Question

若cos2a=a,则sin*4a+cos*4a=?

Answer 1

*4是4次方吗
原式=(sin²a+cos²a)²-2sin²acpos²a
=1²-1/2*(2sinacosa)²
=1-1/2*sin²2a
=1-1/2*(1-cos²2a)
=1-(1-a²)/2
=(a²+1)/2

Answer 2

sin^4a+cos^4a=?
=(sin²a+cos²a)²-2sina²cos²a
=1-1/2(2sinacosa)²
=1-1/2sin²2a
=1-1/2(1-cos²2a)
=1-1/2+a²/2
=1/2+a²/2
=(1+a²)/2