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化简[2cos^4(x)-2cos²x+1/2]/[2tan(π/4-x)·sin²（π/4+x)]
题目详情
化简[2cos^4(x)-2cos²x+1/2]/[2tan(π/4-x)·sin²（π/4+x)]
▼优质解答
答案和解析
[2cos^4(x)-2cos²x+1/2]/[2tan(π/4-x)·sin²（π/4+x)]
=1/2*[4cos^4(x)-4cos²x+1]/{2tan[(π/2-π/4)-x]·sin²（π/4+x)}
=1/2*[2cos²x-1]²]/{2tan[π/2-(π/4+x)]·sin²（π/4+x)}
=1/2*[2cos²x-1]²]/{2cot(π/4+x)·sin²（π/4+x)}
=1/2*[2cos²x-1]²]/{2cos(π/4+x)·sin（π/4+x)}
=1/2*(cos2x)²/sin[2(π/4+x)]
=1/2*(cos2x)²/sin[π/2+2x]
=1/2*(cos2x)²/(cos2x)
=1/2*cos2x
=1/2*[4cos^4(x)-4cos²x+1]/{2tan[(π/2-π/4)-x]·sin²（π/4+x)}
=1/2*[2cos²x-1]²]/{2tan[π/2-(π/4+x)]·sin²（π/4+x)}
=1/2*[2cos²x-1]²]/{2cot(π/4+x)·sin²（π/4+x)}
=1/2*[2cos²x-1]²]/{2cos(π/4+x)·sin（π/4+x)}
=1/2*(cos2x)²/sin[2(π/4+x)]
=1/2*(cos2x)²/sin[π/2+2x]
=1/2*(cos2x)²/(cos2x)
=1/2*cos2x
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