What would be the sum of Trigonometric Functions

Question

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) ?

Answer 1

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

$$\sin x - \cos x = \sqrt2\sin\left(\theta - \frac{\pi}{4}\right)$$

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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 $a\sin\theta + b\cos\theta$.

Answer 2

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,

$$\sin(x)-\cos(x)=\sqrt{2}\sin(x-\pi/4)$$

Answer 3

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 $y=x$.