Question

Помогииитееее
срооочноо

Answer 1

2)cos2α+2sin²α=cos²α-sin²α+sin²α+sin²α=
=(cos²α+sin²α)-sin²α+sin²α=1;
1-sin2α·cosα/2sinα=1-2sinα·cosα·cosα/2sinα=
=1-cos²α=sin²α;
3)sin40⁰+cos70⁰-cos10⁰=0
sin40⁰+(-2sin[(70⁰+10⁰)/2]sin[(70⁰-10⁰)/2])=
=sin40⁰-2sin40⁰sin30⁰=sin40⁰-2sin40⁰·1/2=sin40⁰-sin40⁰=0
4)sinα-sin(π/3-α)=2sin(α-π/3+α)/2·cos(α+π/3-α)/2=
=2sin(α-π/6)·cosπ/6=2·sin(α-π/6)·√3/2=√3·sin(α-π/6)