若sin(3π/2-2x)=3/5,则tan^2 x等于？ 若sin(3π2?2x)＝35，则tan2x等于 ____...
sin(3\u03c0/2-2x)=3/5 \u6c42tan²x\u89e3\uff1a
\u6839\u636e\u8bf1\u5bfc\u516c\u5f0f\u5f97
sin(3\u03c0/2-2x)=-cos2x=3/5
cos2x=2cos²x-1=-3/5
2cos²x=2/5
cos²x=1/5
sin²x=1-cos²x=4/5
\u6240\u4ee5
tan²x=sin²x/cos²x=4/5 / 1/5=4
sin(3\u03c02?2x)=sin[\u03c0+\uff08\u03c02-2x\uff09]=-sin\uff08\u03c02-2x\uff09=-cos2x=35\uff0c\u6240\u4ee5cos2x=-35\uff0c\u53732cos2x-1=-35\uff0c\u5219cos2x=15\uff0c\u6240\u4ee5tan2x=sec2x-1=1cos2x-1=5-1=4\uff0e\u6545\u7b54\u6848\u4e3a\uff1a4
sin（3π/2-2x）=3/5，所以cos2x=-3/5,根据万能公式
cos2x=(1-(tanx)^2)/(1+(tanx)^2)
所以 （tanx）^2=4
sin(3π/2-2x)=3/5
sin3π/2cos2x-cos3π/2sin2x=3/5
cos2x=-3/5
cos2x=2cos^2(x)-1
cos^2(x)=1/5
sec^2(x)=5
1+tan^2(x)=sec^2(x)=5
tan^2 x=4
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