Given: cos^6x - sin^6x = m/16 × cos2x + n/16 × cos6x. Find m + n

Question

The answers are 15,16,-16 or-14. Help pls

Answer 1

* cos6x= cos^2(3x) - sin^2(3x) = (cos3x+sin3x) × (cos3x-sin3x) = [(4cos^3x-3cosx)+(3sinx-4sin^3x)] [(4cos^3x-3cosx)-(3sinx-4sin^3x )] = [4(cos^3x-sin^3x)-3(cosx-sinx)]× [4(cos^3+sin^3x)-3(sinx+cosx)] = [4(cosx-sinx)(cos^2x+cosxsinx+sin^2x) -3(cosx-sinx)] × [4(cosx+sinx)(cos^2x-cosxsinx+sin^2x) -3(cosx+sinx)] = [(cosx-sinx)(1+4cosxsinx)]× [(cosx+sinx)(1-4cosxsinx)] = (cos^2x-sin^2x)(1-16cos^2xsin^2x) = cos2x . (1-4.4cos^2xsin^2x) = cos2x . [1-4sin^2(2x)] = cos2x - 4sin^2(2x)cos2x ** cos^6x - sin^6x = (cos^3x)^2 - (sin^3x)^2 = (cos^3x+sin^3x)(cos^3x-sin^3x) = [(cosx+sinx)(cos^2x-cosxsinx+sin^2x)] × [(cosx-sinx)(cos^2x+cosxsinx+sin^2x)] = (cos^2x-sin^2x)(1-cos^2xsin^2x) = cos2x (1- 1/4 ×4cos^2xsin^2x) = cos2x (1- 1/4 ×sin^2(2x)) = cos2x - 1/4 (sin^2(2x)cos2x) = 1/16 (16cos2x - 16/4sin^2(2x)cos2x) = 1/16 (15cos2x + cos2x - 4sin^2(2x)cos2x)(*) = 1/16 (15cos2x + cos6x) =15/16cos2x + 1/16cos6x -> m=15, n=1, m+n=16