cosα= -4/5 ,α∈(π/2,π)求cos(π/3+α) 那个,再不好意思地问一下另一道.已知sinα=-2/3,α∈(π,3π/2),cosβ=3/4.β∈(3π/2 ,2π)求cos(β-α)
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![cosα= -4/5 ,α∈(π/2,π)求cos(π/3+α) 那个,再不好意思地问一下另一道.已知sinα=-2/3,α∈(π,3π/2),cosβ=3/4.β∈(3π/2 ,2π)求cos(β-α)](/uploads/image/z/8574158-38-8.jpg?t=cos%CE%B1%3D+-4%2F5+%2C%CE%B1%E2%88%88%28%CF%80%2F2%2C%CF%80%29%E6%B1%82cos%28%CF%80%2F3%2B%CE%B1%29+%E9%82%A3%E4%B8%AA%2C%E5%86%8D%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D%E5%9C%B0%E9%97%AE%E4%B8%80%E4%B8%8B%E5%8F%A6%E4%B8%80%E9%81%93.%E5%B7%B2%E7%9F%A5sin%CE%B1%3D-2%2F3%2C%CE%B1%E2%88%88%EF%BC%88%CF%80%2C3%CF%80%2F2%EF%BC%89%2Ccos%CE%B2%3D3%2F4.%CE%B2%E2%88%88%EF%BC%883%CF%80%2F2+%2C2%CF%80%EF%BC%89%E6%B1%82cos%EF%BC%88%CE%B2-%CE%B1%EF%BC%89)
cosα= -4/5 ,α∈(π/2,π)求cos(π/3+α) 那个,再不好意思地问一下另一道.已知sinα=-2/3,α∈(π,3π/2),cosβ=3/4.β∈(3π/2 ,2π)求cos(β-α)
cosα= -4/5 ,α∈(π/2,π)求cos(π/3+α)
那个,再不好意思地问一下另一道.已知sinα=-2/3,α∈(π,3π/2),cosβ=3/4.β∈(3π/2 ,2π)求cos(β-α)
cosα= -4/5 ,α∈(π/2,π)求cos(π/3+α) 那个,再不好意思地问一下另一道.已知sinα=-2/3,α∈(π,3π/2),cosβ=3/4.β∈(3π/2 ,2π)求cos(β-α)
cos(π/3+α)=cosπ/3*cosα-sinα*sinπ/3
因为 α∈(π/2,π)所以sinα=3/5
所以式子等于(带进去就行了,原谅我跳步)
=-(4+3根号3)/10
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