若0<x<pai⼀2,-pai⼀2<y<0,cos(pai⼀4+x)=1⼀3,COS(pai⼀4-Y⼀2)=根号3⼀3,求cos(x+y⼀2)=

2026年09月28日 13:26
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0-π/4π/4<π/4+x<3π/4 π/4<π/4-y/2<0
cos(π/4+x)=1/3,sin(π/4+x)=2√2/3 COS(π/4-y/2)= √3/3,sin(π/4-y/2)=√6/3
cos(x+y/2)=cos[(π/4+x)-(π/4-y/2)]
=cos(π/4+x)cos(π/4-y/2)+sin(π/4+x)sin(π/4-y/2)
=1/3* √3/3+2√2/3 * √6/3
=5√3/9