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# 3.2: Sum and Difference Formulas

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We will now derive identities for the trigonometric functions of the sum and difference of two angles. For the sum of any two angles $$A$$ and $$B$$, we have the addition formulas:

$\sin\;(A+B) ~=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B\label{eqn:sumsin}$

$\cos\;(A+B) ~=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B\label{eqn:sumcos}$

To prove these, first assume that $$A$$ and $$B$$ are acute angles. Then $$A+B$$ is either acute or obtuse, as in Figure 3.2.1. Note in both cases that $$\angle\,QPR = A$$, since

\nonumber \begin{align*} \angle\,QPR ~&=~ \angle\,QPO - \angle\,OPM ~=~ (90^\circ - B) - (90^\circ - (A+B)) ~=~ A ~~\text{in Figure 3.2.1(a), and}\\[4pt] \nonumber \angle\,QPR ~&=~ \angle\,QPO + \angle\,OPM ~=~ (90^\circ - B) + (90^\circ - (180^\circ - (A+B))) ~=~ A ~~\text{in Figure 3.2.1(b).} \end{align*}

Thus,

\nonumber \begin{align} \sin\;(A+B) ~&=~ \frac{MP}{OP} ~=~ \frac{MR+RP}{OP} ~=~ \frac{NQ+RP}{OP} ~=~ \frac{NQ}{OP} ~+~ \frac{RP}{OP} \\[4pt] \nonumber &=~ \frac{NQ}{OQ}\,\cdot\,\frac{OQ}{OP} ~+~ \frac{RP}{PQ}\,\cdot\,\frac{PQ}{OP}\notag\\[4pt] &=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B ~,\label{eqn:sinsumproof} \end{align}

and

\nonumber \begin{align} \cos\;(A+B) ~&=~ \frac{OM}{OP} ~=~ \frac{ON-MN}{OP} ~=~ \frac{ON-RQ}{OP} ~=~ \frac{ON}{OP} ~-~ \frac{RQ}{OP} \\[4pt] \nonumber &=~ \frac{ON}{OQ}\,\cdot\,\frac{OQ}{OP} ~-~ \frac{RQ}{PQ}\,\cdot\,\frac{PQ}{OP} \\[4pt] &=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B ~.\label{eqn:cossumproof} \end{align}

So we have proved the identities for acute angles $$A$$ and $$B$$. It is simple to verify that they hold in the special case of $$A=B=0^\circ$$. For general angles, we will need to use the relations we derived in Section 1.5 which involve adding or subtracting $$90^\circ$$:

\nonumber \begin{alignat*}{4} \sin\;(\theta + 90^\circ) ~ &= ~ \phantom{-}\cos\;\theta &\qquad\quad \sin\;(\theta - 90^\circ) ~ &= ~ -\cos\;\theta\\[4pt] \nonumber \cos\;(\theta + 90^\circ) ~ &= ~ -\sin\;\theta &\qquad\quad \cos\;(\theta - 90^\circ) ~ &= ~ \phantom{-}\sin\;\theta \end{alignat*}

These will be useful because any angle can be written as the sum of an acute angle (or $$0^\circ$$) and integer multiples of $$\pm 90^\circ$$. For example, $$155^\circ = 65^\circ + 90^\circ$$, $$222^\circ = 42^\circ + 2(90^\circ)$$, $$-77^\circ = 13^\circ - 90^\circ$$, etc. So if we can prove that the identities hold when adding or subtracting $$90^\circ$$ to or from either $$A$$ or $$B$$, respectively, where $$A$$ and $$B$$ are acute or $$0^\circ$$, then the identities will also hold when repeatedly adding or subtracting $$90^\circ$$, and hence will hold for all angles. Replacing $$A$$ by $$A+90^\circ$$ and using the relations for adding $$90^\circ$$ gives

\nonumber \begin{align*} \sin\;((A+90^\circ) + B) ~&=~ \sin\;((A+B) + 90^\circ) ~=~ \cos\;(A+B)~,\\[4pt] \nonumber &=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B ~~\text{(by Equation \ref{eqn:cossumproof})}\\[4pt] \nonumber &=~ \sin\;(A + 90^\circ)~\cos\;B ~+~ \cos\;(A + 90^\circ)~\sin\;B ~,\end{align*}
so the identity holds for $$A+90^\circ$$ and $$B$$ (and, similarly, for $$A$$ and $$B+90^\circ$$). Likewise,

\nonumber \begin{align*} \sin\;((A-90^\circ) + B) ~&=~ \sin\;((A+B) - 90^\circ) ~=~ -\cos\;(A+B)~,\\[4pt] \nonumber &=~ -(\cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B) \\[4pt] \nonumber &=~ (-\cos\;A) ~ \cos\;B ~+~ \sin\;A ~ \sin\;B\\[4pt] \nonumber &=~ \sin\;(A - 90^\circ)~\cos\;B ~+~ \cos\;(A - 90^\circ)~\sin\;B ~, \end{align*}

so the identity holds for $$A-90^\circ$$ and $$B$$ (and, similarly, for $$A$$ and $$B+90^\circ$$). Thus, the addition Equation \ref{eqn:sumsin} for sine holds for all $$A$$ and $$B$$. A similar argument shows that the addition Equation \ref{eqn:sumcos} for cosine is true for all $$A$$ and $$B$$. QED

Replacing $$B$$ by $$-B$$ in the addition formulas and using the relations $$\sin\;(-\theta) = -\sin\;\theta$$ and $$\cos\;(-\theta) = \cos\;\theta$$ from Section 1.5 gives us the subtraction formulas:

$\sin\;(A-B) ~=~ \sin\;A ~ \cos\;B ~-~ \cos\;A ~ \sin\;B\label{eqn:diffsin}$

$\cos\;(A-B) ~=~ \cos\;A ~ \cos\;B ~+~ \sin\;A ~ \sin\;B\label{eqn:diffcos}$

Using the identity $$\tan\;\theta = \frac{\sin\;\theta}{\cos\;\theta}$$, and the addition formulas for sine and cosine, we can derive the addition formula for tangent:

\require{cancel} \nonumber \begin{align*} \tan\;(A+B) ~&=~ \frac{\sin\;(A+B)}{\cos\;(A+B)}\\[4pt] \nonumber &=~ \frac{\sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B}{\cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B}\\[4pt] \nonumber &=~ \frac{\dfrac{\sin\;A ~ \cos\;B}{\cos\;A ~ \cos\;B} ~+~ \dfrac{\cos\;A ~ \sin\;B}{\cos\;A ~ \cos\;B}}{\dfrac{\cos\;A ~ \cos\;B}{\cos\;A ~ \cos\;B} ~-~ \dfrac{\sin\;A ~ \sin\;B}{\cos\;A ~ \cos\;B}}\quad\text{(divide top and bottom by $$\cos\;A ~ \cos\;B$$)}\\[4pt] \nonumber &=~ \frac{\dfrac{\sin\;A}{\cos\;A} \;\cdot\; \cancel{\dfrac{\cos\;B}{\cos\;B}} ~+~ \cancel{\dfrac{\cos\;A}{\cos\;A}} \;\cdot\; \dfrac{\sin\;B}{\cos\;B}}{1 ~-~ \dfrac{\sin\;A}{\cos\;A} \;\cdot\; \dfrac{\sin\;B}{\cos\;B}} ~=~ \frac{\tan\;A ~+~ \tan\;B}{1 ~-~ \tan\;A ~ \tan\;B} \end{align*}

This, combined with replacing $$B$$ by $$-B$$ and using the relation $$\tan\;(-\theta) = -\tan\;\theta$$, gives us the addition and subtraction formulas for tangent:

$\tan\;(A+B) ~=~ \frac{\tan\;A ~+~ \tan\;B}{1 ~-~ \tan\;A ~ \tan\;B}\label{eqn:sumtan}$

$\tan\;(A-B) ~=~ \frac{\tan\;A ~-~ \tan\;B}{1 ~+~ \tan\;A ~ \tan\;B}\label{eqn:difftan}$

Example 3.8

Given angles $$A$$ and $$B$$ such that $$\sin\;A = \frac{4}{5}$$, $$\cos\;A = \frac{3}{5}$$, $$\sin\;B = \frac{12}{13}$$, and $$\cos\;B = \frac{5}{13}$$, find the exact values of $$\sin\;(A+B)$$, $$\cos\;(A+B)$$, and $$\tan\;(A+B)$$.

Solution:

Using the addition formula for sine, we get:

\nonumber \begin{align*} \sin\;(A+B) ~&=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B\\[4pt] \nonumber &=~ \frac{4}{5} \;\cdot\; \frac{5}{13} ~+~ \frac{3}{5} \;\cdot\; \frac{12}{13} \quad\Rightarrow\quad \boxed{\sin\;(A+B) ~=~ \frac{56}{65}} \end{align*}

Using the addition formula for cosine, we get:

\nonumber \begin{align*}\cos\;(A+B) ~&=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B\\[4pt] \nonumber &=~ \frac{3}{5} \;\cdot\; \frac{5}{13} ~-~ \frac{4}{5} \;\cdot\; \frac{12}{13} \quad\Rightarrow\quad \boxed{\cos\;(A+B) ~=~ -\frac{33}{65}}\end{align*}

Instead of using the addition formula for tangent, we can use the results above:

$\nonumber \tan\;(A+B) ~=~ \frac{\sin\;(A+B)}{\cos\;(A+B)} ~=~ \frac{\frac{56}{65}}{-\frac{33}{65}} \quad\Rightarrow\quad \boxed{\tan\;(A+B) ~=~ -\frac{56}{33}}$

Example 3.9

Prove the following identity:

$\nonumber \sin\;(A+B+C) ~=~ \sin\;A~\cos\;B~\cos\;C \;+\; \cos\;A~\sin\;B~\cos\;C \;+\; \cos\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\sin\;C$

Solution:

Treat $$A+B+C$$ as $$(A+B)+C$$ and use the addition formulas three times:

\nonumber \begin{align*} \sin\;(A+B+C) ~&=~ \sin\;((A+B)+C)\\[4pt] \nonumber &=~ \sin\;(A+B)~\cos\;C \;+\; \cos\;(A+B)~\sin\;C\\[4pt] \nonumber &=~ (\sin\;A ~ \cos\;B \;+\; \cos\;A ~ \sin\;B)~\cos\;C \;+\; (\cos\;A ~ \cos\;B \;-\; \sin\;A ~ \sin\;B)~\sin\;C\\[4pt] \nonumber &=~ \sin\;A~\cos\;B~\cos\;C \;+\; \cos\;A~\sin\;B~\cos\;C \;+\; \cos\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\sin\;C \end{align*}

Example 3.10

For any triangle $$\triangle\,ABC$$, show that $$\tan\;A + \tan\;B + \tan\;C = \tan\;A~\tan\;B~\tan\;C$$.

Solution:

Note that this is not an identity which holds for all angles; since $$A$$, $$B$$, and $$C$$ are the angles of a triangle, it holds when $$A$$, $$B$$, $$C$$ $$> 0^\circ$$ and $$A + B + C = 180^\circ$$. So using $$C = 180^\circ - (A+B)$$ and the relation $$\;\tan\;(180^\circ - \theta) = -\tan\;\theta\;$$ from Section 1.5, we get:

\nonumber \begin{align} \tan\;A \;+\; \tan\;B \;+\; \tan\;C ~&=~ \tan\;A \;+\; \tan\;B \;+\; \tan\;(180^\circ - (A+B))\\[4pt] \nonumber &=~ \tan\;A \;+\; \tan\;B \;-\; \tan\;(A+B)\\[4pt] \nonumber &=~ \tan\;A \;+\; \tan\;B \;-\; \frac{\tan\;A + \tan\;B}{1 - \tan\;A ~ \tan\;B} \\[4pt] \nonumber &=~ (\tan\;A \;+\; \tan\;B)~\left( 1 \;-\; \dfrac{1}{1 - \tan\;A ~ \tan\;B} \right) \\[4pt] \nonumber &=~ (\tan\;A \;+\; \tan\;B)~\left( \dfrac{1 - \tan\;A ~ \tan\;B}{1 - \tan\;A ~ \tan\;B} \;-\; \dfrac{1}{1 - \tan\;A ~ \tan\;B} \right ) \\[4pt] \nonumber &=~ (\tan\;A \;+\; \tan\;B)\;\cdot\;\left ( \frac{-\tan\;A ~ \tan\;B}{1 - \tan\;A ~ \tan\;B} \right ) \\[4pt] \nonumber &=~ \tan\;A ~ \tan\;B \;\cdot\; \left( -\frac{\tan\;A \;+\; \tan\;B}{1 - \tan\;A ~ \tan\;B} \right )\\[4pt] \nonumber &=~ \tan\;A ~ \tan\;B \;\cdot\; (-\tan\;(A+B))\\[4pt] \nonumber &=~ \tan\;A ~ \tan\;B \;\cdot\; (\tan\;(180^\circ - (A+B)))\\[4pt] \nonumber &=~ \tan\;A ~ \tan\;B ~ \tan\;C \end{align}

Example 3.11

Let $$A$$, $$B$$, $$C$$, and $$D$$ be positive angles such that $$A+B+C+D=180^\circ$$. Show that

$\nonumber \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~=~ \sin\;(A+C)~\sin\;(B+C) ~.$

Solution:

It may be tempting to expand the right side, since it appears more complicated. However, notice that the right side has no $$D$$ term. So instead, we will expand the left side, since we can eliminate the $$D$$ term on that side by using $$D=180^\circ - (A+B+C)$$ and the relation

$\sin\;(180^\circ -(A+B+C)) ~=~ \sin\;(A+B+C).\nonumber$

So since $$\;\sin\;D = \sin\;(A+B+C)$$, we get

\nonumber \begin{align} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~&=~ \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;(A+B+C) ~,~\text{so by Example 3.9 we get}\\[4pt] \nonumber &=~ \sin\;A~\sin\;B ~+~ \sin\;C~(\sin\;A~\cos\;B~\cos\;C \;+\; \cos\;A~\sin\;B~\cos\;C\\[4pt] \nonumber &\quad +\; \cos\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\sin\;C)\\[4pt] \nonumber &=~ \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;A~\cos\;B~\cos\;C ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\[4pt] \nonumber &\quad +~ \sin\;C~\cos\;A~\cos\;B~\sin\;C ~-~ \sin\;C~\sin\;A~\sin\;B~\sin\;C ~. \end{align}
It may not be immediately obvious where to go from here, but it is not completely guesswork. We need to end up with $$\sin\;(A+C)~\sin\;(B+C)$$, and we know that $$\sin\;(B+C) = \sin\;B~\cos\;C + \cos\;B~\sin\;C$$. There are two terms involving $$\;\cos\;B~\sin\;C$$, so group them together to get
\nonumber \begin{align} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~ &=~ \sin\;A~\sin\;B ~-~ \sin\;C~\sin\;A~\sin\;B~\sin\;C ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\[4pt] \nonumber &\quad+~ \cos\;B~\sin\;C~(\sin\;A~\cos\;C ~+~ \cos\;A~\sin\;C)\\[4pt] \nonumber &=~ \sin\;A~\sin\;B~(1 - \sin^2 \;C) ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\[4pt] \nonumber &\quad +~ \cos\;B~\sin\;C~\sin\;(A+C)\\[4pt] \nonumber &=~ \sin\;A~\sin\;B~\cos^2 \;C ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\[4pt] \nonumber &\quad +~ \cos\;B~\sin\;C~\sin\;(A+C)~. \end{align}
We now have two terms involving $$\;\sin\;B~\cos\;C$$, which we can factor out:
\nonumber \begin{align} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~ &=~ \sin\;B~\cos\;C~(\sin\;A~\cos\;C + \cos\;A~\sin\;C~)\\[4pt] \nonumber &\quad +~ \cos\;B~\sin\;C~\sin\;(A+C)\\[4pt] \nonumber &=~ \sin\;B~\cos\;C~\sin\;(A+C) ~+~ \cos\;B~\sin\;C~\sin\;(A+C)\\[4pt] \nonumber &=~ \sin\;(A+C)~(\sin\;B~\cos\;C + \cos\;B~\sin\;C)\\[4pt] \nonumber &=~ \sin\;(A+C)~\sin\;(B+C) \\[4pt] \end{align}

Example 3.12

In the study of the propagation of electromagnetic waves, Snell's law gives the relation

$n_1 ~\sin\;\theta_1 ~=~ n_2 ~\sin\;\theta_2\label{eqn:snell} ~,$

where $$\theta_1$$ is the angle of incidence at which a wave strikes the planar boundary between two mediums, $$\theta_2$$ is the angle of transmission of the wave through the new medium, and $$n_1$$ and $$n_2$$ are the indexes of refraction of the two mediums. The quantity

$r_{1\;2\;s} ~=~ \frac{n_1 ~\cos\;\theta_1 ~-~ n_2 ~\cos\;\theta_2}{n_1 ~\cos\;\theta_1 ~+~ n_2 ~\cos\;\theta_2}\label{3.21}$

is called the Fresnel coefficient for normal incidence reflection of the wave for s-polarization. Show that this can be written as:

$r_{1\;2\;s} ~=~ \frac{\sin\;(\theta_2 - \theta_1)}{\sin\;(\theta_2 + \theta_1)} \nonumber$

Solution:

Multiply the top and bottom of $$r_{1\;2\;s}$$ by $$\;\sin\;\theta_1 ~ \sin\;\theta_2\;$$ to get:

\nonumber \begin{align*} r_{1\;2\;s} ~&=~ \frac{n_1 ~\cos\;\theta_1 ~-~ n_2 ~\cos\;\theta_2}{n_1 ~\cos\;\theta_1 ~+~ n_2 ~\cos\;\theta_2} \;\cdot\; \frac{\sin\;\theta_1 ~ \sin\;\theta_2}{\sin\;\theta_1 ~ \sin\;\theta_2}\\[4pt] \nonumber &=~ \frac{(n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_1 ~-~ (n_2 ~\sin\;\theta_2)~\cos\;\theta_2 ~\sin\;\theta_1}{ (n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_1 ~+~ (n_2 ~\sin\;\theta_2)~\cos\;\theta_2 ~\sin\;\theta_1}\\[4pt] \nonumber &=~ \frac{(n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_1 ~-~ (n_1 ~\sin\;\theta_1)~\cos\;\theta_2 ~\sin\;\theta_1}{ (n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_1 ~+~ (n_1 ~\sin\;\theta_1)~\cos\;\theta_2 ~\sin\;\theta_1} \qquad\text{(by Snell's law)}\\[4pt] \nonumber &=~ \frac{\sin\;\theta_2 ~\cos\;\theta_1 ~-~ \cos\;\theta_2 ~\sin\;\theta_1}{ \sin\;\theta_2 ~\cos\;\theta_1 ~+~ \cos\;\theta_2 ~\sin\;\theta_1}\\[4pt] \nonumber &=~ \frac{\sin\;(\theta_2 - \theta_1)}{\sin\;(\theta_2 + \theta_1)} \end{align*}

The last two examples demonstrate an important aspect of how identities are used in practice: recognizing terms which are part of known identities, so that they can be factored out. This is a common technique.