How to derive cosine difference formula?

Question

I was asked on a take home quiz to "Derive the Cosine Difference Formula". I've looked for an hour but I've gotten different results this is kind of my last chance.

Answer 1

And here is a totally inappropriate way.

$e^{i(a-b)} = \cos(a-b)+i\sin(a-b) $ and

$\begin{array}\\ e^{i(a-b)} &=e^{i(a)}e^{i(-b)}\\ &=(\cos(a)+i\sin(a))(\cos(-b)+i\sin(-b))\\ &=(\cos(a)+i\sin(a))(-i\sin(b))\\ &=\cos(a)\cos(b)+\sin(a)\sin(b)+i(\sin(a)\cos(b)-\cos(a)\sin(b))\\ \end{array} $

Equating real and imaginary parts, $\cos(a-b) =\cos(a)\cos(b)+\sin(a)\sin(b) $ and $\sin(a-b) =\sin(a)\cos(b)-\cos(a)\sin(b) $.

As is often the case, there is absolutely nothing original here.

Answer 2

Here's an easy way: take two angles $A$ and $B$ and assume $B\geq A$

Look at the unit circle and take the following two points on it:

$(\cos A,\sin A)$ and $(\cos (B, \sin B)$ so that $A$ and $B$ are angles swept out from the $x-$ axis to the line segments connecting the origin to the points $(\cos A, \sin A)$ and $(\cos B, \sin B)$, resp., on the circle. . Now, compute the distance between these two points.

Next, rotate the line segment connecting the two points through an angle $-A$, so that the new points on the circle are $(1, 0)$ and $(\cos (B-A),\sin (B-A))$. Now, compute the distance between these two points.

The two distances you computed must be the same, so equate them. The formula will follow directly.

Answer 3

Here is a very different approach that verifies the trig formulas if either are true.

$$\frac d{dx}\cos(x+a)=\frac d{dx}\cos(x)\cos(a)-\sin(x)\sin(a)$$

Trying to differentiate each side individually, we get:

$$-\sin(x+a)=-\sin(x)\cos(a)-\cos(x)\sin(a)$$

Which simplifies down into the sum formula for sine.

Thus, one being true proves the other to be true.

From here, I would think that proving such definitions would be most easily done through the complex extension of our exponential function:

$$e^{\phi i}=\cos(\phi)+i\sin(\phi)$$

$$e^{(x+a)i}=\cos(x+a)+i\sin(x+a)$$

$$e^{(x+a)i}=e^{xi}e^{ai}=(\cos(x)+i\sin(x))(\cos(a)+i\sin(a))$$

Use exponent rules, then apply the complex exponential formula with $\phi=x,a$.

Now factor:

$$=\cos(x)\cos(a)+i\sin(x)\cos(a)+i\cos(x)\sin(a)-\sin(x)\sin(a)$$

Looking back, this must be equal to $\cos(x+a)+i\sin(x+a)$.

Since trig functions only produce real outputs for real inputs, then the real part of $\cos(x)\cos(a)+i\sin(x)\cos(a)+i\cos(x)\sin(a)-\sin(x)\sin(a)$ must be equal to the real part of $\cos(x+a)+i\sin(x+a)$. Similarly, the imaginary parts must be equal.

$$\cos(x+a)=\cos(x)\cos(a)-\sin(x)\sin(a)$$

$$\sin(x+a)=\sin(x)\cos(a)+\cos(x)\sin(a)$$

These are our trigonometric identities. Oddly, they also work for complex inputs, which must be proven using $\cos(\phi)=\frac{e^{\phi i}+e^{-\phi i}}2$ and $\sin(\phi)=\frac{e^{-\phi i}-e^{\phi i}}{2i}$. You can see that it works if you write it all out and multiply all of the stuff.

Another direction would to use the Taylor series of cosine and sine and binomial expansion, though I think it would be very messy.

EDIT

Another way is the inclusion of $\cos^2+\sin^2=1$.

Since the sum of angles formula for cosine must work if the sine formula also works, we will assume that they are true and attempt to use the pythagorean identity to simplify.

$$cos^2(x+a)=(\cos(x)\cos(a)-\sin(x)\sin(a))^2=\cos^2(x)\cos^2(a)-2\cos(x)\cos(a)\sin(x)\sin(a)+\sin^2(x)\sin^2(a)$$

$$\sin^2(x+a)=(\sin(x)\cos(a)+\cos(x)\sin(a))^2=\sin^2(x)\cos^2(a)+2\sin(x)\cos(a)\cos(x)\sin(a)+\cos^2(x)\sin^2(a)$$

Add the two and you get a true statement.