Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $ Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $
Can someone please explain to me how to go about doing this? 
 A: you can write $$\sqrt 3 \sin t - \cos t = 2\left(\frac{\sqrt 3}2 \sin t - \frac 12 \cos t\right) = 2\left(\sin t\sin \left(\frac{\pi}3 \right) - \cos\left(\frac {\pi} 3\right) \cos t\right)= -2\cos \left(t + \frac{\pi}3\right) = 2 \cos \left(t - \frac{2\pi}3\right) $$
A: The moment you see $a$ and $b$ as coefficients of $\cos \theta , \sin \theta$,multiply and divide by $\sqrt { a^2 + b^2} $ so that the latter can be absorbed into $\theta $ additively. Now since $ \sqrt{ 3 +1} =2, $
$$\sqrt 3 \sin t - \cos t = 2\left(\sin t \frac{\sqrt 3}2  -  \cos t \;\frac 12\right) =  2 \sin ( t - \pi/6).$$
A: If 
\begin{align*}
\cos\varphi & = \frac{b}{\sqrt{a^2 + b^2}}\\
\sin\varphi & = \frac{a}{\sqrt{a^2 + b^2}}
\end{align*}
then 
$$a\sin\theta - b\cos\theta = -\sqrt{a^2 + b^2}\cos(\theta + \varphi)$$
since 
\begin{align*}
-\sqrt{a^2 + b^2}\cos(\theta + \varphi) & = -\sqrt{a^2 + b^2}(\cos\theta\cos\varphi - \sin\theta\sin\varphi)\\
& = -\sqrt{a^2 + b^2}\left(\cos\theta \cdot \frac{b}{\sqrt{a^2 + b^2}} - \sin\theta \cdot \frac{a}{\sqrt{a^2 + b^2}}\right)\\
& = -b\cos\theta + a\sin\theta\\
& = a\sin\theta - b\cos\theta
\end{align*}
In the expression, $\sqrt{3}\sin\theta - \cos\theta$, $a = \sqrt{3}$ and $b = 1$, so $$a^2 + b^2 = \sqrt{3}^2 + 1^2 = 3 + 1 = 4$$  Thus, $\sqrt{a^2 + b^2} = \sqrt{4} = 2$.  Hence, 
\begin{align*}
\cos\varphi & = \frac{b}{\sqrt{a^2 + b^2}} = \frac{1}{2}\\
\sin\varphi & = \frac{a}{\sqrt{a^2 + b^2}} = \frac{\sqrt{3}}{2}
\end{align*}
Therefore, 
$$\varphi = \arccos\left(\frac{1}{2}\right) = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$$
so we obtain 
$$\sqrt{3}\sin\theta - \cos\theta = -2\cos\left(\theta + \frac{\pi}{3}\right)$$
Note:  We could also express the answer in the form 
$$\sqrt{3}\sin\theta - \cos\theta = \sin\left(\theta - \frac{\pi}{6}\right)$$
if we use the formula
$$a\sin\theta - b\cos\theta = \sqrt{a^2 + b^2}\sin(\theta - \varphi)$$
where 
\begin{align*}
\cos\varphi & = \frac{a}{\sqrt{a^2 + b^2}}\\
\sin\varphi & = \frac{b}{\sqrt{a^2 + b^2}}
\end{align*}
with $a = \sqrt{3}$, $b = 1$, and 
$$\varphi = \arcsin\left(\frac{1}{2}\right) = \arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$$
A: You have to get an expression in the form of $a \space cos \space(\theta + \alpha)$ which can be expanded to $a \space cos\space\theta \space cos\space\alpha - a \space sin\space\theta\space\sin\space\alpha $. That is the form you need to get. Now look at what you have on the left:
$\sqrt3\space sin\space\theta \space - \cos\space\theta$ ...now does it look like the expansion of the cosine above? Yes! Only the expressions are in the wrong order.
Rearranging:
$- \cos\space\theta + \sqrt3\space sin\space\theta = -1.\space\cos\theta + \sqrt3\space sin\space\theta $
By comparison:
$-a\space cos\alpha = 1 \space$ and, $\space -a \space sin\space\alpha = \sqrt3$
Dividing the sine expression by the cos expression give you:
$tan\space\alpha = \sqrt3 \implies \alpha = \pi/3, ...$
Substituting this values in the sine expression gives you:
$-a\space sin (\pi/3) = \sqrt 3 \implies a = -\frac{\sqrt 3}{\sqrt 3/2} = -2$
Hence, your $\alpha$ could be $\pi/3$ and your a is -2.
A: $% Predefined Typography
 \newcommand{\paren} [1]{\left({#1}\right)}
 \newcommand{\polar}[2] {\left(#1 ~\angle~ #2\right)}
$

$$\sqrt{3}\sin(\theta) - \cos(\theta)$$

First write as $\cos() + \cos()$:
$$\sqrt{3}\sin(\theta) - \cos(\theta) = \sqrt{3}\cos\paren{\theta - \frac{\pi}2} + \cos(\theta + \pi)$$
Then add (polar coordinates) $\polar{\sqrt{3}}{-\frac{\pi}{2}} + \polar{1}{\pi}$ to get:
$$\polar{\sqrt{3}}{-\frac{\pi}{2}} + \polar{1}{\pi} = \polar{2}{-\frac{2\pi}{3}}$$
So
$$\sqrt{3}\cos\paren{\theta - \frac{\pi}2} + \cos(\theta + \pi) = 2\cos\paren{\theta - \frac{2\pi}{3}}$$
