Prove that sin 3x sin^3x + cos 3x cos^3x = cos^3 2x
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Step-by-step explanation:
sin3x=3sinx−4sin 3 x
∴sin 3 x= 1/4 (3sinx−sin3x)
Similarly, cos 3x=1 /4(3cosx+cos3x)
L.H.S.= 1/4 [sin3x(3sinx−sin3x+cos3x(3cosx+cos3x)
=1/4 [3cos(3x−x)+(cos 2 3x−sin 2 3x)]
= 1/4 [3cos2x+cos6x]
=1/4 [3cosA+cos3A]
=cos 3A,by(2)=cos 3 2x
∵A=2x
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