(cos10°-cos50°)÷(sin10°-sin50°)
问题描述:
(cos10°-cos50°)÷(sin10°-sin50°)
答
cos α-cos β=-2sin[(α+β)/2]·sin[(α-β)/2]
所以cos10°-cos50=2sin[30°]·sin[20°]
sin α-sin β=2cos[(α+β)/2]·sin[(α-β)/2]
所以sin10°-sin50°)=-2cos[30°]·sin[20°]
所以(cos10°-cos50°)÷(sin10°-sin50°)
=-2sin[30°]·sin[20°] /2cos[30°]·sin[20°]
=-tan30°=-√3/3
答
原式=[cos(30-20)-cos(30+20)]÷[sin(30-20)-sin(30+20)]=(cos30cos20+sin30sin20-cos30cos20+sin30sin20)÷(sin30cos20-cos30sin20-sin30cos20-cos30sin20)=2sin30sin20÷(-2cos30sin20)=-sin30/cos30=-√3/3