Rút gọn biểu thức a. sin2x.sin6x.cos4x+ $\frac{1}{4}$cos12x b.sin4x.sin10x-sin11x.sin3x-sin7x.sinx

Đáp án:

a;sin2x.sin6x.cos4x +$\frac{1}{4}$ có12x 

=($\frac{1}{2}$) sin(2.6)x.cos4x+$\frac{1}{4}$cos12x

=($\frac{1}{8}$)(cos12x-cos8x)+$\frac{1}{4}$cos12x

=$\frac{3}{8}$ cos12x-$\frac{1}{8}$cos8x

sin4x.sin10x - sin11.x.sin3.x - sin7.x.sin.x

= [cos(10-4)x - cos(10+4)x]:2 - [cos(11+3)x-cos(11-3)x]:2 - [cos(7+x)- cos(x-7)] : 2

= [sin6xcos(-14)+sin14xcos(-6)] : 2 - [sin14xcos(-18)+sin18xcos(-14)] : 2 - [sincos(x-7)-sincos(x+7)]: 2

= [(0.5)(0.5)(sin20)+0]-(0+(0.5)(0.5)(sincos24))-(sincos(x+7)-(sincos(x-7))

= (0.25)sin20 - (0.25)sincos24 - sincos(x+7) + sincos(x-7)

                                               nhớ vote và ctlhn nhé