Question

Помогите, пожалуйста, с 4 заданием, с решением!

Answer 1

2) =7(cos²α +sin²α) -5 =7*1 -5 =2.
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3) (cos²62° +cos²28°)/4 -(1/2)*cos60° =( (sin(90°-28°))² +cos²28) /4 -(1/2)*(1/2)=
 =( sin²28° +cos²28) /4 -1/4 = 1/4-1/4 =0.
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4) sin²17°*sin²73°/sin²34° = sin²17°*sin²(90°-17°/sin²34° =
sin²17°*cos²17°/sin²34° =(sin17°*cos17°)²/sin²34° =((1/2)*sin34°)²/sin²34° =
(1/4)*sin²34°/sin²34° =1/4.
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5) 2cos²α -cos2α = 2cos²α  -(2cos²α -1) =1.

т.к.  cos2α =cos²α -sin²α = cos²α -(1-cos²α) = cos²α -1+cos²α= 2cos²α -1.