y=cos(a+π/4)+sin2a化简详解
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y=cos(a+π/4)+sin2a化简详解
y=cos(a+π/4)+sin2a化简详解
y=cos(a+π/4)+sin2a化简详解
y=cos(a+π/4)+sin2a
=√2/2(cosa-sina)+sin2a
令cosa-sina=t,则两边平方得1-sin2a=t²
所以y=√2t/2+1-t²
y=cos(a+π/4)-cos(2a+π/2)
=cos(a+π/4)-cos[2(a+π/4)]
=cos(a+π/4)-1+2cos²(a+π/4)
=2cos²(a+π/4)+cos(a+π/4)-1
若设:cos(a+π/4)=x
则:y=2x²+x-1 【转化为闭区间上的二次函数】
y=cos(a+π/4)+sin2a化简详解
化简1/2sin2a-cos^2(π/4-a)
化简1-sin2a-2cos^2(a-π/4)
cos(a-π/4)=1/4,则sin2a
1-sin2a-cos^2(a-π/4)
sin2a-2cos^2 a/sin(a-π/4)=?化简
化简cos^2(π/4-a)-sin^2(π/4-a)得到?A.sin2a B.-sin2a C.cos2a D.-cos2a
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证明:2sin(π+a)cos(π-a)=sin2a
已知tan(π/4+a=1/2)求sin2a-cos²a/2cos²a+sin²a
化简sin2a-2cos²a / sin(a-π/4)
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tana=-1/3 (2sin^a+sin2a)/cos(a-π/4)=?-π/2
已知cos(π/4+a)=4/5,求(sin2a-2sin^2a)(1-tana)
tan(a+π/4)=2,则(sin2a-cos²a)/(1+cos2a)=?
A为三角形的内角且sin2A=-3/5,则cos(A+π/4)=
已知tan(π/4+a)=1/2求cos2a/sin2a+cos²a
tan(π/4+a)=1/2 求(sin2a-cos平方a)÷(1+cos2a)