求cos(a+b)的值已知cos
已知cos(a-b/2)=-1/9,sin(a/2-b)=2/3且π/2<a<π,0<b<π/2,求cos(a+b)的值
π/2<a<π,0<b<π/2
--->π/4<a-b/2<π,∵cos(a-b/2)=-1/9<0--->π/2<a-b/2<π
--->sin(a-b/2)=4√5/9
同时:-π/4<a/2-b<π/2,∵sin(a/2-b)>0--->0<a/2-b<π/2
--->cos(a/2-b)=√5/3
cos(a/2+b/2) = cos[(a-b/2)-(a/2-b)]
= cos(a-b/2)cos(a/2-b)+sin(a-b/2)sin...全部
已知cos(a-b/2)=-1/9,sin(a/2-b)=2/3且π/2<a<π,0<b<π/2,求cos(a+b)的值
π/2<a<π,0<b<π/2
--->π/4<a-b/2<π,∵cos(a-b/2)=-1/9<0--->π/2<a-b/2<π
--->sin(a-b/2)=4√5/9
同时:-π/4<a/2-b<π/2,∵sin(a/2-b)>0--->0<a/2-b<π/2
--->cos(a/2-b)=√5/3
cos(a/2+b/2) = cos[(a-b/2)-(a/2-b)]
= cos(a-b/2)cos(a/2-b)+sin(a-b/2)sin(a/2-b)
= (-1/9)(√5/3)+(4√5/9)(2/3)
= 7√5/27
--->cos(a+b) = 2cos²(a/2+b/2)-1 = -239/729。
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