Question

Prove that 2(sin 6 theta +cos 6 theta ) -3(sin 4 theta +cos 4 theta) +1 =0

Answer 1

2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1
=2[(sin²θ)³+(cos²θ)³]-3sin⁴θ-3cos⁴θ+1
=2(sin²θ+cos²θ)[(sin²θ)²+(cos²θ)²-sinθ(cosθ)]-3sin⁴θ-3cos⁴θ+1
=2(1)[sin⁴θ+cos⁴θ-sinθ(cosθ)]-3sin⁴θ-3cos⁴θ+1                               (sin²θ+cos²θ=1)
=2sin⁴θ+2cos⁴θ-2sinθ(cosθ)-3sin⁴θ-3cos⁴θ+1
=-sin⁴θ-cos⁴θ-2sinθ(cosθ)+1
=1-[sin⁴θ+cos⁴θ+2sinθ(cosθ)]
=1-[(sin²θ)²+(cos²θ)²+2sinθ(cosθ)]
=1-(sin²θ+cos²θ)²
=1-(1)²
1-1=0
Thus,proved 2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1=0

Answer 2

letr theta=A
 2(sin6A +cos6A)= 2((sin2A)3 + (cos2A)3)
 = 2(sin2A+cos2A)( sin4A+cos4A -sin2Acos2A)
 =2(sin4A+cos4A -sin2Acos2A)
 = 2sin4A+2cos4A -2sin2Acos2A
 -3(sin4A+cos4A) = -3sin4A-3cos4A
so 2(sin6A+cos6A)-3(sin4A+cos4A) = 2sin4A +2cos4A -3sin4A -3sin4A -2sin2Acos2A
 = -(sin4A+cos4A+2sin2Acos2A)
 =-(sin2A+cos2A)2
= -1
 so-1+1 =0