Question

2(sin*6 theta + cos*6theta)-3(sin *4 theta +cos*4theta )+1 equal to....tell me fast its urgent​

Answer 1

Answer:

zero (0)

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

=-1+1

=0