3.4: Other Identities

Though the identities in this section fall under the category of "other'', they are perhaps (along with $\cos^2 \;\theta + \sin^2 \;\theta = 1$) the most widely used identities in practice. It is very common to encounter terms such as $\;\sin\;A + \sin\;B\;$ or $\;\sin\;A~\cos\;B\;$ in calculations, so we will now derive identities for those expressions. First, we have what are often called the product-to-sum formulas:

$$\begin{align}\sin\;A~\cos\;B ~&=~ \phantom{-}\tfrac{1}{2}\;(\sin\;(A+B) ~+~ \sin\;(A-B))\label{eqn:p2ssincos}\\ \cos\;A~\sin\;B ~&=~ \phantom{-}\tfrac{1}{2}\;(\sin\;(A+B) ~-~ \sin\;(A-B))\label{eqn:p2scossin}\\ \cos\;A~\cos\;B ~&=~ \phantom{-}\tfrac{1}{2}\;(\cos\;(A+B) ~+~ \cos\;(A-B))\label{eqn:p2scoscos}\\ \sin\;A~\sin\;B ~&=~ -\tfrac{1}{2}\;(\cos\;(A+B) ~-~ \cos\;(A-B))\label{eqn:p2ssinsin} \end{align} \nonumber$$

We will prove the first formula; the proofs of the others are similar (see Exercises 1-3). We see that
$$\nonumber \require{cancel} \begin{align*} \sin\;(A+B) ~+~ \sin\;(A-B) ~&=~ (\sin\;A~\cos\;B ~+~ \cancel{\cos\;A~\sin\;B}) ~+~ (\sin\;A~\cos\;B ~-~ \cancel{\cos\;A~\sin\;B})\\ \nonumber &=~ 2\;\sin\;A~\cos\;B ~, \end{align*} \nonumber$$
so Equation $\ref{eqn:p2ssincos}$ follows upon dividing both sides by $2$. Notice how in each of the above identities a product (e.g. $\sin\;A~\cos\;B$) of trigonometric functions is shown to be equivalent to a sum (e.g. $\tfrac{1}{2}\;(\sin\;(A+B) ~+~ \sin\;(A-B))$) of such functions. We can go in the opposite direction, with the sum-to-product formulas:

$$\begin{align} \sin\;A ~+~ \sin\;B ~&=~ \phantom{-}2\;\sin\;\tfrac{1}{2}(A+B)~ \cos\;\tfrac{1}{2}(A-B)\label{eqn:s2psinpsin}\\ \sin\;A ~-~ \sin\;B ~&=~ \phantom{-}2\;\cos\;\tfrac{1}{2}(A+B)~ \sin\;\tfrac{1}{2}(A-B)\label{eqn:s2psinmsin}\\ \cos\;A ~+~ \cos\;B ~&=~ \phantom{-}2\;\cos\;\tfrac{1}{2}(A+B)~ \cos\;\tfrac{1}{2}(A-B)\label{eqn:s2pcospcos}\\ \cos\;A ~-~ \cos\;B ~&=~ -2\;\sin\;\tfrac{1}{2}(A+B)~\sin\;\tfrac{1}{2}(A-B)\label{eqn:s2pcosmcos} \end{align} \nonumber$$

These formulas are just the product-to-sum formulas rewritten by using some clever substitutions: let $x=\frac{1}{2}(A+B)$ and $y=\frac{1}{2}(A-B)$. Then $x+y=A$ and $x-y=B$. For example, to derive Equation $\ref{eqn:s2pcospcos}$, make the above substitutions in Equation $\ref{eqn:p2scoscos}$ to get
$$\nonumber \begin{align*} \cos\;A ~+~ \cos\;B ~&=~ \cos\;(x+y) ~+~ \cos\;(x-y)\\ \nonumber &=~ 2\;\cdot\;\tfrac{1}{2}(\cos\;(x+y) ~+~ \cos\;(x-y))\\ \nonumber &=~ 2\;\cos\;x~\cos\;y\qquad\qquad\text{(by Equation \ref{eqn:p2scoscos})}\\ \nonumber &=~ 2\;\cos\;\tfrac{1}{2}(A+B)~\cos\;\tfrac{1}{2}(A-B) ~. \end{align*} \nonumber$$
The proofs of the other sum-to-product formulas are similar (see Exercises 4-6).

Example 3.16

We are now in a position to prove Mollweide's equations, which we introduced in Section 2.3: For any triangle $\triangle\,ABC$,

$$\nonumber \frac{a-b}{c} ~=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C} \qquad\text{and}\qquad \frac{a+b}{c} ~=~ \frac{\cos\;\frac{1}{2}(A-B)}{\sin\;\frac{1}{2}C} ~. \nonumber$$

First, since $C=2\;\cdot\;\tfrac{1}{2}C$, by the double-angle formula we have $\;\sin\;C = 2\;\sin\;\tfrac{1}{2}C~\cos\;\tfrac{1}{2}C$. Thus,

$$\nonumber \require{cancel} \begin{align*} \frac{a-b}{c} ~&=~ \frac{a}{c} ~-~ \frac{b}{c} ~=~ \frac{\sin\;A}{\sin\;C} ~-~ \frac{\sin\;B}{\sin\;C}\quad\text{(by the Law of Sines)}\\ \nonumber &=~ \frac{\sin\;A ~-~ \sin\;B}{\sin\;C} ~=~ \frac{\sin\;A ~-~ \sin\;B}{2\;\sin\;\tfrac{1}{2}C~\cos\;\tfrac{1}{2}C}\\ \nonumber &=~ \frac{2\;\cos\;\tfrac{1}{2}(A+B)~\sin\;\tfrac{1}{2}(A-B)}{2\;\sin\;\tfrac{1}{2}C~ \cos\;\tfrac{1}{2}C}\quad\text{(by Equation \ref{eqn:s2psinmsin})}\\ \nonumber &=~ \frac{\cos\;\tfrac{1}{2}(180^\circ - C)~\sin\;\tfrac{1}{2}(A-B)}{\sin\;\tfrac{1}{2}C~ \cos\;\tfrac{1}{2}C}\quad\text{(since \(A+B=180^\circ - C\))}\\ \nonumber &=~ \frac{\cancel{\cos\;(90^\circ - \tfrac{1}{2}C)}~\sin\;\tfrac{1}{2}(A-B)}{ \cancel{\sin\;\tfrac{1}{2}C}~\cos\;\tfrac{1}{2}C}\\ \nonumber &=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C}\quad\text{(since \(\;\cos\;(90^\circ - \tfrac{1}{2}C) = \sin\;\tfrac{1}{2}C\))}~. \end{align*} \nonumber$$

This proves the first equation. The proof of the other equation is similar (see Exercise 7).

Example 3.17

Using Mollweide's equations, we can prove the Law of Tangents: For any triangle $\triangle\,ABC$,

$$\nonumber \frac{a-b}{a+b} ~=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} ~,\quad \frac{b-c}{b+c} ~=~ \frac{\tan\;\frac{1}{2}(B-C)}{\tan\;\frac{1}{2}(B+C)} ~,\quad \frac{c-a}{c+a} ~=~ \frac{\tan\;\frac{1}{2}(C-A)}{\tan\;\frac{1}{2}(C+A)} ~. \nonumber$$

We need only prove the first equation; the other two are obtained by cycling through the letters. We see that

$$\nonumber \begin{align*} \frac{a-b}{a+b} ~&=~ \dfrac{\dfrac{a-b}{c}}{\dfrac{a+b}{c}} ~=~ \dfrac{\dfrac{\sin\;\tfrac{1}{2}(A-B)}{\cos\;\tfrac{1}{2}C}}{ \dfrac{\cos\;\tfrac{1}{2}(A-B)}{\sin\;\tfrac{1}{2}C}}\quad\text{(by Mollweide's equations)}\\ \nonumber &=~ \dfrac{\sin\;\tfrac{1}{2}(A-B)}{\cos\;\tfrac{1}{2}(A-B)} \;\cdot\; \dfrac{\sin\;\tfrac{1}{2}C}{\cos\;\tfrac{1}{2}C}\\ \nonumber &=~ \tan\;\tfrac{1}{2}(A-B) \;\cdot\; \tan\;\tfrac{1}{2}C ~=~ \tan\;\tfrac{1}{2}(A-B) \;\cdot\; \tan\;(90^\circ - \tfrac{1}{2}(A+B)) \quad\text{(since \(C=180^\circ - (A+B)\))}\\ \nonumber &=~ \tan\;\tfrac{1}{2}(A-B) \;\cdot\; \cot\;\tfrac{1}{2}(A+B)\quad\text{(since \(\tan\;(90^\circ - \tfrac{1}{2}(A+B)) = \cot\;\tfrac{1}{2}(A+B) \), see Section 1.5)}\\ \nonumber &=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} ~.\quad\textbf{QED} \end{align*} \nonumber$$

Example 3.18

For any triangle $\triangle\,ABC$, show that
$$\nonumber \cos\;A ~+~ \cos\;B ~+~ \cos\;C ~=~ 1 ~+~ 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~. \nonumber$$

Solution

Since $\;\cos\;(A+B+C) = \cos\;180^\circ = -1$, we can rewrite the left side as

$$\nonumber \begin{align*} \cos\;A \;+\; \cos\;B \;+\; \cos\;C ~&=~ 1 \;+\; (\cos\;(A+B+C) \;+\; \cos\;C) \;+\; (\cos\;A \;+\; \cos\;B)~~\text{, so by Equation \ref{eqn:s2pcospcos}}\\ \nonumber &=~ 1 \;+\; 2\;\cos\;\tfrac{1}{2}(A+B+2C)~\cos\;\tfrac{1}{2}(A+B) \;+\; 2\;\cos\;\tfrac{1}{2}(A+B)~\cos\;\tfrac{1}{2}(A-B)\\ \nonumber &=~ 1 \;+\; 2\;\cos\;\tfrac{1}{2}(A+B)~\left( \cos\;\tfrac{1}{2}(A+B+2C) \;+\; \cos\;\tfrac{1}{2}(A-B) \right) ~~\text{, so}\\ \nonumber &=~ 1 \;+\; 2\;\cos\;\tfrac{1}{2}(A+B)\;\cdot\;2\;\cos\;\tfrac{1}{2}(A+C)~ \cos\;\tfrac{1}{2}(B+C)~~\text{by Equation \ref{eqn:s2pcospcos},}\end{align*} \nonumber$$
since $\tfrac{1}{2}\left( \tfrac{1}{2}(A+B+2C) + \tfrac{1}{2}(A-B) \right) = \tfrac{1}{2}(A+C)$ and $\tfrac{1}{2}\left( \tfrac{1}{2}(A+B+2C) - \tfrac{1}{2}(A-B) \right) = \tfrac{1}{2}(B+C)$. Thus,}
$$\nonumber \begin{align*} \cos\;A \;+\; \cos\;B \;+\; \cos\;C ~&=~ 1 \;+\; 4\;\cos\;(90^\circ - \tfrac{1}{2}C)~\cos\;(90^\circ - \tfrac{1}{2}B)~ \cos\;(90^\circ - \tfrac{1}{2}A)\\ \nonumber &=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}C~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}A ~~,\text{ so rearranging the order gives}\\ \nonumber &=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~. \end{align*} \nonumber$$

Example 3.19

For any triangle $\triangle\,ABC$, show that $\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~ \sin\;\tfrac{1}{2}C \;\le\; \frac{1}{8}\;$.

Solution

Let $u=\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C$. Apply Equation $\ref{eqn:p2ssinsin}$ to the first two terms in $u$ to get

$$\nonumber u ~=~ -\tfrac{1}{2}\;(\cos\;\tfrac{1}{2}(A+B) \;-\; \cos\;\tfrac{1}{2}(A-B))~ \sin\;\tfrac{1}{2}C ~=~ \tfrac{1}{2}\;(\cos\;\tfrac{1}{2}(A-B) \;-\; \cos\;\tfrac{1}{2}(A+B))~\cos\;\tfrac{1}{2}(A+B) ~, \nonumber$$

since $\;\sin\;\tfrac{1}{2}C = \cos\;\tfrac{1}{2}(A+B)$, as we saw in Example 3.18. Multiply both sides by $2$ to get
$$\nonumber \cos^2 \;\tfrac{1}{2}(A+B) ~-~ \cos\;\tfrac{1}{2}(A-B)~\cos\;\tfrac{1}{2}(A+B) ~+~ 2u ~=~ 0 ~, \nonumber$$
after rearranging the terms. Notice that the expression above is a quadratic equation in the term $\;\cos\;\tfrac{1}{2}(A+B)$. So by the quadratic formula,
$$\nonumber \cos\;\tfrac{1}{2}(A+B) ~=~ \frac{\cos\;\tfrac{1}{2}(A-B) \;\pm\; \sqrt{\cos^2 \;\tfrac{1}{2}(A-B) - 4(1)(2u)}}{2} ~~, \nonumber$$
which has a real solution only if the quantity inside the square root is nonnegative. But we know that $\;\cos\;\tfrac{1}{2}(A+B)\;$ is a real number (and, hence, a solution exists), so we must have
$$\nonumber \cos^2 \;\tfrac{1}{2}(A-B) \;- \; 8u ~\ge~ 0 \quad\Rightarrow\quad u ~\le~ \tfrac{1}{8}\; \cos^2 \;\tfrac{1}{2}(A-B) ~\le~ \tfrac{1}{8} \quad\Rightarrow\quad \;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~\le~ \tfrac{1}{8} ~. \nonumber$$

Example 3.20

For any triangle $\triangle\,ABC$, show that $\;1 ~<~ \cos\;A + \cos\;B + \cos\;C ~\le~ \tfrac{3}{2}\;$.

Solution

Since $0^\circ < A,\; B,\; C < 180^\circ$, the sines of $\tfrac{1}{2}A$, $\tfrac{1}{2}B$, and $\tfrac{1}{2}C$ are all positive, so

$$\nonumber \cos\;A \;+\; \cos\;B \;+\; \cos\;C ~=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~ > ~ 1 \nonumber$$

by Example 3.18. Also, by Examples 3.18 and 3.19 we have

$$\cos\;A \;+\; \cos\;B \;+\; \cos\;C ~=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~\le~ 1 \;+\; 4\;\cdot\;\tfrac{1}{8} ~=~ \tfrac{3}{2} ~.\nonumber$$

Hence, $\;1 ~<~ \cos\;A + \cos\;B + \cos\;C ~\le~ \tfrac{3}{2}\;$.

Example 3.21

Recall Snell's law from Example 3.12 in Section 3.2: $n_1 ~\sin\;\theta_1 = n_2 ~\sin\;\theta_2$. Use it to show that the p-polarization transmission Fresnel coefficient defined by

$$t_{1\;2\;p} ~=~ \frac{2\;n_1 ~\cos\;\theta_1}{n_2 ~\cos\;\theta_1 ~+~ n_1 ~\cos\;\theta_2}\label{3.45}$$

can be written as:

$$t_{1\;2\;p} ~=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{\sin\;(\theta_1 + \theta_2)~ \cos\;(\theta_1 - \theta_2)} ~. \nonumber$$

Solution

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

$$\nonumber \begin{align*} t_{1\;2\;p} ~&=~ \frac{2\;n_1 ~\cos\;\theta_1}{n_2 ~\cos\;\theta_1 ~+~ n_1 ~\cos\;\theta_2} \;\cdot\; \frac{\sin\;\theta_1 ~ \sin\;\theta_2}{\sin\;\theta_1 ~ \sin\;\theta_2}\\ \nonumber &=~ \frac{2\;(n_1 ~\sin\;\theta_1)~\cos\;\theta_1 ~\sin\;\theta_2}{ (n_2 ~\sin\;\theta_2)~\sin\;\theta_1 ~\cos\;\theta_1 ~+~ (n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_2}\\ \nonumber &=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{ \sin\;\theta_1 ~\cos\;\theta_1 ~+~ \sin\;\theta_2 ~\cos\;\theta_2} \qquad\text{(by Snell's law)}\\ \nonumber &=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{ \tfrac{1}{2}\;(\sin\;2\,\theta_1 ~+~ \sin\;2\theta_2)} \qquad\text{(by the double-angle formula)}\\ \nonumber &=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{ \tfrac{1}{2}\;(2\;\sin\;\tfrac{1}{2}(2\theta_1 + 2\theta_2)~ \cos\;\tfrac{1}{2}(2\theta_1 - 2\theta_2))} \qquad\text{(by formula \ref{eqn:s2psinpsin})}\\ \nonumber &=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{\sin\;(\theta_1 + \theta_2)~ \cos\;(\theta_1 - \theta_2)} \end{align*} \nonumber$$

Example 3.22

In an AC electrical circuit, the instantaneous power $p(t)$ delivered to the entire circuit in the \emph{sinusoidal steady state} at time $t$ is given by

$$\nonumber p(t) ~=~ v(t)\;i(t) ~, \nonumber$$

where the voltage $v(t)$ and current $i(t)$ are given by

$$\nonumber \begin{align*} v(t) ~&=~ V_m \;\cos\;\omega t ~,\\ \nonumber i(t) ~&=~ I_m \;\cos\;(\omega t + \phi)~, \end{align*} \nonumber$$

for some constants $V_m$, $I_m$, $\omega$, and $\phi$. Show that the instantaneous power can be written as

$$\nonumber p(t) ~=~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;\phi ~+~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;(2\omega t + \phi) ~. \nonumber$$

Solution

By definition of $p(t)$, we have

$$\nonumber \begin{alignat*}{2} p(t) ~&=~ V_m \;I_m \;\cos\;\omega t~\cos\;(\omega t + \phi)\\ \nonumber &=~ V_m \;I_m \;\cdot\;\tfrac{1}{2}(\cos\;(2\omega t + \phi) \;+\; \cos\;(-\phi)) \qquad&&\text{(by Equation \ref{eqn:p2scoscos})}\\ \nonumber &=~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;\phi ~+~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;(2\omega t + \phi) \qquad&&\text{(since \(\cos\;(-\phi) = \cos\;\phi\))}~. \end{alignat*} \nonumber$$