sinα+cosα=4/5,
平方得
1+2sinαcosα=16/25
則sin2α=-9/25sin65°+sin15°sin10°/sin25°-cos25°cos80°=log2為底cosπ/9+log2為底cos2π/9+log2為底cos4π/9=問(wèn)我這么多,好。疑似錯(cuò)了。已經(jīng)修正sin65°+sin15°sin10°/sin25°-cos15°cos80°=(sin(75°-10°)+cos75°sin10°)/(cos65°-cos15°cos80°)=sin75°cos10°/sin15°sin80°=tan75°=tan(45°+30°)=2+根號(hào)3log2為底cosπ/9+log2為底cos2π/9+log2為底cos4π/9=log2為底(cosπ/9cos2π/9cos4π/9)因?yàn)閏osπ/9cos2π/9cos4π/9=(sinπ/9cosπ/9cos2π/9cos4π/9)/sinπ/9=(sin8π/9)/(8sinπ/9)=1/8所以原式=-3
已知sinα+cosα=4/5,則sin2α=
已知sinα+cosα=4/5,則sin2α=
數(shù)學(xué)人氣:898 ℃時(shí)間:2020-05-10 11:47:16
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