Question

2 (sin⁶ θ + cos⁶ θ) − 3 (sin⁴ θ + cos⁴ θ) is equal to 

Options

•  0

•  1

•  −1

• None of these

Answer 1

The given expression is `2(sin^6θ+cos^6θ)-3(sin^4θ+cos^4θ)` 

Simplifying the given expression, we have

`2(sinθ+cos^6θ)-3(sin^4θ+cos^4θ)` 

= `2sin^6θ+2cos^6θ-3sin^4θ-3cos^4θ`

=`(2 sin^6 θ-3sin^4θ)+(2 cos^6-3 cos^4θ)`

=`sin^4θ(2sin^2θ-3)+cos^4θ(2 cos^2θ-3)`

`=sin^4θ{2(1-cos^2)-3}+cos^4θ{2(1-sin^2 θ)-3)` 

`= sin^4θ(2-2cos^2θ-3)+cos^4θ(2-2sin^2 θ-3) `

`=sin^4θ(-1-2cos^θ)+cos^4θ(1-2sin^2θ)` 

`= -sin^4θ-2 sin^4θ cos^2θ-cos^4θ-2cos^4 θ sin^2θ`

`=sin^4θ-cos^4θ-2 cos^4 θ sin^2θ-2 sin^4 θcos^2θ`

`=-sin^4θ-cos^4θ-2cos^2θ sin^2(cos^2+sin^2θ)`

`=-sin^4θ-cos^4θ-2cos^2θsin^2θ(1)`

`=-sin^4θ-cos^4θ-2cos^2sin^2θ`

`=(sin^4θ+cos^4 θ+2 cos^2 θ sin^2 θ)`

`=-{(sin^2θ)^2+(cos^2θ)^2+2 sin^2 θ cos^2θ}`

` =-(sin^2θ+cos^2θ)^2` 

`=-(1)^2`

`=-1`