sin(-11π/6)+cos12π/5·tan3π+cos(-π/4)

sin(-11/6π)+cos12/5π×tan4π化简 求值:sin(-11π/6)+cos12π/5*tan4π+1/cos7π/3 sin(-11π/6)+cos12π/5·tan3π+cos(-π/4) 计算 1:sin（-11π/6）+cos12π/5 *tan4π 2：sin11ts40*cos（-690）+tan1845第二题是度数1140度 sin12/π-cos12/π= sin12分之25πcos6分之11-cos12分之11πsin6分之5π等于? sin12分之π等于 cos12分之5π嘛 化简下列各式：（1）a^2 sin(-1350度)+b^2 tan405度-（a-b）^2 cot765度-2ab cos(-1080度)（2）sin(-11/6兀)+cos12/5兀*tan4兀-sec13/3兀 （cos12分之5π+sin12分之5π）（cos12分之5π-sin12分之5π）的值为 （sin12分之25π乘以cos6分之11π)减去(cos12分之11π乘以sin6分之5π)的值是多少 sin12分之π*cos12分之π=具体点, 1.sin42-cos12+sin54=_________2.sin40(1+2sin10)=___________3.y=3sin(x+20)+5sin(x+80)的最大值是____ sin 11π/6 8(cos12分之7π+isin12分之7π)÷4(cos3分之π+isin3分之π) sin0=0 sin(1/6π)=1/2sin(1/3π)=√3/2sin(1/2π)=1sin(2/3π)=√3/2sin(5/6π)=1/2sin(π)=0sin(-π)=0sin(-5/6π)=-1/2sin(-2/3π)=-√3/2sin(-1/2π)=-1sin(-1/3π)=-√3/2sin(-1/6π)=-1/2sin(7/6π)=-1/2sin(4/3π)=-√3/2sin(5/3π)=-√3/2sin(11/6π)=-1/ 计算,sin(π/12)-sin(5π/12)+2sin(π/8)sin(3π/8) 已知sin(x+π/6)=1/4,求sin(5π/6-x)+sin^2（11π/6-x）如题, 已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x)