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:
We will prove the first formula; the proofs of the others are similar (see Exercises
1-3). We see that
\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})\\
&=~ 2\;\sin\;A~\cos\;B ~,
\end{align*}
so formula (3.37) 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:
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 formula (3.43), make the above substitutions in formula
(3.39) to get
\begin{align*}
\cos\;A ~+~ \cos\;B ~&=~ \cos\;(x+y) ~+~ \cos\;(x-y)\\
&=~ 2\;\cdot\;\tfrac{1}{2}(\cos\;(x+y) ~+~ \cos\;(x-y))\\
&=~ 2\;\cos\;x~\cos\;y\qquad\qquad\text{(by formula (\ref{eqn:p2scoscos}))}\\
&=~ 2\;\cos\;\tfrac{1}{2}(A+B)~\cos\;\tfrac{1}{2}(A-B) ~.
\end{align*}
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 \(\triangle\,ABC\),
\begin{displaymath}
\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} ~.
\end{displaymath}
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,
\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)}\\
&=~ \frac{\sin\;A ~-~ \sin\;B}{\sin\;C} ~=~
\frac{\sin\;A ~-~ \sin\;B}{2\;\sin\;\tfrac{1}{2}C~\cos\;\tfrac{1}{2}C}\\
&=~ \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 formula (\ref{eqn:s2psinmsin}))}\\
&=~ \frac{\cos\;\tfrac{1}{2}(180\Degrees - C)~\sin\;\tfrac{1}{2}(A-B)}{\sin\;\tfrac{1}{2}C~
\cos\;\tfrac{1}{2}C}\quad\text{(since $A+B=180\Degrees - C$)}\\
&=~ \frac{\cancel{\cos\;(90\Degrees - \tfrac{1}{2}C)}~\sin\;\tfrac{1}{2}(A-B)}{
\cancel{\sin\;\tfrac{1}{2}C}~\cos\;\tfrac{1}{2}C}\\
&=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C}\quad\text{(since $\;\cos\;(90\Degrees -
\tfrac{1}{2}C) = \sin\;\tfrac{1}{2}C$)}~.
\end{align*}
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 \(\triangle\,ABC\),
\begin{displaymath}
\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)} ~.
\end{displaymath}
We need only prove the first equation; the other two are obtained by cycling through the letters.
We see that
\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)}\\
&=~ \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}\\
&=~ \tan\;\tfrac{1}{2}(A-B) \;\cdot\; \tan\;\tfrac{1}{2}C ~=~
\tan\;\tfrac{1}{2}(A-B) \;\cdot\; \tan\;(90\Degrees - \tfrac{1}{2}(A+B))
\quad\text{(since $C=180\Degrees - (A+B)$)}\\
&=~ \tan\;\tfrac{1}{2}(A-B) \;\cdot\; \cot\;\tfrac{1}{2}(A+B)\quad\text{(since $\tan\;(90\Degrees
- \tfrac{1}{2}(A+B)) = \cot\;\tfrac{1}{2}(A+B)$, see Section 1.5)}\\
&=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} ~.\quad\text{qed}
\end{align*}
Example 3.18
For any triangle \(\triangle\,ABC\), show that
\begin{displaymath}
\cos\;A ~+~ \cos\;B ~+~ \cos\;C ~=~ 1 ~+~
4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~.
\end{displaymath}
Solution: Since \(\;\cos\;(A+B+C) = \cos\;180\Degrees = -1\), we can rewrite the left side as
\begin{align*}
\cos\;A \;+\; \cos\;B \;+\; \cos\;C ~&=~ 1 \;+\; (\cos\;(A+B+C) \;+\; \cos\;C) \;+\; (\cos\;A
\;+\; \cos\;B)~~\text{, so by formula (\ref{eqn:s2pcospcos})}\\
&=~ 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)\\
&=~ 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}\\
&=~ 1 \;+\; 2\;\cos\;\tfrac{1}{2}(A+B)\;\cdot\;2\;\cos\;\tfrac{1}{2}(A+C)~
\cos\;\tfrac{1}{2}(B+C)~~\text{by formula (\ref{eqn:s2pcospcos}),}\\
\intertext{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,}
\cos\;A \;+\; \cos\;B \;+\; \cos\;C ~&=~
1 \;+\; 4\;\cos\;(90\Degrees - \tfrac{1}{2}C)~\cos\;(90\Degrees - \tfrac{1}{2}B)~
\cos\;(90\Degrees - \tfrac{1}{2}A)\\
&=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}C~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}A
~~,\text{ so rearranging the order gives}\\
&=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~.
\end{align*}
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 formula (3.40) to the first two terms in \(u\) to get
\begin{displaymath}
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) ~,
\end{displaymath}
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
\begin{displaymath}
\cos^2 \;\tfrac{1}{2}(A+B) ~-~ \cos\;\tfrac{1}{2}(A-B)~\cos\;\tfrac{1}{2}(A+B) ~+~ 2u ~=~ 0 ~,
\end{displaymath}
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,
\begin{displaymath}
\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} ~~,
\end{displaymath}
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
\begin{displaymath}
\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} ~.
\end{displaymath}
Example 3.20
For any triangle \(\triangle\,ABC\), show that \(\;1 ~<~ \cos\;A + \cos\;B + \cos\;C ~\le~
\tfrac{3}{2}\;\).
Solution: Since \(0\Degrees < A,\; B,\; C < 180\Degrees\), the sines of
\(\tfrac{1}{2}A\), \(\tfrac{1}{2}B\), and \(\tfrac{1}{2}C\) are all positive, so
\begin{displaymath}
\cos\;A \;+\; \cos\;B \;+\; \cos\;C ~=~ 1 \;+\;
4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~ > ~ 1
\end{displaymath}
by Example 3.18. Also, by Examples 3.18 and 3.19
we have
\begin{displaymath}
\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} ~.
\end{displaymath}
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
\begin{equation}
t_{1\;2\;p} ~=~ \frac{2\;n_1 ~\cos\;\theta_1}{n_2 ~\cos\;\theta_1 ~+~ n_1 ~\cos\;\theta_2}
\tag{3.45}\end{equation}
can be written as:
\begin{displaymath}
t_{1\;2\;p} ~=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{\sin\;(\theta_1 + \theta_2)~
\cos\;(\theta_1 - \theta_2)} ~.
\end{displaymath}
Solution: Multiply the top and bottom of \(t_{1\;2\;p}\) by \(\;\sin\;\theta_1
~\sin\;\theta_2\;\) to get:
\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}\\[7pt]
&=~ \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}\\[7pt]
&=~ \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)}\\[7pt]
&=~ \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)}\\[7pt]
&=~ \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}))}\\[7pt]
&=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{\sin\;(\theta_1 + \theta_2)~
\cos\;(\theta_1 - \theta_2)}
\end{align*}
Example 3.22
In an AC electrical circuit, the instantaneous power \(p(t)\) delivered to the entire circuit
in the sinusoidal steady state at time \(t\) is given by
\begin{displaymath}
p(t) ~=~ v(t)\;i(t) ~,
\end{displaymath}
where the voltage \(v(t)\) and current \(i(t)\) are given by
\begin{align*}
v(t) ~&=~ V_m \;\cos\;\omega t ~,\\
i(t) ~&=~ I_m \;\cos\;(\omega t + \phi)~,
\end{align*}
for some constants \(V_m\), \(I_m\), \(\omega\), and \(\phi\). Show that the instantaneous power can be
written as
\begin{displaymath}
p(t) ~=~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;\phi ~+~
\tfrac{1}{2}\,V_m \; I_m \;\cos\;(2\omega t + \phi) ~.
\end{displaymath}
Solution: By definition of \(p(t)\), we have
\begin{alignat*}{2}
p(t) ~&=~ V_m \;I_m \;\cos\;\omega t~\cos\;(\omega t + \phi)\\
&=~ V_m \;I_m \;\cdot\;\tfrac{1}{2}(\cos\;(2\omega t + \phi) \;+\; \cos\;(-\phi))
\qquad&&\text{(by formula (\ref{eqn:s2pcospcos}))}\\
&=~ \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*}
Exercises
3
Prove formula (3.38).
Prove formula (3.39).
Prove formula (3.40).
3
Prove formula (3.41).
Prove formula (3.42).
Prove formula (3.44).
Prove Mollweide’s second equation: For any triangle \(\triangle\,ABC\),
\(~\dfrac{a+b}{c} ~=~ \dfrac{\cos\;\tfrac{1}{2}(A-B)}{\sin\;\tfrac{1}{2}C}\) .
Continuing Example 3.21, use Snell’s law to show that the
p-polarization reflection Fresnel coefficient
\begin{equation}
r_{1\;2\;p} ~=~ \frac{n_2 ~\cos\;\theta_1 ~-~ n_1 ~\cos\;\theta_2}{n_2 ~\cos\;\theta_1 ~+~
n_1 ~\cos\;\theta_2}
\tag{3.46}\end{equation}
can be written as:
\begin{displaymath}
r_{1\;2\;p} ~=~ \frac{\tan\;(\theta_1 - \theta_2)}{\tan\;(\theta_1 + \theta_2)}
\end{displaymath}
There is a more general form for the instantaneous power \(p(t) = v(t)\;i(t)\) in an electrical
circuit than the one in Example 3.22. The voltage \(v(t)\) and current \(i(t)\) can be
given by
\begin{align*}
v(t) ~&=~ V_m \;\cos\;(\omega t + \theta)~,\\
i(t) ~&=~ I_m \;\cos\;(\omega t + \phi)~,
\end{align*}
where \(\theta\) is called the phase angle. Show that \(p(t)\) can be written as
\begin{displaymath}
p(t) ~=~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;(\theta - \phi) ~+~
\tfrac{1}{2}\,V_m \; I_m \;\cos\;(2\omega t + \theta + \phi) ~.
\end{displaymath}
For Exercises 10-15, prove the given identity or inequality
for any triangle \(\triangle\,ABC\).
\(\sin\;A \;+\; \sin\;B \;+\; \sin\;C ~=~
4\;\cos\;\tfrac{1}{2}A~\cos\;\tfrac{1}{2}B~\cos\;\tfrac{1}{2}C\)
(Hint: Mimic Example
3.18 using \((\sin\;A \;+\; \sin\;B) \;+\; (\sin\;C \;-\; \sin\;(A+B+C))\).)
\(\cos\;A \;+\; \cos\;(B-C) ~=~ 2\;\sin\;B~\sin\;C\)
\(\sin\;2A \;+\; \sin\;2B \;+\; \sin\;2C ~=~ 4\;\sin\;A~\sin\;B~\sin\;C\)
(Hints: Group
\(\sin\;2B\) and \(\sin\;2C\) together, use the double-angle formula for \(\sin\;2A\), use Exercise
11.)
\(\dfrac{a-b}{a+b} ~=~ \dfrac{\sin\;A \;-\; \sin\;B}{\sin\;A \;+\; \sin\;B}\)
\(\cos\;\tfrac{1}{2}A ~=~ \sqrt{\dfrac{s\;(s-a)}{bc}}~~\) and
\(~~\sin\;\tfrac{1}{2}A ~=~ \sqrt{\dfrac{(s-b)\;(s-c)}{bc}}\;\), where
\(s=\tfrac{1}{2}(a+b+c)\)
(Hint: Use the Law of Cosines to show that \(2bc\;(1 + \cos\;A) ~=~
4s\;(s-a)\).)
\(\tfrac{1}{2}\;(\sin\;A \;+\; \sin\;B) ~\le~
\sin\;\tfrac{1}{2}(A+B)\)
(Hint: Show that \(\sin\;\tfrac{1}{2}(A+B) \;-\;
\tfrac{1}{2}\;(\sin\;A \;+\; \sin\;B) \;\ge\; 0\).)
In Example 3.20, which angles \(A\), \(B\), \(C\) give the maximum value of
\(\cos\;A \;+\; \cos\;B \;+\; \cos\;C\;\)?