设向量a=(1,sin0),b=(3sin0,1)且a‖b,则cos20=
\u5df2\u77e5\u5411\u91cfa=(cos\u03b8\uff0c1)\uff0cb=(\u221a3\uff0csin\u03b8)\uff0c\u4e14a\u5782\u76f4\u4e8eb\uff0c\u5176\u4e2d \u03b8\u2208(0\uff0c\u6d3e)\u3002(1)\u6c42\uff081\uff09a\u5782\u76f4\u4e8eb===>a*b=0===>(Cos\u03b8\uff0c1)*(\u221a3\uff0csin\u03b8)=\u221a3cos\u03b8+sin\u03b8=0===>2sin(\u03b8+pai/3)=0
\u6240\u4ee5 \u03b8=2pai/3=120\u00b0\uff0c
\uff082\uff09tan( \u03b8-\u6d3e/4)=tan(120\u00b0-45\u00b0\uff09=tan75\u00b0=\uff081+TAN30\u00b0/(1-tan30\u00b0\uff09= 2+\u221a3\u3002
那么1*1-sinθ*3sinθ=0
所以sin²θ=1/3
所以cos2θ=1-2sin²θ=1-2/3=1/3
∵a‖b
∴1：3sino=sino:1
3(sino)^2=1
(sino)^2=1/3
(coso)^2=1-1/3=2/3
cos20=(coso)^2-(sin0)^2=2/3-1/3=1/3
a∥b
则：1=3sin²θ
得：sin²θ=1/3
由倍角公式：cos2θ=1-2sin²θ=1/3
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可得：a=tb 即：
1=3tsinx
sinx=t
解得：sin²x=1/3
因：sin²x+cos²x=1 所以有：cos²x=2/3
cos2x=cos²x-sin²x
=2/3-1/3
=1/3
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