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Grade: 11
        
Prove that cos^6 A - sin^6 A = cos 2A(1-1sin^2 2A)........
one month ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
2671 Points
							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
one month ago
Arun
21951 Points
							
2a)2a/4 )= cos2a(1-1/4sin2a*sin2a )= (cos2a)(1 – 4cos2a*sin2a)= (cos2a)(1 – cos2a*sin2a + cos2a*sin2a)= (cos2a)(1 – 2 cos2a*sin2a + cos4a + sin4a )(cos2a – sin2= (cos3a)2 – (sin3a)2cos
one month ago
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