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# Sum and Difference angle formulas

Alignments to Content Standards: F-TF.C.9

In this task, you will show how all of the sum and difference angle formulas can be derived from a single formula when combined with relations you have already learned.

For the following task, assume that the sum angle formula for sine is true. Namely, $$\sin(\theta+\phi)= \sin\theta \cos\phi +\cos\theta \sin\phi.$$

1. To derive the difference angle formula for sine, write $\sin(\theta-\phi)$ as $\sin(\theta+(-\phi))$ and apply the sum angle formula for sine to the angles $\theta$ and $-\phi$. Use the fact that sine is an odd function while cosine is even function to simplify your answer. Conclude that $$\sin(\theta-\phi) = \sin(\theta) \cos(\phi)-\cos(\theta)\sin(\phi).$$
2. To derive the sum angle formula for cosine, use what what you learned in (a) to show that $$\cos(\theta+\phi)= \cos\theta \cos\phi -\sin\theta \sin\phi.$$ You may want to start with an exploration of $\sin\left(\frac{\pi}{2} -(\theta+\phi)\right)$.
3. Derive the difference angle formula for cosine, $$\cos(\theta-\phi)= \cos\theta \cos\phi +\sin\theta \sin\phi.$$
4. Derive the sum angle formula for tangent, $$\tan(\theta+\phi) = \frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}.$$
5. Derive the difference angle formula for tangent, $$\tan(\theta-\phi) = \frac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}.$$

## IM Commentary

The goal of this task is to have students derive the addition and subtraction formulas for cosine and tangent, and the subtraction formula for cosine, from the sum formula for sine. The task provides varying levels of scaffolding, pointing out possible relations to use early on, but leaving more creative work for the student later. In addition, the task assumes the sum angle formula for sine and shows how the other sum and difference formulas must follow.

This text of this problem and its solution assumes familiarity with the Greek letters theta $(\theta)$ and phi $(\phi)$. However, some teachers or books will use alpha $(\alpha)$ and beta $(\beta)$. Still others use Latin letters like $u$ and $v$ or $A$ and $B$. Instructors should feel free to change the letters to match those of their source, as the choice of letters is not important; it is the relationships the letters represent that are useful.

Before embarking on this task, students should be aware that sine is odd (hence $\sin(-\theta)=-\sin(\theta))$ and cosine is even (hence $\cos(-\theta)=\cos(\theta))$ as found in standard F-TF.4. Students should know the relations between sine, cosine and tangent found in standard G-SRT.6. In addition, students should know the relation between trigonometric values of "complementary" angles found in standard G-SRT.7, ($\sin(\theta) = \cos(\pi/2-\theta)$, etc.).

The emphasis of this task is to show how one result can be extended to a family of results using known relations. This is a central strategy in mathematical thinking, and illustrates Standards for Mathematical Practices 7 and 8, looking for structure and making use of repeated reasonin. Along these lines, solutions other than the ones given here are also viable -- for example, students might prove part (e) from part (d) by making the "substitution" $\phi\to -\phi$ (perhaps not in that language).

## Solution

We assume the sum angle formula for sine is correct: $$\sin(\theta+\phi)= \sin\theta \cos\phi +\cos\theta \sin\phi.$$

1. To derive the difference angle formula for sine, we write $\sin(\theta-\phi)$ as $\sin(\theta+(-\phi))$ to get \begin{eqnarray*} \sin(\theta-\phi)&=& \sin(\theta+(-\phi))\\ &=& \sin\theta \cos(-\phi) +\cos\theta \sin(-\phi) \end{eqnarray*} Since cosine is even and sine is odd, we can simplify $\cos(-\phi)=\cos(\phi)$ and $\sin(-\phi)=-\sin(\phi)$. Substituting, we have \begin{eqnarray*} \sin(\theta-\phi)&=& \sin\theta \cos(-\phi) +\cos\theta \sin(-\phi)\\ &=& \sin\theta\cos\phi - \cos\theta\sin\phi \end{eqnarray*} as desired. That is, $$\sin(\theta-\phi) = \sin(\theta) \cos(\phi)-\cos(\theta)\sin(\phi).$$
2. To derive the sum angle formula for cosine, use the angle relations $\cos(A) = \sin\left(\frac{\pi}{2}-A\right)$ and $\sin(A)=\cos\left(\frac{\pi}{2}-A\right)$ for any angle $A$. We begin by observing that we can write $\cos(\theta+\phi)$ as $\sin(\pi/2-(\theta+\phi))$. Moreover, we can write $$\sin\left(\frac{\pi}{2}-(\theta+\phi)\right) = \sin\left(\left(\frac{\pi}{2}-\theta\right)-\phi\right)$$ Putting these equalities together we have $$\cos(\theta+\phi) = \sin\left(\left(\frac{\pi}{2}-\theta\right)-\phi\right)$$ Using the difference angle formula for sine we have \begin{eqnarray*} \cos(\theta+\phi) &=& \sin\left(\left(\frac{\pi}{2}-\theta\right)-\phi\right)\\ &=&\sin\left(\frac{\pi}{2} - \theta\right)\cos\phi-\cos\left(\frac{\pi}{2}-\theta\right)\sin\phi\\ &=& \cos\theta\cos\phi-\sin\theta\sin\phi. \end{eqnarray*} We conclude that $$\cos(\theta+\phi)= \cos\theta \cos\phi -\sin\theta \sin\phi.$$
3. To derive the difference angle formula for cosine, we apply the strategy from (a) and write \begin{eqnarray*} \cos(\theta-\phi)&=&\cos(\theta+(-\phi))\\ &=& \cos\theta \cos(-\phi) -\sin\theta \sin(-\phi). \end{eqnarray*} Using the fact that sine is odd and cosine is even we reduce further to get \begin{eqnarray*} \cos(\theta-\phi)&=& \cos\theta \cos(-\phi) -\sin\theta \sin(-\phi)\\ &=& \cos\theta\cos\phi+\sin\theta\sin\phi \end{eqnarray*} We conclude that $$\cos(\theta-\phi)= \cos\theta \cos\phi +\sin\theta \sin\phi.$$
4. To derive the sum angle formula for tangent, we can write $$\tan(\theta+\phi) = \frac{\sin(\theta+\phi)}{\cos(\theta+\phi)}$$ Applying the sum angles for sine and cosine, we have $$\tan(\theta+\phi) = \frac{\sin\theta \cos\phi +\cos\theta \sin\phi}{\cos\theta \cos\phi -\sin\theta \sin\phi}$$ In order get tangent into the right hand side, we can substitute the equations $\sin\theta=\tan\theta\cos\theta$ and $\sin\phi=\tan\phi \cos\phi$. With this substitution, we have \begin{eqnarray*} \frac{\sin\theta \cos\phi +\cos\theta \sin\phi}{\cos\theta \cos\phi -\sin\theta \sin\phi} &=& \frac{\tan\theta\cos\theta \cos\phi +\cos\theta \tan\phi \cos\phi}{\cos\theta \cos\phi -\tan\theta\cos\theta \tan\phi \cos\phi}\\ &=& \frac{(\cos\theta\cos\phi)(\tan\theta+\tan\phi)}{(\cos\theta\cos\phi)(1-\tan\theta\tan\phi)}\\ &=& \frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi} \end{eqnarray*} In the last two lines, we factored the numerator and denominator and the canceled the common factors. We conclude that $$\tan(\theta+\phi) = \frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}.$$ As an alternate route, at the point of $$\tan(\theta+\phi) = \frac{\sin\theta \cos\phi +\cos\theta \sin\phi}{\cos\theta \cos\phi -\sin\theta \sin\phi}$$ we could divide each term on the right hand side by $\cos\theta\cos\phi$ to achieve the same result.
5. To derive the difference angle formula for tangent, we apply the sum angle formula for tangent to $\theta+(-\phi)$. \begin{eqnarray*} \tan(\theta-\phi) &=& \tan(\theta+(-\phi))\\ &=& \frac{\tan\theta+\tan(-\phi)}{1-\tan\theta\tan(-\phi)}. \end{eqnarray*} Since tangent is odd, we can simplify $\tan(-\phi)=-\tan\phi$. With this substitution, we have \begin{eqnarray*} \tan(\theta-\phi) &=& \frac{\tan\theta+\tan(-\phi)}{1-\tan\theta\tan(-\phi)}\\ &=& \frac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi} \end{eqnarray*} We conclude that $$\tan(\theta-\phi) = \frac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}.$$