能告诉关于正余弦的sin（π/2+x），π-x，π/2-x，π+x，3π/2+x，3π/2-x的所有正确公式吗 非常感谢
(1)\u82e5arcsin2x=\u6d3e/3,\u5219x=\uff1f(2)\u4e24\u89d2\u548c\u7684\u6b63\u5f26\u516c\u5f0f\u662f\u5728\u786e\u5b9a\u4e86\u4e24\u89d2\u5dee\u7684\u4f59\u5f26\u516c\u5f0f\u540e\u63a8\u5bfc\u7684\uff0c\u65e2sin(a+\u88ab\u4ed6)=__...(1)\u82e5arcsin2x=\u6d3e/3,\u5219x=\u221a3/4
(2)sin(a+\u03b2)=
cos[90-(a+\u03b2)]
=cos[(90-a)-\u03b2]
=cos(90-a)cos\u03b2+sin(90-a)sin\u03b2
=sinacos\u03b2+cosasin\u03b2
sin(\u03b1+\u03c0/4)=4/5, cos(\u03b1+\u03c0/4)=-3/5
cos\u03b1=cos[(\u03b1+\u03c0/4)-\u03c0/4]=cos(\u03b1+\u03c0/4)cos\u03c0/4+sin(\u03b1+\u03c0/4)sin\u03c0/4
=(\u221a2/2)(-3/5)+(4/5)(\u221a2/2)
= \u221a2/10
cos(π/2+x)=-sinx
sin(π-x)=sinx
cos(π-x)=-cosx
sin(π/2-x)=cosx
cos(π/2-x)=sinx
sin(π+x)=-sinx
cos(π+x)=-cosx
sin(3π/2+x)=-cosx
cos(3π/2+x)=sinx
sin(3π/2-x)=-cosx
cos(3π/2-x)=-sinx
sin(π/2+x)=cos(x);
sin(π-x)=sin(x);
sin(π/2-x)=cos(x);
sin(π+x)=-sin(x);
sin(3π/2+x)=-cos(x);
sin(3π/2-x)=cos(x);
cos(π/2+x)=-sin(x);
cos(π-x)=-cos(x);
cos(π/2-x)=sin(x);
cos(π+x)=-cos(x);
cos(3π/2+x)=sin(x);
cos(3π/2-x)=-sin(x);
sin(π/2+x)=-cos(x-π/2)
sin(π-x)=sin(x)
sin(π/2-x)=-cos(π/2+x)
sin(π+x)=sinx
sin(3π/2+x)=sin(π/2+x)
sin(3π/2-x)=sin(π/2-x)
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