tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)
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![tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)](/uploads/image/z/2561818-58-8.jpg?t=tan%28%CE%B1%2B%CF%80%2F6%29%3D1%2F2%2Ctan%28%CE%B2-7%2F6%CF%80%29%3D1%2F3%2C%E6%B1%82tan%28%CE%B1%2B%CE%B2%29)
tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)
tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)
tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)
tan(β-7/6π)=tan(β-π-π/6)=tan(β-π/6)
tan(α+β)=tan[(α+π/6)+(β-π/6)]
=[tan(α+π/6)+tan(β-7/6π)]/[1-tan(α+π/6)*tan(β-7/6π)]
=[1/2+1/3]/[1-(1/2)*(1/3)]
=1
由tan(β-7/6π)=1/3得tan(β-1/6π)=1/3
故tan(α+β)=(1/2+1/3)/(1-1/6)=1