若sin(x+5π/12)的平方-sin(x-5π/12的平方)=-根号3/4 且x∈(3π/4,π),求tanx
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![若sin(x+5π/12)的平方-sin(x-5π/12的平方)=-根号3/4 且x∈(3π/4,π),求tanx](/uploads/image/z/8857503-63-3.jpg?t=%E8%8B%A5sin%EF%BC%88x%2B5%CF%80%2F12%EF%BC%89%E7%9A%84%E5%B9%B3%E6%96%B9-sin%28x-5%CF%80%2F12%E7%9A%84%E5%B9%B3%E6%96%B9%EF%BC%89%3D-%E6%A0%B9%E5%8F%B73%2F4+%E4%B8%94x%E2%88%88%EF%BC%883%CF%80%2F4%2C%CF%80%EF%BC%89%2C%E6%B1%82tanx)
若sin(x+5π/12)的平方-sin(x-5π/12的平方)=-根号3/4 且x∈(3π/4,π),求tanx
若sin(x+5π/12)的平方-sin(x-5π/12的平方)=-根号3/4 且x∈(3π/4,π),求tanx
若sin(x+5π/12)的平方-sin(x-5π/12的平方)=-根号3/4 且x∈(3π/4,π),求tanx
sin^2(x+5π/12) - sin^2(x-5π/12) =- √3/4
{ sin(x+5π/12) + sin(x-5π/12) } { sin(x+5π/12) - sin(x-5π/12) } =- √3/4
2sin[(x+5π/12+x-5π/12)/2] cos[(x+5π/12-x+5π/12)/2] * 2cos[(x+5π/12+x-5π/12)/2] sin[(x+5π/12-x+5π/12)/2] =- √3/4
2sinx cos(5π/12) * 2cosx sin(5π/12) =- √3/4
2sinx cosx * 2 sin(5π/12)cos(5π/12) =- √3/4
sin(2x) * sin(5π/6) =- √3/4
sin(2x) * √3/2=- √3/4
sin(2x) = -1/2
x ∈(3π/4,π)
2x ∈(3π/2,2π)
cos(2x) > 0
cos(2x) = √{1-sin^2(2x)} = √(1-1/4) = √3/2
tanx = [1-cos(2x)]/sin(2x) = (1-√3/2)/(-1/2) = √3 - 2
先利用倍角公式求出sin2x,则可求cos2x,然后利用tanx=(1-cos2x)/sin2x即可求出,方法多样,自己思考相信是可以做出来的。