Question

sin20°cos50°=1/2(sin70°-sin30°)

这是为毛啊

Answer 1

证明：sin20°cos50°=1/2(sin70°-sin30°)，
sin20°cos50°-1/2(sin70°-sin30°)= sin(30°-10°)cos(60°-10°)-1/2[sin(60°+10°)-sin30°]
=( sin30°cos10°- cos30° sin10°)( cos60°cos 10°+ sin60°sin10°)-1/2(sin60°cos10°+ cos60°sin10°-1/2)
=(1/2* cos10°-√3/2* sin10°)( 1/2* cos10°+√3/2* sin10°)- 1/2(√3/2*cos10°+ 1/2*sin10°-1/2)
=1/4* cos10°^2-3/4* sin10°^2-√3/4*cos10°-1/4*sin10°+1/4
=1/4*( cos10°^2-3 sin10°^2-√3*cos10°-sin10°+1)
=1/4*( cos10°^2-3 sin10°^2-√3*cos10°-sin10°+ cos10°^2+ sin10°^2)
=1/4*( 2cos10°^2-2 sin10°^2-√3*cos10°-sin10°)
=1/2*( cos10°^2- sin10°^2-√3/2*cos10°-1/2*sin10°)
=1/2*( cos20°-√3/2*cos10°-1/2*sin10°)
=1/2*( cos20°- cos30°*cos10°- sin30*sin10°)
=1/2*( cos20°- cos(30°-10°))
=1/2*( cos20°- cos20°)=0
所以，上式成立。