5 (sin^8 A- cos ^8 )=(2sin^2A-1)(1-2sin^2 A cos ^2 A)
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Ques. is wrong.
Remove 5 from the beginning
The only problem you will find then is to substitute the value of(sin4A + cos4A) by (1-2sin2A.cos2A)
Formula used is a2+b2=(a+b)2 -2.a.b
Here a=sin2A
b=cos2A
Remove 5 from the beginning
The only problem you will find then is to substitute the value of(sin4A + cos4A) by (1-2sin2A.cos2A)
Formula used is a2+b2=(a+b)2 -2.a.b
Here a=sin2A
b=cos2A
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