求解三角函数化简
\u4e09\u89d2\u51fd\u6570\u5316\u7b80,\u6c42\u89e3\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002 \u9898\u9519\u4e86\u5427 \u662f\u4e0d\u662ftan\u3002\u3002\u3002\uff1f
\u7b2c\u4e00\u6b65\u662f\u5427\uff0c\u8fd9\u4e2a\u6bd4\u8f83\u7b80\u5355\u7684\uff1a
2cosx^4-2cosx^2+1/2
=2cosx^2(cosx^2-1)+1/2
=1/2-2cosx^2*sinx^2
=1/2-2*(2sinxcosx)^2/4-----------\u62ec\u53f7\u5185\uff1a2sinxcosx=sin2x
=1/2-sin2x^2/2
---------------------------------
\u4e0b\u4e00\u6b65\uff1a\u5206\u5b50\u5206\u6bcd\u540c\u4e58\u4ee5cos(\u03c0/4-x)
\u524d\u9762=((1/2)cos2x^2*cos(\u03c0/4-x))/(2sin(x+\u03c0/4)^2*sin(\u03c0/4-x))
\u5229\u7528\u8bf1\u5bfc\u516c\u5f0f\uff1a
sin(\u03c0/2+2x)=cos2x\uff0csin(\u03c0/4+x)=cos(\u03c0/4-x)\uff0csin(\u03c0/4-x)=cos(\u03c0/4+x)
-------------------------------------
\u5012\u6570\u7b2c\u4e8c\u6b65\uff1a\u500d\u89d2\u516c\u5f0f
2sin(\u03c0/4+x)cos(\u03c0/4+x)=sin(\u03c0/2+2x)
----------------------------
\u6700\u540e\u4e00\u6b65\uff1a\u8bf1\u5bfc\u516c\u5f0f
∵cosx=cos²(x/2)-sin²(x/2)
=[cos(x/2)+sin(x/2)][cos(x/2)-sin(x/2)]
1-sinx=sin²(x/2)-2sin(x/2)cos(x/2)+cos²(x/2)
=[cos(x/2)-sin(x/2)]²
∴左边
=[cos(x/2)+sin(x/2)]/[cos(x/2)-sin(x/2)]
=[1+tan(x/2)]/[1-tan(x/2)]
=[1+tanπ/4*tan(x/2)]/[tanπ/4-tan(x/2)]
=1/tan(π/4-x/2)
=cot(π/4-x/2)
设a=x/2,则
左边
=（1-2sin²a)/(cos²a-2sinacosa+sin²a)
=（cos²a+sin²a-2sin²a)/(cos²a-2sinacosa+sin²a)
=(cos²a-sin²a)/(cosa-sina)²
=(cosa+sina)/(cosa-sina)
=(1+tana)/(1-tana)
=[tan(π/4)+tana]/(1-tanπ/4tana)
=tan(π/4+a)
=cot(π/4-a)
=右边。
证明:
左边=[(cosx/2)^2-(sinx/2)^2]/[(sinx/2-cosx/2)^2=(cosx/2+sinx/2)*(cosx/2-sinx/2)/[(cosx/2-sinx/2)^2=(cosx/2+sinx/2)/(cosx/2-sinx/2)
=(1+tanx/2)/(1-tanx/2),(分子 分母同除以cosx/2)
=tan(π/4+X/2)
=cot(π/2-π/4-X/2)
=cot(π/4-x/2)
=右边
右边，
cot(π/4-x/2)
=cos(π/4-x/2) / sin(π/4-x/2)
=[cosπ/4*cosx/2 + sinπ/4*sinx/2] / [sinπ/4cosx/2 - sinx/2cosπ/4]
=√2/2*[cosx/2 + sinx/2] /√2/2*[cosx/2 - sinx/2]
=[cosx/2+sinx/2 ] /[cosx/2-sinx/2]
=[cosx/2 + sinx]^2 / [cosx/2 - sinx/2]*[cosx + sinx]
=[1+ 2sinx/2cosx/2] / [cosx/2^2 -s inx/2^2]
=[1+sinx]/cosx
=tanx+1/cosx
左边，
cosx*(1 + sinx) / [1 - sinx^2]
=[cosx +s inx*cosx] / coxs^2
=tanx+1/cosx
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