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  • (cos80°-cos20°)/sin80°+sin20°)=

    (cos80°-cos20°)/sin80°+sin20°)=
    順便再回答個(gè)
    cos20°+cos100°+cos140°=
    數(shù)學(xué)人氣:263 ℃時(shí)間:2020-04-13 21:43:43
    優(yōu)質(zhì)解答
    (cos80°-cos20°)/(sin80°+sin20°)
    =[cos(50°+30°)-cos(50°-30°)]/[sin(50°+30°)+sin(50°-30°)]
    =[(cos50°cos30°-sin50°sin30°)- (cos50°cos30°+sin50°sin30°)]/[ (sin50°cos30°+cos50°sin30°)+ (sin50°cos30°-cos50°sin30°)]
    =(-2sin50°sin30°)/ (2sin50°cos30°)
    = -2sin30°/ cos30°
    = -tan30°
    = -√3/3
    cos20°+cos100°+cos140°
    = cos (80°-60°)+cos100°+cos(80°+60°)
    = (cos80°cos60°+sin80°sin60°)+cos100°+ (cos80°cos60°-sin80°sin60°)
    = 2cos80°cos60°+cos100°
    = cos80°+cos100°
    = cos80°-cos80°
    =0
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