1)докажите тождество 2cos²(45°+4α)+sin8α=1 2)вычислите cos70°+sin140°-cos10°

1)2cos²(45°+4α)+sin8α=1
1+cos(2*(45°+4α))+sin8α=1
1+cos(90°+8α)+sin8α=1
1+cos(π/2+8α)+sin8α=1
1-sin8α+sin8α=1
1≡1

2)cos70°+sin140°-cos10°=(cos70°-cos10°)+sin140°=
-2sin(70°+10°)/2*sin(70°-10°)/2+sin140°=-2sin40°sin30°+sin140°=-2*1/2sin40°+sin140°=-sin40°+sin140°=sin140°-sin40°=2sin(140°-40°)/2*cos(140°+40°)/2=
2sin50°*cos90°=2sin50°*0=0