Question

помогите решить 4 и 5 номер

Answer 1

3)√6*sin63π/4*tq64π/3 =√6*sin(16π-π/4)*tq(22π-π/3) =-√6*s(-sinπ/4)*(-tqπ/3)=
√6*1/√2/ *√3 =3.
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4) ( 1-2sin²54°)/(8tq9°sin²99°) =cos2*54°/((8tq9°* sin²(90° +9°))
 =cos108°/(8tq9°*cos²9°)=cos(90°+18°)/(8sin9°*cos9°) = -sin18°/4sin18° == -1/4.
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5) √2*(sin40°cos5°-sin(180°+50°)sin5°)/(sin25°sin35° -sin(90°+25°)cos35°)=
= -√2*sin45°/*cos60° = -√2*1/√2/1(1/2) = -2.