化简√3cos2xtan(x+π/4)-2(sinx)2
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咨询记录 · 回答于2023-06-12
化简√3cos2xtan(x+π/4)-2(sinx)2
您好，√3cos2xtan(x+π/4)-2(sinx)2化简为：√3cos2xtan(x+π/4)-2(sinx)2=√3cos2x*tanx+2cos2x+√3cos2x-2。首先，将√3cos2xtan(x+π/4)展开：√3cos2xtan(x+π/4)=√3cos2x*(tanx+tan(π/4))=√3cos2x*(tanx+1)=√3cos2x*tanx+√3cos2x接下来，将2(sinx)2化简：2(sinx)2=2(sin2x)=2(1-cos2x)=2-2cos2x将以上两个式子相加，得到：√3cos2xtan(x+π/4)-2(sinx)2=√3cos2x*tanx+√3cos2x-2+2cos2x=√3cos2x*tanx+2cos2x+√3cos2x-2因此，最终化简后的结果为：√3cos2xtan(x+π/4)-2(sinx)2=√3cos2x*tanx+2cos2x+√3cos2x-2
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