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Dear bibek khadka try this and if satisfied reply me
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)
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