Prove that cos^6 A - sin^6 A = cos 2A(1-1sin^2 2A)........

Question

Saurabh Koranglekar · Answer

Dear student

LHS

a^ 6 - b^ 6 = ( a^ 2 - b^ 2)(a^ 4 + b^4 + (ab)^2)
here LHS = cos2A ( sin^4A + cos^4A + ( sinA cos A)^2)
Adjusting second bracket a^ 2 + b^2 + 2ab -2ab + ab = (a+b)^2 -ab

We get
Cos 2A*( 1- sin^2(2A)) = RHS

Regards

Arun · Answer

2a) 2 a/4 )= cos2a(1-1/4sin 2 a*sin 2 a )= (cos2a)(1 – 4cos 2 a*sin 2 a)= (cos2a)(1 – cos 2 a*sin 2 a + cos 2 a*sin 2 a)= (cos2a)(1 – 2 cos 2 a*sin 2 a + cos 4 a + sin 4 a )(cos 2 a – sin 2 = (cos 3 a) 2 – (sin 3 a) 2 cos

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Prove that cos^6 A - sin^6 A = cos 2A(1-1sin^2 2A)........

Prove that cos^6 A - sin^6 A = cos 2A(1-1sin^2 2A)........

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Prove that cos^6 A - sin^6 A = cos 2A(1-1sin^2 2A)........

Prove that cos^6 A - sin^6 A = cos 2A(1-1sin^2 2A)........

2 Answers

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