3.2 Sum and Difference Formulas

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:

\begin{equation} \sin\;(A+B) ~=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B \tag{3.12}\end{equation}

\begin{equation} \cos\;(A+B) ~=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B \tag{3.13}\end{equation}

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

\begin{align*} \angle\,QPR ~&=~ \angle\,QPO - \angle\,OPM ~=~ (90\Degrees - B) - (90\Degrees - (A+B)) ~=~ A ~~\text{in Figure \ref{fig:anglesum}(a), and}\\ \angle\,QPR ~&=~ \angle\,QPO + \angle\,OPM ~=~ (90\Degrees - B) + (90\Degrees - (180\Degrees - (A+B))) ~=~ A ~~\text{in Figure \ref{fig:anglesum}(b).} \end{align*}

tikzpicture[scale=0.7,every node/.style=font=] [dashed] (2.536,0) -- (2.536,9.464); [dashed] (2.536,6) -- (6,6); (1.2,0) arc (0:45:1.2); (45:1.8) arc (45:75:1.8); ([shift=(2.536,9.464)] -90:1.2) arc (-90:-45:1.2); [-latex] (3,0) arc (0:75:3); [linecolor] (5.6,0) -- (5.6,0.4) -- (6,0.4); [linecolor,rotate around=45:(6,6)] (5.6,6) -- (5.6,6.4) -- (6,6.4); (2.136,0) -- (2.136,0.4) -- (2.536,0.4); (2.936,6) -- (2.936,6.4) -- (2.536,6.4); [linecolor,line width=1.5pt] (0,0) -- (6,0) -- (6,6) -- (2.536,9.464) -- cycle; [linecolor,line width=1.5pt] (0,0) -- (6,6) node [black,sloped,above,pos=0.17] B; [left] at (0,0) O; [right] at (6,6) Q; [right] at (6,0) N; [above] at (2.536,9.464) P; [below] at (2.536,0) M; [below right] at (2.536,6) R; at (0.8,0.3) A; [right] at (2.536,8.7) A; [above right] at (75:3) A+B; tikzpicture — Figure 3.2.1$\sin\;(A+B)$ and $\cos\;(A+B)$ for acute $A$ and $B$

tikzpicture[scale=0.7,every node/.style=font=] [dashed] (0,0) -- (-3.564,0) -- (-3.564,9.464); [dashed] (-3.564,3) -- (3,3); (1.2,0) arc (0:45:1.2); (45:1.4) arc (45:109:1.4); [-latex] (2.3,0) arc (0:109:2.3); [-latex] ([shift=(-3.564,9.464)] -90:1.8) arc (-90:-44:1.8); [linecolor] (2.6,0) -- (2.6,0.4) -- (3,0.4); [linecolor,rotate around=45:(3,3)] (2.6,3) -- (2.6,3.4) -- (3,3.4); (-3.164,0) -- (-3.164,0.4) -- (-3.564,0.4); (-3.164,3) -- (-3.164,3.4) -- (-3.564,3.4); [linecolor,line width=1.5pt] (0,0) -- (3,0) -- (3,3) -- (-3.464,9.464) -- cycle; [linecolor,line width=1.5pt] (0,0) -- (3,3) node [black,sloped,above,pos=0.17] B; [below] at (0,0) O; [right] at (3,3) Q; [right] at (3,0) N; [above] at (-3.564,9.464) P; [below] at (-3.564,0) M; [left] at (-3.564,3) R; at (0.8,0.3) A; [right] at (-3.564,7.5) A; [above right] at (80:2.2) A+B; tikzpicture — Figure 3.2.1$\sin\;(A+B)$ and $\cos\;(A+B)$ for acute $A$ and $B$

Thus,

\begin{align} \sin\;(A+B) ~&=~ \frac{MP}{OP} ~=~ \frac{MR+RP}{OP} ~=~ \frac{NQ+RP}{OP} ~=~ \frac{NQ}{OP} ~+~ \frac{RP}{OP}\\ &=~ \frac{NQ}{OQ}\,\cdot\,\frac{OQ}{OP} ~+~ \frac{RP}{PQ}\,\cdot\,\frac{PQ}{OP}\\ &=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B ~, \tag{3.14}\end{align}

and

\begin{align} \cos\;(A+B) ~&=~ \frac{OM}{OP} ~=~ \frac{ON-MN}{OP} ~=~ \frac{ON-RQ}{OP} ~=~ \frac{ON}{OP} ~-~ \frac{RQ}{OP}\\ &=~ \frac{ON}{OQ}\,\cdot\,\frac{OQ}{OP} ~+~ \frac{RQ}{PQ}\,\cdot\,\frac{PQ}{OP}\\ &=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B ~. \tag{3.15}\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 $A=B=0\Degrees$. For general angles, we will need to use the relations we derived in Section 1.5 which involve adding or subtracting $90\Degrees$:

\begin{alignat*}{4} \sin\;(\theta + 90\Degrees) ~ &= ~ \phantom{-}\cos\;\theta &\qquad\quad \sin\;(\theta - 90\Degrees) ~ &= ~ -\cos\;\theta\\ \cos\;(\theta + 90\Degrees) ~ &= ~ -\sin\;\theta &\qquad\quad \cos\;(\theta - 90\Degrees) ~ &= ~ \phantom{-}\sin\;\theta \end{alignat*}

These will be useful because any angle can be written as the sum of an acute angle (or $0\Degrees$) and integer multiples of $\pm90\Degrees$. For example, $155\Degrees = 65\Degrees + 90\Degrees$, $222\Degrees = 42\Degrees + 2(90\Degrees)$, $-77\Degrees = 13\Degrees - 90\Degrees$, etc. So if we can prove that the identities hold when adding or subtracting $90\Degrees$ to or from either $A$ or $B$, respectively, where $A$ and $B$ are acute or $0\Degrees$, then the identities will also hold when repeatedly adding or subtracting $90\Degrees$, and hence will hold for all angles. Replacing $A$ by $A+90\Degrees$ and using the relations for adding $90\Degrees$ gives

\begin{align*} \sin\;((A+90\Degrees) + B) ~&=~ \sin\;((A+B) + 90\Degrees) ~=~ \cos\;(A+B)~,\\ &=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B ~~\text{(by equation (\ref{eqn:cossumproof}))}\\ &=~ \sin\;(A + 90\Degrees)~\cos\;B ~+~ \cos\;(A + 90\Degrees)~\sin\;B ~, \intertext{so the identity holds for $A+90\Degrees$ and $B$ (and, similarly, for $A$ and $B+90\Degrees$). Likewise,} \sin\;((A-90\Degrees) + B) ~&=~ \sin\;((A+B) - 90\Degrees) ~=~ -\cos\;(A+B)~,\\ &=~ -(\cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B) \\ &=~ (-\cos\;A) ~ \cos\;B ~+~ \sin\;A ~ \sin\;B\\ &=~ \sin\;(A - 90\Degrees)~\cos\;B ~+~ \cos\;(A - 90\Degrees)~\sin\;B ~, \end{align*}

so the identity holds for $A-90\Degrees$ and $B$ (and, similarly, for $A$ and $B+90\Degrees$). Thus, the addition formula (3.12) for sine holds for all $A$ and $B$. A similar argument shows that the addition formula (3.13) for cosine is true for all $A$ and $B$. $\text{qed}$

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:

\begin{equation} \sin\;(A-B) ~=~ \sin\;A ~ \cos\;B ~-~ \cos\;A ~ \sin\;B \tag{3.16}\end{equation}

\begin{equation} \cos\;(A-B) ~=~ \cos\;A ~ \cos\;B ~+~ \sin\;A ~ \sin\;B \tag{3.17}\end{equation}

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:

\begin{align*} \tan\;(A+B) ~&=~ \frac{\sin\;(A+B)}{\cos\;(A+B)}\\[5pt] &=~ \frac{\sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B}{\cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B}\\[5pt] &=~ \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 $\cos\;A ~ \cos\;B$)}\\[5pt] &=~ \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:

\begin{equation} \tan\;(A+B) ~=~ \frac{\tan\;A ~+~ \tan\;B}{1 ~-~ \tan\;A ~ \tan\;B} \tag{3.18}\end{equation}

\begin{equation} \tan\;(A-B) ~=~ \frac{\tan\;A ~-~ \tan\;B}{1 ~+~ \tan\;A ~ \tan\;B} \tag{3.19}\end{equation}

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:

\begin{align*} \sin\;(A+B) ~&=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B\\ &=~ \frac{4}{5} \;\cdot\; \frac{5}{13} ~+~ \frac{3}{5} \;\cdot\; \frac{12}{13} \quad\Rightarrow\quad \boxed{\sin\;(A+B) ~=~ \frac{56}{65}}\\ \intertext{Using the addition formula for cosine, we get:} \cos\;(A+B) ~&=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B\\ &=~ \frac{3}{5} \;\cdot\; \frac{5}{13} ~-~ \frac{4}{5} \;\cdot\; \frac{12}{13} \quad\Rightarrow\quad \boxed{\cos\;(A+B) ~=~ -\frac{33}{65}}\\ \intertext{Instead of using the addition formula for tangent, we can use the results above:} \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}} \end{align*}

Example 3.9

Prove the following identity:

\begin{displaymath} \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 \end{displaymath}

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

\begin{align*} \sin\;(A+B+C) ~&=~ \sin\;((A+B)+C)\\ &=~ \sin\;(A+B)~\cos\;C \;+\; \cos\;(A+B)~\sin\;C\\ &=~ (\sin\;A ~ \cos\;B \;+\; \cos\;A ~ \sin\;B)~\cos\;C \;+\; (\cos\;A ~ \cos\;B \;-\; \sin\;A ~ \sin\;B)~\sin\;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 \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 $A$, $B$, $C$ $> 0\Degrees$ and $A + B + C = 180\Degrees$. So using $C = 180\Degrees - (A+B)$ and the relation $\;\tan\;(180\Degrees - \theta) = -\tan\;\theta\;$ from Section 1.5, we get:

\begin{align*} \tan\;A \;+\; \tan\;B \;+\; \tan\;C ~&=~ \tan\;A \;+\; \tan\;B \;+\; \tan\;(180\Degrees - (A+B))\\ &=~ \tan\;A \;+\; \tan\;B \;-\; \tan\;(A+B)\\ &=~ \tan\;A \;+\; \tan\;B \;-\; \frac{\tan\;A + \tan\;B}{1 - \tan\;A ~ \tan\;B}\\ &=~ (\tan\;A \;+\; \tan\;B)~\left( 1 \;-\; \dfrac{1}{1 - \tan\;A ~ \tan\;B} \right)\\ &=~ (\tan\;A \;+\; \tan\;B)~\left( \dfrac{1 - \tan\;A ~ \tan\;B}{1 - \tan\;A ~ \tan\;B} \;-\; \dfrac{1}{1 - \tan\;A ~ \tan\;B} \right)\\ &=~ (\tan\;A \;+\; \tan\;B)\;\cdot\;\left( \frac{-\tan\;A ~ \tan\;B}{{1 - \tan\;A ~ \tan\;B}} \right)\\ &=~ \tan\;A ~ \tan\;B \;\cdot\; \left( -\frac{\tan\;A \;+\; \tan\;B}{{1 - \tan\;A ~ \tan\;B}} \right)\\ &=~ \tan\;A ~ \tan\;B \;\cdot\; (-\tan\;(A+B))\\ &=~ \tan\;A ~ \tan\;B \;\cdot\; (\tan\;(180\Degrees - (A+B)))\\ &=~ \tan\;A ~ \tan\;B ~ \tan\;C \end{align*}

Example 3.11

Let $A$, $B$, $C$, and $D$ be positive angles such that $A+B+C+D=180\Degrees$. Show that^[1]

\begin{displaymath} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~=~ \sin\;(A+C)~\sin\;(B+C) ~. \end{displaymath}

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 $D=180\Degrees - (A+B+C)$ and the relation

\[\sin\;(180\Degrees -(A+B+C)) ~=~ \sin\;(A+B+C).\]

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

\begin{align*}\begin{split} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~&=~ \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;(A+B+C) ~,~\text{so by Example \ref{exmp:sumsinabc} we get}\\ &=~ \sin\;A~\sin\;B ~+~ \sin\;C~(\sin\;A~\cos\;B~\cos\;C \;+\; \cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +\; \cos\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\sin\;C)\\ &=~ \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;A~\cos\;B~\cos\;C ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +~ \sin\;C~\cos\;A~\cos\;B~\sin\;C ~-~ \sin\;C~\sin\;A~\sin\;B~\sin\;C ~.\\ \intertext{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} \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\\ &\mathrel{\phantom{=}} {} +~ \cos\;B~\sin\;C~(\sin\;A~\cos\;C ~+~ \cos\;A~\sin\;C)\\ &=~ \sin\;A~\sin\;B~(1 - \sin^2 \;C) ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +~ \cos\;B~\sin\;C~\sin\;(A+C)\\ &=~ \sin\;A~\sin\;B~\cos^2 \;C ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +~ \cos\;B~\sin\;C~\sin\;(A+C)~.\\ \intertext{We now have two terms involving $\;\sin\;B~\cos\;C$, which we can factor out:} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~ &=~ \sin\;B~\cos\;C~(\sin\;A~\cos\;C + \cos\;A~\sin\;C~)\\ &\mathrel{\phantom{=}} {} +~ \cos\;B~\sin\;C~\sin\;(A+C)\end{split}\\ &=~ \sin\;B~\cos\;C~\sin\;(A+C) ~+~ \cos\;B~\sin\;C~\sin\;(A+C)\\ &=~ \sin\;(A+C)~(\sin\;B~\cos\;C + \cos\;B~\sin\;C)\\ &=~ \sin\;(A+C)~\sin\;(B+C) \end{align*}

Example 3.12

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

\begin{equation} n_1 ~\sin\;\theta_1 ~=~ n_2 ~\sin\;\theta_2 ~, \tag{3.20}\end{equation}

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

\begin{equation} 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} \tag{3.21}\end{equation}

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

\begin{displaymath} r_{1\;2\;s} ~=~ \frac{\sin\;(\theta_2 - \theta_1)}{\sin\;(\theta_2 + \theta_1)} \end{displaymath}

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

\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}\\[7pt] &=~ \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}\\[7pt] &=~ \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)}\\[7pt] &=~ \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}\\[7pt] &=~ \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.

Exercises

Verify the addition formulas (3.12) and (3.13) for $A=B=0\Degrees$.

For Exercises 2 and 3, find the exact values of $\sin\;(A+B)$, $\cos\;(A+B)$, and $\tan\;(A+B)$.

$\sin\;A = \frac{8}{17}$, $\cos\;A = \frac{15}{17}$, $\sin\;B = \frac{24}{25}$, $\cos\;B = \frac{7}{25}$
$\sin\;A = \frac{40}{41}$, $\cos\;A = \frac{9}{41}$, $\sin\;B = \frac{20}{29}$, $\cos\;B = \frac{21}{29}$
Use $75\Degrees = 45\Degrees + 30\Degrees$ to find the exact value of $\;\sin\;75\Degrees$.
Use $15\Degrees = 45\Degrees - 30\Degrees$ to find the exact value of $\;\tan\;15\Degrees$.
Prove the identity $\;\sin\;\theta + \cos\;\theta = \sqrt{2}\;\sin\;(\theta + 45\Degrees)\;$. Explain why this shows that

\begin{displaymath} -\sqrt{2} ~\le~ \;\sin\;\theta ~+~ \cos\;\theta ~\le~ \sqrt{2} \end{displaymath}

for all angles $\theta$. For which $\theta$ between $0\Degrees$ and $360\Degrees$ would $\;\sin\;\theta \;+\; \cos\;\theta\;$ be the largest?

For Exercises 7-14, prove the given identity.

$\cos\;(A+B+C) \;=\; \cos\;A~\cos\;B~\cos\;C \;-\; \cos\;A~\sin\;B~\sin\;C \;-\; \sin\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\cos\;C$
$\tan\;(A+B+C) ~=~ \dfrac{\tan\;A \;+\; \tan\;B \;+\; \tan\;C \;-\; \tan\;A~\tan\;B~\tan\;C}{1 \;-\; \tan\;B~\tan\;C \;-\; \tan\;A~\tan\;C \;-\; \tan\;A~\tan\;B}$ 2
$\cot\;(A+B) ~=~ \dfrac{\cot\;A~\cot\;B \;-\; 1}{\cot\;A \;+\; \cot\;B}$
$\cot\;(A-B) ~=~ \dfrac{\cot\;A~\cot\;B \;+\; 1}{\cot\;B \;-\; \cot\;A}$

2
$\tan\;(\theta + 45\Degrees) ~=~ \dfrac{1 \;+\; \tan\;\theta}{1 \;-\; \tan\;\theta}$
$\dfrac{\cos\;(A+B)}{\sin\;A~\cos\;B} ~=~ \cot\;A \;-\; \tan\;B$

$\cot\;A ~+~ \cot\;B ~=~ \dfrac{\sin\;(A+B)}{\sin\;A~\sin\;B}$
$\dfrac{\sin\;(A-B)}{\sin\;(A+B)} ~=~ \dfrac{\cot\;B \;-\; \cot\;A}{\cot\;B \;+\; \cot\;A}$
Generalize Exercise 6: For any $a$ and $b$, $-\sqrt{a^2 + b^2} \;\le\; a\;\sin\;\theta \;+\; b\;\cos\;\theta \;\le\; \sqrt{a^2 + b^2}\;$ for all $\theta$.
Continuing Example 3.12, use Snell’s law to show that the s-polarization transmission Fresnel coefficient

\begin{equation} t_{1\;2\;s} ~=~ \frac{2\;n_1 ~\cos\;\theta_1}{n_1 ~\cos\;\theta_1 ~+~ n_2 ~\cos\;\theta_2} \tag{3.22}\end{equation}

can be written as:

\begin{displaymath} t_{1\;2\;s} ~=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{\sin\;(\theta_2 + \theta_1)} \end{displaymath}

[r]
Suppose that two lines with slopes $m_1$ and $m_2$, respectively, intersect at an angle $\theta$ and are not perpendicular (i.e. $\theta \ne 90\Degrees$), as in the figure on the right. Show that

\begin{displaymath} \tan\;\theta ~=~ \left| \frac{m_1 ~-~ m_2}{1 ~+~ m_1 \; m_2} \right| ~. \end{displaymath}

(Hint: Use Example 1.26 from Section 1.5.)

Use Exercise 17 to find the angle between the lines $y=2x+3$ and $y=-5x-4$.
For any triangle $\triangle\,ABC$, show that $\;\cot\;A~\cot\;B ~+~ \cot\;B~\cot\;C ~+~ \cot\;C~\cot\;A ~=~ 1$.
(Hint: Use Exercise 9 and $C=180\Degrees - (A+B)$.)
For any positive angles $A$, $B$, and $C$ such that $A+B+C=90\Degrees$, show that

\begin{displaymath} \tan\;A~\tan\;B ~+~ \tan\;B~\tan\;C ~+~ \tan\;C~\tan\;A ~=~ 1 ~. \end{displaymath}
Prove the identity $\;\sin\;(A+B)~\cos\;B ~-~ \cos\;(A+B)~\sin\;B ~=~ \sin\;A$.
Note that the right side depends only on $A$, while the left side depends on both $A$ and $B$.
A line segment of length $r > 0$ from the origin to the point $(x,y)$ makes an angle $\alpha$ with the positive $x$-axis, so that $(x,y) = (r\;\cos\;\alpha,r\;\sin\;\alpha)$, as in the figure below. What are the endpoint’s new coordinates $(x',y')$ after a counterclockwise rotation by an angle $\beta\;$? Your answer should be in terms of $r$, $\alpha$, and $\beta$.

$tikzpicture[scale=1.2,every node/.style=font=] [black!60,line width=0.3pt,-latex] (-3,0) -- (3,0) node [right] x; [black!60,line width=0.3pt,-latex] (0,-0.2) -- (0,2.2) node [above] y; [dashed] (1.43394,0) -- (55:2.5); (1.23394,0) -- (1.23394,0.2) -- (1.43394,0.2); [below left] at (0,0) 0; [right] at (40:0.2) ; (0:0.6) arc (0:55:0.6); [-latex] (55:0.8) arc (55:165:0.8); [below] at (110:0.8) ; [linecolor,line width=1.5pt] (0,0) -- (55:2.5) node[black,above,midway] r; [dashed,linecolor,line width=1.5pt] (0,0) -- (165:2.5) node[black,above,midway] r; (0,0) circle (1.66pt); (55:2.5) circle (1.66pt) node[above right] (x,y) = (r\; \; ,r\; \; ); (165:2.5) circle (1.66pt) node[above] (x',y'); tikzpicture$

This is the “trigonometric form” of Ptolemy’s Theorem, which says that a quadrilateral can be inscribed in a circle if and only if the sum of the products of its opposite sides equals the product of its diagonals. ↩