`sin^2Acos^4A=1/16+1/32cos2A-1/16cos4A-1/32cos6A`

Prove that, sin^(2)A cos^(4) A = 1/16 + 1/32 cos2A - 1/16 cos 4 A - 1/32 cos 6 A

sin^4A-cos^4A=2sin^2A-1=1-2cos^2A=sin^2A-cos^2A

Show tha: cos6A=32cos^6A-48cos^4A+18cos^2A-1

Prove that cos6A=32cos^6A-48cos^4A+18cos^2A-1

Prove any one of the following: cos 6A=32 cos^6 A-48 cos^4A+18 cos^2A-1

Prove that: (sin3A cos4A-sin Acos2A)/(sin4A sin A+cos6A cos A)=tan2A

sin(5theta)/sintheta is equal to a.16cos^4theta-12cos^2theta+1, b.16cos^4theta+12cos^2theta+1, c.16cos^4theta-12cos^2theta-1, d.16cos^4theta+12cos^2theta-1

cos6x=32cos^(6)x-48cos^(4)x-18cos^(2)x-1

Prove that: [1/(sec^(2)A-cos^(2)A)+1/("cosec"^(2)A-sin^(2)A)].sin^(2)A.cos^(2)A=(1-sin^(2)Acos^(2)A)/(2+sin^(2)Acos^(2)A)

cos^6A-sin^6A=cos2A(1-1/4sin^2 2A)