(cos10°-cos50°)÷(sin10°-sin50°)
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(cos10°-cos50°)÷(sin10°-sin50°)(cos10°-cos50°)÷(sin10°-sin50°)(cos10°-cos50°)÷(sin10°-sin50°)原式=[cos
(cos10°-cos50°)÷(sin10°-sin50°)
(cos10°-cos50°)÷(sin10°-sin50°)
(cos10°-cos50°)÷(sin10°-sin50°)
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
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
求值:cos10°/cos50°
(cos10°-cos50°)÷(sin10°-sin50°)
cos10°+cos110°-cos50°=
【求解释】(tan10°-根号3)cos50°=-1原式 = cos50°×(tan10°-tan60°) = cos50°×(sin10°cos60°-sin60°cos10°)/cos10°cos60° = cos50°×-sin(50°)/cos10°cos60° (此部以下求解释、算的对么、sin100°咋来的、最后怎么又
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