$$ \begin{align} &如图，设\overrightarrow{OP}=(acosx，asinx)，且\notag\\ &\qquad\overrightarrow{PQ}=(bcos(x+φ_0),bsin(x+φ_0)))\\ &由\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ}，可知\notag\\ &\qquad\overrightarrow{OQ}=(acosx+bcos(x+φ_0），asinx+bsin(x+φ_0)).\\ &先看一个简单的特例：令a=b=1,φ_0=\frac{\pi}{2},则\notag\\ \end{align} $$

$$ \begin{align} &\overrightarrow{OQ}&=&(acosx+bcos(x+φ_0），asinx+bsin(x+φ_0))\notag\\ &&\overset{a=b=1}{\underset{x=\frac{\pi}{2}}{====}}&(cosx+cos(x+\frac{\pi}{2}),sinx+sin(x+\frac{\pi}{2})))\notag\\ &&=&(cosx-sinx,sinx+cosx) \end{align} $$

$$ \begin{aligned}
&注意：\overrightarrow{OP}就是a(cosx,sinx)\notag\\ &\qquad\overrightarrow{OR}就是b(1,1)\notag\\ &\textcolor{red}{由向量的数量积坐标运算法则：}\\ &\textcolor{red}{(x_1,y_1)\centerdot(x_2,y_2)=x_1x_2+y_1y_2}\\ &可知 \\ \end{aligned} $$

$$ \begin{align} &&(cosx,sinx)\centerdot(1,1)\qquad\notag\\ &=&cosx\centerdot 1 + sinx \centerdot 1\qquad\notag\\ &=&cosx + sinx\qquad\qquad \end{align} $$

$$ \begin{align} &\textcolor{red}{另外根据数量积公式:}\notag\\ &\textcolor{red}{a\centerdot b=|a|\centerdot |b|\centerdot cos\theta}\qquad\qquad\notag\\ &得：\notag\\ \end{align} $$

$$ \begin{align} &(cosx,sinx)\centerdot(1,1)\notag\\ &=\sqrt{cos^2{x}+sin^2{x}}\centerdot\sqrt{1^2+1^2}\centerdot cos\theta\notag\\ &=\sqrt{2}\centerdot cos\theta \qquad\qquad\qquad\qquad\quad\quad\\ &\theta为\overrightarrow{OP}和\overrightarrow{OR}的夹角：\notag\\ &图中看出：\theta=x-\frac{\pi}{4} \\ &然后根据(4)和(5)可以得出:\notag\\ &cosx+sinx=\sqrt{2}cos(x-\frac{\pi}{4}) \end{align} $$