数学问题：cos3π/8=sin?
\u6bd4\u8f83sin(sin3\u03c0/8)\u548csin(cos3\u03c0/8)\u7684\u5927\u5c0f\uff0c\u8981\u8fc7\u7a0bsin(3\u03c0/8)=sin(3\u03c0/8);
cos(3\u03c0/8)=sin(\u03c0/2-3\u03c0/8)=sin(\u03c0/8);
\u6240\u4ee5\uff1a
sin(3\u03c0/8)>cos(3\u03c0/8).
\u7531\u4e8e\u6b63\u5f26\u51fd\u6570\u662f\u5728\u7b2c\u4e00\u8c61\u9650\u662f\u589e\u51fd\u6570\uff0c\u6240\u4ee5\uff1a
sin[sin(3\u03c0/8)]>sin[cos(3\u03c0/8)].
cos3\u03c0/8=cos(\u03c0/2-\u03c0/8)=sin\u03c0/8
sin3\u03c0/8=cos\u03c0/8
(cos\u03c0/8+sin\u03c0/8)\uff08cos3\u03c0/8-sin3\u03c0/8\uff09
=(cos\u03c0/8+sin\u03c0/8)\uff08sin\u03c0/8-cos\u03c0/8\uff09
=\uff08sin\u03c0/8)^2-(cos\u03c0/8\uff09^2
=-cos\u03c0/4=-\u221a2/2
cos3π/8
=sin(π/2-3π/8)
=sinπ/8
7/8π
z
sinπ/8
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