Question

高一数学~~

(sin20°-cos50°)/cos80°
化简~~求教~~

Answer 1

sin20-cos50=sin20-sin40=2cos30sin(-10)=-2cos30sin10=-2cos30cos80
除以cos80即-2cos30=负根号3

追问

sin20-sin40=2cos30sin(-10)怎么来的~~？

追答

这是和差化积
sin(a+b)=sinacosb+sinbcosa
sin(a-b)=sinacosb-sinbcosa;
所以sin(a+b)-sin(a-b)=2sinbcosa;
这里把20当做a+b，40当做a-b，
sin20-sin40=2sin[(20-40)/2]cos[(20+40)/2]

Answer 2

(sin20°-cos50°)/cos80°
=(sin20°-sin40°)/sin10°
=(sin20°-2sin20°*cos20°)/sin10°
=sin20°*(1-2cos20°)/sin10°
=2sin10°*cos10°*(1-2cos20°)/sin10°
=2cos10°-2cos10°*cos20°
=2cos10°-2cos10°*[1-2(sin10°)^2]
=4(sin10°)^2

Answer 3