№1
вычислите: (3cos28°cos34°-3cos62°cos56°)/(sin74°cos46°-cos44°sin16°)

(3cos28°cos34°-3cos62°cos56°)/(sin74°cos46°-cos44°sin16°) ==3(cos28cos34-sin28sin34)/(cos16sin44-cos44sin16)==3cos(28+34)/sin(44-16)=3cos62/sin28=3sin28/sin28=3-------------------------------------------------------------------------------------используем формулыsin(π/2-a)=cosacos(π/2-a)=sinacosacosb-sinasinb=cos(a=b)sinacosb-cosasinb=sin(a-b)