cos(a+5π/12)cos(a+π/6)+cos(π/12-a)cos(π/3-a）
=sin(π/2-(a+5π/12))cos(a+π/6)+cos(π/12-a)sin(π/2-(π/3-a）)
=sin(π/12-a)cos(a+π/6)+cos(π/12-a)sin(a+π/6)
=sin(π/12-a+a+π/6)
=sin(π/12+π/6)
=sinπ/4
=√2/2
计算cos(a+5π/12)cos(a+π/6)+cos(π/12-a)cos(π/3-a）
1个回答
相关问题
-
为什么因为（π/6-a）+（5π/6+a）=π,可以得到cos（5π/6+a）=cos【π-（π/6-a）】=-cos（
-
已知cos（5π/12+a）=1/3,且-《a《-π/2,则cos（π/12—a）等于?
-
cos(α+5π/12)cos(α+π/6)+cos(π/12-α)cos(π/3-α) 化简
-
cos(-79π/6) =cos(79π/6) =cos(12π+7π/6) =cos(7π/6) =cos(π+π/6
-
cos(π/12)cos(5π/12)=
-
可以这样做吗 cos(a-π)=-5/13 cos(-π+a)=-5/13 -cos(π-a)=-5/13 cos(π-
-
cos^2(π/3-a)+cos^2(π/6+a)
-
[cos(a+π)sin²（2+3π）]/[tan(a+π)cos³（-a-π）]
-
已知sin(a-π)=2cos(2π-a),求(sin(π-a)+5cos(2π-a))/(3cos(π-a)-sin(
-
求证:tan(2π-a) sin(-2π-a) cos(6π-a) / cos(a-π) sin(5π-a)=-tana