What are some useful tricks/shortcuts for verifying trigonometric identities? What "tricks" are there that could help verify trigonometric identities? 
For example one is:
$$a\cos\theta+b\sin\theta = \sqrt{a^2+b^2}\,\cos(\theta-\phi)$$        
 A: Note that $\cos(x-y)=\cos x\cos y+\sin x\sin y$.  This is obtained from the more familiar formula for $\cos(x+y)$ by replacing $y$ by $-y$. 
Note also that 
$$a\cos\theta+b\sin\theta=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos\theta+   \frac{b}{\sqrt{a^2+b^2}}\sin\theta   \right).$$
So if $\phi$ is the angle whose cosine is $\frac{a}{\sqrt{a^2+b^2}}$ and whose sine is $\frac{b}{\sqrt{a^2+b^2}}$, then in the formula above we can replace  $\frac{a}{\sqrt{a^2+b^2}}$ by $\cos\phi$, and  $\frac{b}{\sqrt{a^2+b^2}}$ with $\sin\phi$, and obtain
$$a\cos\theta+b\sin\theta=\sqrt{a^2+b^2} \cos(\theta-\phi).$$
Remark: As to tricks and shortcuts, mostly it is a question of experience and practice. Already, I am sure, you recognize certain patterns and know how to exploit them. After a while, you will have used most of the common devices a dozen times, and then things get easy.  
A: Expand the cosine of the difference of angles on the right.
$$ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi$$
Now, collect terms so that the expression is a linear combination of $\sin \theta$ and $\cos \theta$, as in the expression on the left.
$$ \begin{align}
\sqrt{a^2 + b^2}\cos(\theta - \phi) &= \sqrt{a^2 + b^2}(\cos \theta \cos \phi + \sin \theta \sin \phi)\\
&=\sqrt{a^2 + b^2}\cos \phi \cdot \cos \theta + \sqrt{a^2 + b^2}\sin \phi \cdot \sin \theta\\
&=a \cos \theta + b \sin \theta
\end{align}$$
This last equality holds as long as we choose $\phi$ such that
$$ \left\{\begin{align} \cos \phi &= \frac{a}{\sqrt{a^2 + b^2}}\\ \sin \phi &=  \frac{b}{\sqrt{a^2 + b^2}} \end{align}\right.$$
This is always possible since those are the coordinates of a point on the unit circle.
A: The best trig trig for verifying identities is to know your rules for exponents, and know that
$$cos(x) = \frac {e^{ix} + e^{-ix}} 2$$
$$sin(x) = \frac {e^{ix} - e^{-ix}} 2$$
Most trivial trig identities follow easily from this.  It doesn't help with this example, because this is more of a "solve for phi" problem than a "verify if this is correct" problem.  But when it comes to just verifying an identity, Euler's formula makes it easy.
