What would be the sum of Trigonometric Functions I was trying to remember the functions provided on the site: http://www.purplemath.com/modules/idents.htm#restatement
From there I came to know about some of the function, basically the Sum of the trigonometric functions. But I was only able to find the Sin(x) +/- Sin(y) and Cos(x) +/- Cos(y) functions.
I wanted to know, is it possible to use +/- operators on Sin(x) and Cos(x) such as: Sin(x) - Cos(x) ? 
 A: In general, we have (this method is known as the R-Alpha method):
$$a\sin\theta - b\cos\theta = \sqrt{a^2 + b^2}\sin\left(\theta - \tan^{-1}\frac{b}{a}\right)$$
For your case, we get
$$\sin x - \cos x = \sqrt2\sin\left(\theta - \frac{\pi}{4}\right)$$

Here's the proof: Let
$$a\sin\theta - b\cos\theta = R\sin(\theta - \alpha)$$
for some real number $R, \alpha$.
By using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we have
$$a{\color{red}\sin}{\color{red}\theta} - b{\color{blue}\cos}{\color{blue}\theta} = R{\color{red}\sin}{\color{red} \theta} \cos \alpha - R{\color{blue}\cos}{\color{blue} \theta} \sin \alpha$$
Equating the coefficients of ${\color{red}\sin}{\color{red}\theta}$ and ${\color{blue}\cos}{\color{blue}\theta}$, we have:
$$a = R\cos\alpha$$
$$b = R\sin\alpha$$
Divide both equations to get:
$$\tan\alpha = \frac{b}{a} \implies \alpha = \tan^{-1}\frac{b}{a}$$
Now, to find $R$, we take $a^2 + b^2$:
$$\begin{align}a^2 + b^2 &= (R\cos\alpha)^2 + (R\sin\alpha)^2 \\&= R^2(\cos^2\alpha + \sin^2\alpha) \\&= R^2\end{align}$$
so we have: $$R = \sqrt{a^2 + b^2}$$
To conclude: 
$$a\sin\theta - b\cos\theta = \sqrt{a^2 + b^2}\sin\left(\theta - \tan^{-1}\frac{b}{a}\right)$$
In fact, you can use the same approach for $a\sin\theta + b\cos\theta$.
A: Yes it is possible, but it is quite difficult to make an exhaustive list. You need to use the standard sum of angles identities. For your example:
$$\sin(x)-\cos(x)$$
We know the identity $$\sin(x-y)=\sin(x)\cos(y)-\cos(x)\sin(y)$$
If we let, $y=\pi/4$
$$\sin(x-\pi/4)=\sin(x)\frac{1}{\sqrt{2}}-\cos(x)\frac{1}{\sqrt{2}}$$
Or,
$$\sin(x)-\cos(x)=\sqrt{2}\sin(x-\pi/4)$$
A: You can use $cos(y)=sin(\frac{\pi}{2}-y)$, so that
$sin(x)+cos(y)$
$=sin(x)+sin(\frac{\pi}{2}-y)$
$=2.sin(\frac{x+\frac{\pi}{2}-y}{2}).cos(\frac{x-\frac{\pi}{2}+y}{2})
$
$=2.sin(\frac{x-y}{2}+\frac{\pi}{4}).cos(\frac{x+y}{2}-\frac{\pi}{4})$
and so on. Of course, you can let $y=x$.
