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## March 15, 2021

### Sum and Difference Formulas

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Textbook Reference: Section 7.2 Sum and Difference Identities

How many times have you made this mistake? $(a+b)^2=a^2+b^2$ I'm pretty sure that everyone who's ever learned math has made this mistake.

I mean, it's so tempting! The square of the sum is the sum of the squares, right?

Except it doesn't hold up to scrutiny if you plug in numbers, like $$a=3$$ and $$b=4$$: \begin{align*} (3+4)^2&\overset{?}=3^2+4^2\\ 49&\neq 25 \end{align*} If it's so easy to see why this formula isn't true, why is it so alluring?

It's because we're spoiled by the Distributive Property: $c(a+b)=ca+cb$ That statement is true. So, being the pattern-seekers we are, we want to believe that any statement that looks like this is true.

But the problem is that many people think of the Distributive Property at first as being about how parentheses relate to addition, when it's really about how multiplication relates to addition.

In fact, given some function $$f$$, it's almost never true that $$f(a+b)=f(a)+f(b)$$ in general.

Instead, each function has its own way of dealing with sums. Here are just a few common algebra mistakes and how to correct them: \begin{align*} &\underline{\text{Incorrect}} &&\underline{\text{Correct}}\\[1.5 ex] (a+b)^2&\neq a^2+b^2 & (a+b)^2&=a^2+2ab+b^2\\[1.5 ex] \dfrac{1}{a+b}&\neq \dfrac1a+\dfrac1b & \dfrac{a+b}{ab}&=\dfrac1a+\dfrac1b\\[1.5 ex] e^{a+b}&\neq e^a+e^b & e^{a+b}&=e^a\cdot e^b\\[1.5 ex] \ln(a+b)&\neq\ln(a)+\ln(b) & \ln(a\cdot b)&=\ln(a)+\ln(b) \end{align*}

## Sines and cosines of sums

So now we need to answer the inevitable question:

What are $$\sin(\alpha+\beta)$$ and $$\cos(\alpha+\beta)$$?

How do you find the sine and cosine of a sum of two angles?

Let's jump right to the answer: \begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\\ \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\\ \end{align*} Whoa. Where'd those come from?

There are a number of different ways to get these formulas; if you're curious to see a particularly elegent one (and I know you are now even if you weren't before), check out the Proof Box below.

### Proof of the sine and cosine sum formulas

Here you go:

That's it. Seriously.

This is what mathematicians usually call a "proof without words." The idea is that the pictures lay out all the relationships necessary in the proof.

What you should do is look at both diagrams and see how this proves the formulas. Ask yourself, why is the diagram set up that way? Where do all the measurements come from?

(Hint: Start with the $$1$$, and use a bunch of SOH-CAH-TOA.)

Let's use this to find where $$75^\circ$$ lands on the unit circle. (Did you ever notice there was a suspicious gap there between $$60^\circ$$ and $$90^\circ$$?)

Since $$75^\circ=30^\circ+45^\circ$$, we can use $$\alpha=30^\circ$$ and $$\beta=45^\circ$$ in our formulas to find the sine and cosine: \begin{align*} \sin(75^\circ) &= \sin(30^\circ)\cos(45^\circ)+\cos(30^\circ)\sin(45^\circ)\\ &= \dfrac12\cdot\dfrac{\sqrt2}2+\dfrac{\sqrt3}2\cdot\dfrac{\sqrt2}2\\ &= \dfrac{1\cdot\sqrt2}4+\dfrac{\sqrt3\cdot\sqrt2}4\\ &= \dfrac{\sqrt2+\sqrt6}4 \end{align*}\\ \begin{align*} \cos(75^\circ) &= \cos(30^\circ)\cos(45^\circ)+\sin(30^\circ)\sin(45^\circ)\\ &= \dfrac{\sqrt3}2\cdot\dfrac{\sqrt2}2-\dfrac{1}2\cdot\dfrac{\sqrt2}2\\ &= \dfrac{\sqrt3\cdot\sqrt2}4-\dfrac{1\cdot\sqrt2}4\\ &= \dfrac{\sqrt6-\sqrt2}4 \end{align*} So the $$75^\circ$$ lands at the point $$\left(\dfrac{\sqrt6-\sqrt2}{4},\dfrac{\sqrt6+\sqrt2}{4}\right)$$, which is about $$(0.966,0.259)$$.

## Other derived formulas

We have similar formulas for how to deal with the difference of two angles: \begin{align*} \sin(\alpha-\beta) &= \sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)\\ \cos(\alpha-\beta) &= \cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)\\ \end{align*} These formulas are so similar to the sum formulas above that we often just write them together: \begin{align*} \sin(\alpha\pm\beta) &= \sin(\alpha)\cos(\beta)\pm\cos(\alpha)\sin(\beta)\\ \cos(\alpha\pm\beta) &= \cos(\alpha)\cos(\beta)\mp\sin(\alpha)\sin(\beta)\\ \end{align*} Note the "$$\mp$$" in the last formula! This means that the sign used in this part of the formula is the opposite of the one at the beginning.

You might be thinking, if we've got formulas for sine and cosine, what about for tangent?

You're in luck — there are formulas for that too! \begin{align*} \tan(\alpha+\beta) &= \dfrac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}\\ \tan(\alpha-\beta) &= \dfrac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)} \end{align*} Or again we can write these as one formula if we'd like: $\tan(\alpha\pm\beta) = \dfrac{\tan(\alpha)\pm\tan(\beta)}{1\mp\tan(\alpha)\tan(\beta)}$ In the questions below, you'll show where most of these formulas come from.

# Preview Activity 15

1. Use the difference formulas to show that $$15^\circ$$ falls at $$\left(\dfrac{\sqrt6+\sqrt2}4,\dfrac{\sqrt6-\sqrt2}4\right)$$ on the unit circle.
2. Find the coordinates of $$195^\circ$$ on the unit circle.
3. Prove the formula $\sin(\alpha-\beta) = \sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)$ by starting with the formula for $$\sin(\alpha+\beta)$$, replacing $$\beta$$ with $$-\beta$$ throughout, and using the even/odd properties of sine and cosine.
4. Prove the formula $\tan(\alpha+\beta) = \dfrac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ as follows:
1. Rewrite $$\tan(\alpha+\beta)$$ using a quotient identity.
2. Use the sum identities in the numerator and denominator.
3. Multiply the entire numerator by $$\dfrac{1}{\cos(\alpha)\cos(\beta)}$$, and do the same in the denominator. Distribute.
4. Simplify the expression using quotient identities again.
5. Answer AT LEAST one of the following questions:
2. What was an "a-ha" moment you had while doing this activity?
3. What was the muddiest point of this activity for you?
4. What question(s) do you have about anything you've done?