7.2: Addition and Subtraction Formulas

We know how to find $\sin(x)$: just find how far you rotate on the unit circle.

That $x$ value can be decomposed into a sum of two values $x = s + t$. Is there a way to find $\sin(x) = \sin(s + t)$ in terms of just $s$ and $t$ plugged in?

The answer is yes:

For sine: \begin{align} \sin(s + t) &= \sin s \cos t + \cos s \sin t\\[0.5em] \sin(s - t) &= \sin s \cos t - \cos s \sin t \end{align} For cosine: \begin{align} \cos(s + t) &= \cos s \cos t - \sin s \sin t \\[0.5em] \cos(s - t) &= \cos s \cos t + \sin s \sin t \end{align} For tangent: \begin{align} \tan(s + t) &= \frac{\tan s + \tan t}{1 - \tan s \tan t}\\[0.5em] \tan(s - t) &= \frac{\tan s - \tan t}{1 + \tan s \tan t} \end{align}

We will prove $\cos(s + t) = \cos s \cos t - \sin s\sin t$ in class.

Find \[\cos 75^\circ \qquad\qquad \cos \frac{\pi}{12} \qquad\qquad \sin20^\circ\cos40^\circ + \cos20^\circ \sin40^\circ\]
Prove \[\frac{1 + \tan x}{1 - \tan x} = \tan\left(\frac{\pi}{4} + x\right)\]
Prove \[\cos\left(\dfrac{\pi}{2} - u\right) = \sin u\]
Prove \[\cos(x + y)\cos(x - y) = \cos^2x - \sin^2y\]

Solution: The solution will be presented like I am solving this problem for the first time.

This will give you a sense of the experimentation process you need to use to solve harder proofs.

First, we need to choose a side. Let's choose the left hand side because it is low-hanging fruit. We can easily apply the additon and subtraction identities.

We have

\begin{align} \text{LHS} &= \cos(x + y)\cos(x - y) \\&= \bigg(\cos(x)\cos(y) - \sin(x)\sin(y)\bigg) \cdot \bigg(\cos(x)\cos(y) + \sin(x)\sin(y)\bigg) \\&= \cos^2(x)\cos^2(y) + \cos(x)\cos(y)\sin(x)\sin(y) - \cos(x)\cos(y)\sin(x)\sin(y) - \sin^2(x)\sin^2(y) \\&= \cos^2(x)\cos^2(y) - \sin^2(x)\sin^2(y) \end{align} Attempt 1:

Now look at the RHS. We need one $\cos^2(x)$ and one $\sin^2(y)$ but in our calculation we have an extra factor $\cos^2(y)$ and extra factor of $\sin^2(x)$.

Let's add zero, and transform the zero into two factors $\sin^2(y)\cos^2(x)$ and use grouping factorization:

\begin{align} \text{LHS} &= \cos^2(x)\cos^2(y) - \sin^2(x)\sin^2(y) \\&= \cos^2(x)\cos^2(y) + {\color{pink} 0} - \sin^2(x)\sin^2(y) \\&= \cos^2(x)\cos^2(y)\ {\color{pink}- \sin^2(y)\cos^2(x) + \sin^2(y)\cos^2(x)} - \sin^2(x)\sin^2(y) \\&= \bigg(\cos^2(x)\cos^2(y) - \sin^2(y)\cos^2(x)\bigg) + \bigg(\sin^2(y)\cos^2(x) - \sin^2(x)\sin^2(y)\bigg) \\&= \underbrace{\cos^2(x) \bigg(\cos^2(y) - \sin^2(y)\bigg)}_{\text{global term}} + \underbrace{\sin^2(y)\bigg(\cos^2(x) - \sin^2(x)\bigg)}_{\text{global term}} \end{align}

Wait..We are stuck because even though we have two global terms, they don't share any common factors. This train of thought seems like a dud.

Attempt 2:

Instead of adding and subtracting $\sin^2(y)\cos^2(x)$, let's swap the positions of $x$ and $y$ and experiment.

Let's add and subtract $\sin^2(x)\cos^2(y)$ and see if we can make more progress.

\begin{align} \text{LHS} &= \cos^2(x)\cos^2(y) - \sin^2(x)\sin^2(y) \\&= \cos^2(x)\cos^2(y) + {\color{pink} 0} - \sin^2(x)\sin^2(y) \\&= \cos^2(x)\cos^2(y) \ {\color{pink}+ \sin^2(x)\cos^2(y) - \sin^2(x)\cos^2(y)} - \sin^2(x)\sin^2(y) \\&= \bigg(\cos^2(x)\cos^2(y) + \sin^2(x)\cos^2(y)\bigg) + \bigg(-\sin^2(x)\cos^2(y) - \sin^2(x)\sin^2(y) \bigg) \\&= \cos^2(y) \bigg(\cos^2(x) + \sin^2(x)\bigg) -\sin^2(x)\bigg(\cos^2(y) + \sin^2(y)\bigg) \\&= \cos^2(y) \cdot 1 - \sin^2(x) \cdot 1 \\&= \cos^2 y - \sin^2 x \end{align}

The RHS is $\cos^2 x - \sin^2 y$ but in our solution the $x$ and $y$ are in the wrong place.

So we should have stuck to our initial strategy of adding and subtracting $\sin^2(y)\cos^2(x)$ because this should, in theory, swap our $x$ and $y$ in the solution.

But we tried it already...think of the possible differences you can make in Attempt 1.

Attempt 3:

The only thing you can change in Attempt 1 is to swap the added and subtracted terms. Add first, then subtract. Let's see if this works:

\begin{align} \text{LHS} &= \cos^2(x)\cos^2(y) - \sin^2(x)\sin^2(y) \\&= \cos^2(x)\cos^2(y) + {\color{pink} 0} - \sin^2(x)\sin^2(y) \\&= \cos^2(x)\cos^2(y)\ {\color{pink}+ \sin^2(y)\cos^2(x) - \sin^2(y)\cos^2(x)} - \sin^2(x)\sin^2(y) \\&= \bigg(\cos^2(x)\cos^2(y) + \sin^2(y)\cos^2(x)\bigg) + \bigg(-\sin^2(y)\cos^2(x) - \sin^2(x)\sin^2(y)\bigg) \\&= \cos^2(x) \bigg(\cos^2(y) + \sin^2(y)\bigg) - \sin^2(y)\bigg(\cos^2(x) + \sin^2(x)\bigg) \\&= \cos^2(x) \cdot 1 - \sin^2(y)\cdot 1 \\&= \cos^2 x - \sin^2 y \\&= \text{RHS} \end{align}

and we have proved the identity. $\blacksquare$

This is the type of experimentation you will need for tougher problems.

Try, fail, think of a new idea, try again, and repeat.