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標題: 三角函數,倍角公式,證明 tanx+tan(x/2)=cscx-2cot(2x) [打印本頁]

作者: weiye    時間: 2008-6-12 23:50     標題: 三角函數,倍角公式,證明 tanx+tan(x/2)=cscx-2cot(2x)

倍角公式與半角公式例題,

試證: tan(x) + tan(x/2) = csc(x) - 2cot(2x)

證明:

tan(x) + tan(x/2)= sin(x)/cos(x) + {1-cos(x)} / sin(x)

         ={sin^2 (x) + cos(x) - cos^2 (x)} /{sin(x) cos(x)}

         ={(1 - cos^2 (x)) + cos(x) - cos^2 (x)} /{sin(x) cos(x)}

         ={cos(x) + 1 - 2 cos^2(x)} /{sin(x) cos(x)}

         ={cos(x) - cos(2x)} /{sin(x) cos(x)}

         = 1/sin(x)  -  cos(2x)/{sin(x) cos(x)}

         = csc(x) - 2 cos(2x)/{2 sin(x) cos(x)}

         = csc(x)  - 2 cos(2x)/sin(2x)

         = csc(x) - 2cot(2x)

證畢.




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