Question

prove that 2(Sin⁶θ+Cos⁶θ)-3(Sin⁴θ+Cos⁴θ)+1 = 0. ​

Answer 1

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identity

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3 + b3 = (a+b)3 - 3ab(a+b) and

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3 + b3 = (a+b)3 - 3ab(a+b) and a2 + b2 = (a+b)2 - 2ab

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3 + b3 = (a+b)3 - 3ab(a+b) and a2 + b2 = (a+b)2 - 2abare used in the above step where

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3 + b3 = (a+b)3 - 3ab(a+b) and a2 + b2 = (a+b)2 - 2abare used in the above step wherea = sin2θ and b = cos2θ.

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3 + b3 = (a+b)3 - 3ab(a+b) and a2 + b2 = (a+b)2 - 2abare used in the above step wherea = sin2θ and b = cos2θ.writing sin2θ + cos2θ = 1, we have

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3 + b3 = (a+b)3 - 3ab(a+b) and a2 + b2 = (a+b)2 - 2abare used in the above step wherea = sin2θ and b = cos2θ.writing sin2θ + cos2θ = 1, we have = 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3 + b3 = (a+b)3 - 3ab(a+b) and a2 + b2 = (a+b)2 - 2abare used in the above step wherea = sin2θ and b = cos2θ.writing sin2θ + cos2θ = 1, we have = 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1= 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 =2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3 + b3 = (a+b)3 - 3ab(a+b) and a2 + b2 = (a+b)2 - 2abare used in the above step wherea = sin2θ and b = cos2θ.writing sin2θ + cos2θ = 1, we have = 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1= 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1= -3+3=0

hope it helps you ❤️

Answer 2

$\huge\boxed{\fcolorbox{red}{aqua}{Answer}}$

2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)

2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}

2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}

2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)

2-6sin²θcos²θ-3+6sin²θcos²θ= -1

hope it helps...