a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求a(b-d)值域,θ∈(0,π)a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求f(θ)=a(b-d)值域,θ∈(0,π)a和b和c都是向量

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a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求a(b-d)值域,θ∈(0,π)a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求f(θ)=a(b-d)值域,θ∈(0,π)a和b和c都是向量
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a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求a(b-d)值域,θ∈(0,π)a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求f(θ)=a(b-d)值域,θ∈(0,π)a和b和c都是向量
a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求a(b-d)值域,θ∈(0,π)
a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求f(θ)=a(b-d)值域,θ∈(0,π)
a和b和c都是向量

a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求a(b-d)值域,θ∈(0,π)a=(cosθ,sinθ),b=(cos2θ,sin2θ),d=(0,1),求f(θ)=a(b-d)值域,θ∈(0,π)a和b和c都是向量
f(θ)=a·(b-d)=（cosθcos2θ+sinθsin2θ）-sinθ=cosθ-sinθ=√2sin(θ+π/4)
值域[-√2,√2]

非线性方程解析解-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos(b)-y0*sin(a)*co cos^B-cos^C=sin^A,三角形的形状 求非线性方程组的“解析解”-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0 -x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos 求证cos(a+b)cos(a－b)=cos^2b－sin^2a 求证cos(a+b)cos(a-b)=cos^2a-sin^2b 求证:cos(a+b)cos(a-b)=cos平方b-sin平方a 求证:cos²a-sin²b=cos(a-b)cos(a+b) 如何证明sin(a+b)=sin(a)cos(b)+cos(α)sin(b) sin a sin b +cos a cos b =0,则sin a cos a+sin b cos b的值 矩阵A=(cosθ -sinθ)(sinθ cosθ),求A的伴随矩阵 Cos(a+b)*cos(a-b)=1/5 求cos ^2-sin^2 由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证：1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a) cos(a-b)cos(b-c)+sin(a-b)sin(b-c)= cos^2a-cos^2b=m,那么sin(a+b)sin(a-b)=? cos²a-cos²b=c,则sin(a+b)sin(a-b)= cos∧a-cos∧b=t则sin(a+b)sin(a-b)= 证明sin(A+B)sin(A-B)=cos^2B-cos^2A sin(a-b)*cos a-cos(a-b)*sin a=1/5,则cos 2b的值是