Question

2 into sin 6 theta + cos 6 theta into minus 3 into sin 4 theta + cos 4 theta into + 1=0​

Answer 1

Answer:

Step-by-step explanation:

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 (1-3sin²θcos²θ) -3 (1-2sin²θcos²θ) + 1

=2 - 6sin²θcos²θ -3 + 6sin²θcos²θ + 1

= 2-2

=0

Hence proved.