Question

Помогите, пожалуйста
Номер 1 и 3

Answer 1

1. 2*sin(π/2-α)*sinα=2*cosα*sinα=2*sinα*cosα=sin(2α).

3. sin⁴α+cos⁴α=sin⁴α+2sin²αcos²α+cos⁴α-2sin²αcos²α=

=(sin²α+cos²α)²-4sin²αcos²α/2=1²-(2sinαcosα)²/2=

=1-sin²(2α)/2=(2-sin²(2α))/2=(1+1-sin²(2α))/2=(1+cos²(2α))/2.