數學問題一問..(thx)萬分感激!!!!!－wszspj2的部落格

標題:

數學問題一問..(thx)萬分感激!!!!!

發問:

三角恒等式解題: 1.1-sin^2x除cosx 2.tan^2x+2 3.(1-sinx)(2+2sinx) 4.1-cos^2x除1-sin^2x 5.sin^2x+sin^2x除tan^2x

最佳解答:

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其他解答:

1.1-sin^2x=cos^2x(因為cos^2x+sin^2x=1) 1-sin^2x/cosx =cos^2x/cosx =cosx 2.tan^2x=sin^2x/cos^2x tan^2x+2 =(sin^2x/cos^2x)+2 =(2sin^2x/cos^2x)+(2cos^2x/cos^2x)............[[通份母]] =(2sin^2x + 2cos^2x)/cos^2x =2(sin^2x+cos^2x)/cos^2x...........[[抽2]] =2/cos^2x................[[sin^2x+cos^2x=1]] 3.(1-sinx)2(1+sinx)..............[[抽2]]即係2乘(1-sinx)乘(1+sinx) =2(1-sinx)(1+sinx) =2(1^2 - sin^2x)........[[(a+b)(a-b)=a^2 - b^2]] =2(1-sin^2x) =2cos^2x 4.(1-cos^2x)/(1-sin^2x) =sin^2x/cos^2x =tan^2x 5.(sin^2x+sin^2x)/tan^2x =(2sin^2x)/(sin^2x/cos^2x) =(2sin^2x)(cos^2x/sin^2x).....[[a/(b/c)=a(c/b)(除b/c會變做乘c/b]] =2cos^2x.....[[約左sin^2x]]