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# cos^2a.sin^4a=1/32(cos6a-2cos4a-cos2a+2)

9 years ago

1/32(cos6a-cos2a-2cos4a+2)

let us find the value of  cos6a-cos2a-2cos4a+2

by applying the formula cos c-cos d = 2sin(c+d/2)sin(d-c/2)

=2sin4a(sin(-2a)-2cos4a+2

=-2sin4asin2a-2cos4a+2

=2(1-cos4a-sin4asin2a)

=2(cos^2(2a)+sin^2(2a)-(cos^2(2a)-sin^2(2a))-2sin2acos2a(sin2a)

(since cos4a=cos2(2a)=cos^2(2a)+sin^2(2a)) &sin4a=2sin2acos2a & 1=cos^2(2a)+sin^2(2a))

=2(cos^2(2a) +sin^2(2a) - cos^2(2a) +sin^2(2a)-2sin^2(2a)cos2a)

=2(2sin^(2a)-2sin^2(2a)cos2a)

=2(2sin^2(2a)(1-cos2a)

=4sin^2(2a)(2sin^2(a))

=8sin^2(a)(2sinacosa)^2

=8sin^2(a)(4sin^2(a)cos^2(a))

=32(sin^4(a)cos^2(a))

therefore1/32(cos6a-cos2a-2cos4a+2 )=sin^4(a)cos^2(a)