【已知tanθ=1,求sinθcosθsin^2θ—2cos^2θ,和1sin^2θ—sinθcosθ—cos^2θ】
<p>问题：【已知tanθ=1,求sinθcosθsin^2θ—2cos^2θ,和1sin^2θ—sinθcosθ—cos^2θ】<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">刘亚林的回答：<div class="content-b">网友采纳 tanθ=1-->θ=k*pi+pi/4,2θ=2k*pi+pi/21.sinθ*cosθ=sin(2θ)/2=1/2;sin^2θ=(1-cos2θ)/2=1/2;cos^2θ=(1+cos2θ)/2=1/2所以sinθcosθsin^2θ—2cos^2θ=1/2/(1/2-2*1/2)=-...
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