Question #02a6f

Question

986 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-06T14:02:36

To be contd.............

We use, #a^3+b^3=(a+b)(a^2-ab+b^2)# to get,

#(costheta)^6+(sintheta)^6=cos^6theta+sin^6theta#

#=(cos^2theta+sin^2theta)(cos^4theta-cos^2thetasin^2theta+sin^4theta)#

#=(1){(cos^2theta+sin^2theta)^2-2sin^2thetacos^2theta-cos^2thetasin^2theta}#

#=1-3cos^2thetasin^2theta#

#=1-3/4(2sinthetacostheta)^2#

#=1-3/4(sin2theta)^2#

#=1-3/4(sin^2(2theta))#

Recall that, #1-cos2A=2sin^2A#.

Hence, the Exp.#=1-3/8{2sin^2(2theta)}#

#=1-3/8(1-cos4theta)#

#=1-3/8+3/8cos4theta#

#=5/8+3/8cos4theta#