Question

cos cube A• cos3A + sin cube A• sin3A=cos cobe 2A

Answer 1

Cos^3A.cos3A + sin3A.sin^3A=cos^3 2A;=Cos^3A.cos(2A+A) + sin^3A.sin(2A+A);=cos^3A(cos2A.cosA - sin2A.sinA) + sin^3A(sin2A.cosA+sinA.cos2A);[Opening brackets and multiplying]=Cos2A.cos^4A - sin2A.sinA.cos^3A + sin^3A.sin2A.cosA + sin^4A.cos2A;=cos2A(cos^4A+ sin^4A) - sinA.cosA.sin2A(cos^2A-sin^2A);[Cos^2A-sin^2A=cos2A][sinA.cosA=(sin2A)%2][Taking cos2A common]=Cos2A(cos^4A +sin^4A -(sin^2 2A)%2);[Cos^4A+sin^4A=(cos^2A-sin^2A)^2+2sin^2Acos^2A ••>(cos2A)^2+( 2sin^2 2A)%4 ••>cos^2 2A+ (sin^2 2A)%2 ][Substituting the above value in the eqn]=Cos2A(cos^2 2A +( sin^2 2A)%2 -( sin^2 2A)%2);=cos2A(cos^2 2A)=cos^3 2AHENCE PROVED