sin平方π/6=1/4 为什么？ 求过程。 已知sin(π/6-α）=1/4,则sin(π/6+2α）=...
sin\u5e73\u65b9(\u03c0/6-x)\u600e\u4e48\u5316\u7b80(1/2)(1 + cos2x) + (cosx)[cosxcos(\u03c0/3) \u2013 sinxsin(\u03c0/3)] + (1/2)[1\u2013 cos(\u03c0/3 \u2013 2x)]
= 1/2 + (1/2)cos2x+ (1/2)cos 2 x\u2013 (\u221a3/2)cosxsinx+ 1/2 \u2013 (1/2)cos(\u03c0/3 \u2013 2x)
= 1 +(1/2)cos2x + (1/2)*(1/2)(1 + cos2x) \u2013 (\u221a3/4)sin2x \u2013 (1/2)[cos(\u03c0/3)cos2x+ sin(\u03c0/3)sin2x]
= 5/4 + (3/4)cos2x\u2013 (\u221a3/4)sin2x \u2013 (1/4)cos2x \u2013 (\u221a3/4)sin2x
= 5/4 + (1/2)cos2x \u2013 (\u221a3/2)sin2x
= 5/4 + sin(2x + 5\u03c0/6)
sin(\u03c0/6-\u03b1\uff09=1/4,
\u5219cos(\u03c0/3-2\u03b1)=cos[2(\u03c0/6-\u03b1)]
=1-2 sin²(\u03c0/6-\u03b1)
=7/8.
\u6240\u4ee5cos(2\u03b1-\u03c0/3) =7/8.
sin(\u03c0/6+2\u03b1)= sin((2\u03b1-\u03c0/3)+ \u03c0/2)
= cos(2\u03b1-\u03c0/3)==7/8.
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~\u5982\u679c\u4f60\u8ba4\u53ef\u6211\u7684\u56de\u7b54\uff0c\u8bf7\u53ca\u65f6\u70b9\u51fb\u3010\u91c7\u7eb3\u4e3a\u6ee1\u610f\u56de\u7b54\u3011\u6309\u94ae
~~\u624b\u673a\u63d0\u95ee\u7684\u670b\u53cb\u5728\u5ba2\u6237\u7aef\u53f3\u4e0a\u89d2\u8bc4\u4ef7\u70b9\u3010\u6ee1\u610f\u3011\u5373\u53ef\u3002
~\u4f60\u7684\u91c7\u7eb3\u662f\u6211\u524d\u8fdb\u7684\u52a8\u529b
~~O(\u2229_\u2229)O\uff0c\u8bb0\u5f97\u597d\u8bc4\u548c\u91c7\u7eb3\uff0c\u4e92\u76f8\u5e2e\u52a9
sinπ/6=1/2
(1/2)平方=1/4
绛旓細绠鍗曞垎鏋愪竴涓嬶紝璇︽儏濡傚浘鎵绀
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