#2(sin^6A+cos^6A)-3(sin^4A+cos^4A)=-1#?

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Answer 1 · 2018-03-19T15:23:28

Please see explanation section.

We know that,
#color(blue)((1) .x^4+y^4=(x^2+y^2)^2-2x^2y^2)#
#color(red)((2) .x^6+y^6=(x^2+y^2)^3-3x^2y^2(x^2+y^2))#
#(3).sin^2theta+cos^2theta=1#

Here, we use above formula # color(red)((2)) and color(blue)((1))# respectively.

#LHS=2(sin^6A+cos^6A)-3(sin^4A+cos^4A)#

#=2[color(red)((sin^2A+cos^2A)^3-3sin^2Acos^2A(sin^2A+cos^2A))]-3color(blue)([(sin^2A+cos^2A)^2-2sin^2Acos^2A)]#
Using (3),we get

#=2[(1)^3-3sin^2Acos^2A(1)]-3[(1)^2-2sin^2Acos^2A]#

#=2-6sin^2Acos^2A-3+6sin^2Acos^2A#

#=2-3#

#=-1=RHS#