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Mathematics LibreTexts

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:

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


We will prove the first formula; the proofs of the others are similar (see Exercises 1-3). We see that
\[\nonumber \require{cancel} \begin{align*}
 \sin\;(A+B) ~+~ \sin\;(A-B) ~&=~ (\sin\;A~\cos\;B ~+~ \cancel{\cos\;A~\sin\;B}) ~+~
  (\sin\;A~\cos\;B ~-~ \cancel{\cos\;A~\sin\;B})\\ \nonumber 
 &=~ 2\;\sin\;A~\cos\;B ~,
\end{align*}\]
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:

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


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
\[\nonumber \begin{align*}
 \cos\;A ~+~ \cos\;B ~&=~ \cos\;(x+y) ~+~ \cos\;(x-y)\\ \nonumber
 &=~ 2\;\cdot\;\tfrac{1}{2}(\cos\;(x+y) ~+~ \cos\;(x-y))\\ \nonumber
 &=~ 2\;\cos\;x~\cos\;y\qquad\qquad\text{(by Equation \ref{eqn:p2scoscos})}\\ \nonumber
 &=~ 2\;\cos\;\tfrac{1}{2}(A+B)~\cos\;\tfrac{1}{2}(A-B) ~.
\end{align*}\]
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 \(\triangle\,ABC \),

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

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*}\]

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 \(\triangle\,ABC \),

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

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*}\]

Example 3.18

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

Solution:

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

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

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

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

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
\[\nonumber
  \cos^2 \;\tfrac{1}{2}(A+B) ~-~ \cos\;\tfrac{1}{2}(A-B)~\cos\;\tfrac{1}{2}(A+B) ~+~ 2u ~=~ 0 ~,
\]
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,
\[\nonumber
  \cos\;\tfrac{1}{2}(A+B) ~=~ \frac{\cos\;\tfrac{1}{2}(A-B) \;\pm\;
   \sqrt{\cos^2 \;\tfrac{1}{2}(A-B) - 4(1)(2u)}}{2} ~~,
\]
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
\[\nonumber
  \cos^2 \;\tfrac{1}{2}(A-B) \;- \; 8u ~\ge~ 0 \quad\Rightarrow\quad u ~\le~ \tfrac{1}{8}\;
  \cos^2 \;\tfrac{1}{2}(A-B) ~\le~ \tfrac{1}{8} \quad\Rightarrow\quad
  \;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~\le~ \tfrac{1}{8} ~.
\]
 

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

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

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:

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

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

\[\nonumber p(t) ~=~ v(t)\;i(t) ~,\]

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

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

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

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

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*}\]

 

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