Question

2(sin6θ+cos6θ)−3(sin4θ+cos4θ)
​

Answer 1

LHS=2(sin

6

θ+cos

6

θ)−3(sin

4

θ+cos

4

θ)+1

=2{(sin

2

θ+cos

2

θ)

3

−3sin

2

θcos

2

θ(sin

2

θ+cos

2

θ)}−3(sin

2

θ+cos

2

θ)

2

−2(sin

2

θcos

2

θ)}+1

We know, [sin²x+cos²x=1]

=2{1−3sin

2

θcos

2

θ}−3{1−2sin

2

θcos

2

θ}+1

=2−6sin

2

θcos

2

θ−3+6sin

2

θcos

2

θ+1

=0

=RHS

Answer 2

Answer:

LHS=2(sin

6

θ+cos

6

θ)−3(sin

4

θ+cos

4

θ)+1

=2{(sin

2

θ+cos

2

θ)

3

−3sin

2

θcos

2

θ(sin

2

θ+cos

2

θ)}−3(sin

2

θ+cos

2

θ)

2

−2(sin

2

θcos

2

θ)}+1

We know, [sin²x+cos²x=1]

=2{1−3sin

2

θcos

2

θ}−3{1−2sin

2

θcos

2

θ}+1

=2−6sin

2

θcos

2

θ−3+6sin

2

θcos

2

θ+1

=0

=RHS

Step-by-step explanation: