测试md

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

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

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

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

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

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

$$

\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}

$$