若0<α<π/2,-π/2<β<0,cos(π/4+α)=1/3,cos(π/4-β/2)=√3/3则cos(α+β/2
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若0<α<π/2,-π/2<β<0,cos(π/4+α)=1/3,cos(π/4-β/2)=√3/3则cos(α+β/2)是多少谢...
若0<α<π/2,-π/2<β<0,cos(π/4+α)=1/3,cos(π/4-β/2)=√3/3则cos(α+β/2)是多少谢谢求过程
若0<α<π/2,-π/2<β<0,cos(π/4+α)=1/3,cos(π/4-β/2)=√3/3则cos(α+β/2)是多少谢谢求过程
因为0<a<π/2,-π/2<b<0,cos(π/4+a)=1/3,cos(π/4-b/2)=√3/3,
所以
π/4+a,π/4-b/2为锐角
即sin(π/4+a)=2√2/3,sin(π/4-b/2)=√6/3
所以
cos(a+b/2)=cos[(π/4+a)-(π/4-b/2)]
=cos(π/4+a)cos(π/4-b/2)+sin(π/4+a)sin(π/4-b/2)
=1/3*√3/3+2√2/3*√6/3
=(√3+4√3)/9
=5√3/9
所以
π/4+a,π/4-b/2为锐角
即sin(π/4+a)=2√2/3,sin(π/4-b/2)=√6/3
所以
cos(a+b/2)=cos[(π/4+a)-(π/4-b/2)]
=cos(π/4+a)cos(π/4-b/2)+sin(π/4+a)sin(π/4-b/2)
=1/3*√3/3+2√2/3*√6/3
=(√3+4√3)/9
=5√3/9
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