin2a+cos2a)3
-3sin2acos2a(sin2a+cos2a)]
-3[(sin2a+cos2a)2
-2sin2acos2a]+1
=2[1-3sin2acos2a]
-3[1-2sin2acos2a]+1
=2-6sin2acos2a
-3+6sin2acos2a+1=0
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