怎么用和差化积公式算呢

baxiannv

3．已知向量 $a=\left(\cos 75^{\circ}, \sin 75^{\circ}\right), b=\left(\cos 15^{\circ}, \sin 15^{\circ}\right)$ ，则 $|a-b|$ 的值为
A．$\frac{1}{2}$
B． 1
C． 2
D． 3

战巡

回复 1# baxiannv


为啥要和差化积......

\[|\vec{a}-\vec{b}|^2=(\cos(75\du)-\cos(15\du))^2+(\sin(75\du)-\sin(15\du))^2\]
\[=\cos^2(75\du)+\cos^2(15\du)-2\cos(75\du)\cos(15\du)+\sin^2(75\du)+\sin^2(15\du)-2\sin(75\du)\sin(15\du)\]
\[=2-2\cos(75\du-15\du)=...\]

baxiannv

[因为我和差化积算来算去都是根号2b]回复 2# 战巡 [/b]

战巡

回复 3# baxiannv


怎么可能呢？

\[\cos(75\du)-\cos(15\du)=2\sin(\frac{75\du+15\du}{2})\sin(\frac{15\du-75\du}{2})\]
\[=2\sin(45\du)\sin(-30\du)=2\cdot\frac{1}{\sqrt{2}}\cdot(-\frac{1}{2})=-\frac{1}{\sqrt{2}}\]

\[\sin(75\du)-\sin(15\du)=2\sin(\frac{75\du-15\du}{2})\cos(\frac{75\du+15\du}{2})\]
\[=2\sin(30\du)\cos(45\du)=\frac{1}{\sqrt{2}}\]

那么
\[|\vec{a}-\vec{b}|=\sqrt{(-\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}})^2}=1\]

力工

用几何意义，就是正三角形了。楼主何苦自己逼自己走河汊滑鸡呢。

isee

回复 5# 力工


现状是几何（中学生）都很弱，想到单位圆上都不易了，再去化归平面向量的模更难了，虽然可秒~