sin的平方A+cos的平方A=1变形
\u4e3a\u4ec0\u4e48 sin\u7684\u5e73\u65b9A+COS\u7684\u5e73\u65b9A=1sin\u7684\u5e73\u65b9A=a/c
COS\u7684\u5e73\u65b9A=b/c
a^2+b^2=c^2
sin\u7684\u5e73\u65b9A+COS\u7684\u5e73\u65b9A=1
\u7531\u4e8c\u500d\u89d2\u516c\u5f0f\uff0c
cos(a)=1-2sin²(a/2)
\u6240\u4ee5sin²(a/2)=[1-cos(a)]/2=1/4.
同学您好，如果问题已解决，记得采纳哦~~~您的采纳是对我的肯定~谢谢哦
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