Sum and Difference Identities

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Introduction:

In this section, we are going to explore identities involving the sum and difference of two angles and apply to the fundamental trignometric functions,i.e., \(\sin (A \pm B)\), \(\cos (A \pm B)\), \(\tan (A \pm B)\). These identities are known as sum and difference identities.

Expansions of these Identities:

  1. \(\sin (A + B) = \sin A\cos B + \cos A\sin B\)
  2. \(\sin (A - B) = \sin A\cos B - \cos A\sin B\)
  3. \(\cos (A + B) = \cos A\cos B - \sin A\sin B\)
  4. \(\cos (A - B) = \cos A\cos B + \sin A\sin B\)
  5. \(\tan (A + B) = \frac{{\tan\ A + \tan\ B}}{{1 - \tan\ A \tan\ B}}\)
  6. \(\tan (A - B) = \frac{{\tan\ A - \tan\ B}}{{1 + \tan\ A \tan\ B}}\)

These expansions of identities are very useful and can help to simplify many trigonometric expressions and equations.. Let's see their proves and applications.


Proof of \(\sin (A + B)\):

Observe the following figure carefully :

Sum and Difference Identities

Let us take a few moments to understand the various aspects of this figure.

We can note the following:

  • \(\angle VPQ = A\)
  • \(\angle RPV = B\)
  • From \(R\), \(RT \bot PV\) has been drawn onto \(PV\)
  • From \(T\), \(TU \bot RS\) has been drawn onto \(RS\)

Now, It is easy to prove that \(\angle 2 = {90^\circ} - A\) and \(\angle 1 = A\).

[use the facts that \(UT\parallel PQ\), \(\angle UTP = \;A\), \(\angle UTP = 90^\circ  - \angle 2\)]

So,

\[\begin{align} \sin\ (A + B) &= \frac{{RS}}{{PR}} = \frac{{RU + US}}{{PR}}\\  \qquad\qquad\quad &= \frac{{RU + TW}}{{PR}}\left(∵ {US = TW} \right) \end{align}\]

Also, we make the following observations on cos A, sin B, sin A, cos B:

  • \(\frac{{RU}}{{RT}} = \cos\; (\angle 1) = \cos A \Rightarrow RU = RT\;\cos A\)

  • \(\frac{{RT}}{{PR}} = \sin\; B \Rightarrow RT = PR \sin\ B \)

  • \(\frac{{TW}}{{PT}} = \sin\ A \Rightarrow TW = PT \sin\ A\)

  • \(\frac{{PT}}{{PR}} = \cos\; B \Rightarrow PT = PR\cos\; B\)

Now, we make use of all these facts:

\[\begin{align}&\sin\; (A + B) = \frac{{RU + TW}}{{PR}}\\ &\qquad\qquad\;\;\;= \frac{{RT\;\cos A + PT\;\sin A}}{{PR}}\\ &\qquad\qquad\;\;\;= \frac{{\left( {PR\;\sin B} \right){\mathop{\rm cos\;A}\nolimits} + (PR \cos\ B)\sin\ A}}{{PR}}\\ &\qquad\qquad\;\;\;= \frac{{PR\;({\mathop{\rm \sin\ A}\nolimits} {\;\mathop{\rm cos\;B}\nolimits} + {\mathop{\rm cos\;A\;sin\;B}\nolimits} )}}{{PR}}\end{align}\]

Thus,

\[\fbox{sin  (A + B) =  sin A cos B + cos A sin B}\]


Proof of \(cos(A + B)\):

Let’s determine an analogous relation for \(\cos\ (A+B)\) by using the same figure.

Sum and Difference Identities

We have,

\[\begin{align} \cos\,(A + B)\, &= \frac{{PS}}{{PR}}\, = \,\frac{{PW - SW}}{{PR}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \qquad\qquad\quad\;\;&= \frac{{PW - UT}}{{PR}}\,\,\,\,\,\,\,(∵ SW = UT) \end{align}\]

Note the following:

  • \(\begin{align}
      PW &= PT\cos A = \left( {PR\cos B} \right)\cos A \hfill \\
       &= PR\cos A\cos B \hfill \\ 
    \end{align} \)
  • \(\begin{align}
      UT &= RT\sin A = \left( {PR\sin B} \right)\sin A \hfill \\
       &= PR\sin A\sin B \hfill \\ 
    \end{align} \)

Thus,

\[\begin{align}
  \cos \;(A + B) &= \frac{{PW - UT}}{{PR}} \hfill \\
  \qquad \qquad \quad  &= \frac{{PR\;\cos \;A\cos \;B - PR\sin \;A\sin \;B}}{{PR}} \hfill \\
  \qquad \qquad \quad  &= \frac{{PR\;(\cos \;A\cos \;B - \sin A\;\sin B)}}{{PR}} \hfill \\ 
\end{align} \]

To conclude,

\[\fbox{ cos  (A + B) = cos A cos B -  sin A sin B}\]


What about \(\tan(A + B)\)?

Well, now we can make use of the expressions for \(\begin{align}\sin\left( {A + B} \right){\text{ }}and{\text{ }}cos\left( {A + B} \right):\end{align}\)

\[\begin{align}&\tan \;(A + B) = \frac{{\sin\; (A + B)}}{{\cos \;(A + B)}}\\ &\qquad\qquad\quad= \frac{{\sin A\;\cos B + \cos A\;\sin B}}{{\cos A\;\cos B - \sin A\;\sin B}}\end{align}\]

If we divide the numerator and the denominator on the right by cos A cos B,

We have:

\[\fbox{$\displaystyle tan  (A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$}\]

yesChallenge 1: Similarly, can you determine expression for \(\sin\; (A - B)\), \(\cos\; (A - B)\) and \(\tan \;(A - B)\) ?

We will obtain:

  •      \(\sin (A - B) = \sin A\cos B - \cos A\sin B\)
  •      \(\cos (A - B) = \cos A\cos B + \sin A\sin B\)
  •     \(\begin{align}\tan (A - B) = \frac{{\tan A - \tan B}}{{1 + \tan A{\mathop{\rm tan B}\nolimits}}}\end{align}\)

Now, Let us see some solved examples with some challenge questions.


Solved Examples:

Example 1:  Find the values of \(\sin {75^\circ}\) and \(\tan {75^\circ}\).

Solution: Taking \(A = {45^\circ }\)and\(B = {30^\circ }\) ,we have:

\[\begin{align}
  \sin 75^\circ  &= \sin (45^\circ  + 30^\circ ) \hfill \\
   &= \sin 45^\circ \cos 30^\circ  + \cos 45^\circ \sin 30^\circ  \hfill \\
   &= \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} \times \frac{1}{2} \hfill \\
   &\Rightarrow \boxed{\sin 75^\circ  = \frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}} \hfill \\ 
\end{align} \]

Now,

\[\begin{align}
  \tan 75^\circ  &= \tan (45^\circ  + 30^\circ ) \hfill \\
   &= \frac{{\tan 45^\circ  + \tan 30^\circ }}{{1 - \tan 45^\circ \tan 30^\circ }} \hfill \\
   &= \frac{{1 + \frac{1}{{\sqrt 3 }}}}{{1 - 1 \times \frac{1}{{\sqrt 3 }}}} \hfill \\
   &\Rightarrow \boxed{\tan 75^\circ  = \frac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}}} \hfill \\ 
\end{align} \]

yesChallenge 2: Find the value of \(\cos {75^\circ }\).


Example 2: Prove the following double angle formulae:

  1. \(\sin 2\theta = 2\sin \theta\; \cos \theta \)

  2. \(\cos 2\theta = 2{\cos ^2}\theta - 1\)   

           \(\begin{align}&\quad = 1 - 2{\sin ^2}\theta \\&\quad = {\cos ^2}\theta - {\sin ^2}\theta \end{align}\)

Solution: We apply the sum identities for \(A = \theta \) and \(B = \theta \):

  1. \(\sin 2\theta = \sin\; (\theta + \theta )\)
    \(\begin{align}& \qquad\;= \sin\; \theta \cos\; \theta + \cos\; \theta \sin \;\theta \\ &\qquad\;= 2 \sin \theta \cos \theta \end{align}\)

  2. \(\cos 2\theta = \cos (\theta + \theta )\)
    \(\begin{align}& \qquad\;\;= \cos \theta \cos \theta - \sin \theta \sin \theta \\ &\qquad\;\;= {\cos ^2}\theta - {\sin ^2}\theta\qquad\qquad\qquad \cdot \cdot \cdot (i) \end{align}\)

If we substitute \({\sin ^2}\theta = 1 - {\cos ^2}\theta \) in \((i)\)

We obtain

\(\cos 2\theta = 2{\cos ^2}\theta - 1\)

On the other hand, if we substitute \(\cos ^2\theta = 1 - {\sin ^2}\theta \) in \((i)\),we obtain \(\cos 2\theta = 1 - 2{\sin ^2}\theta \)

yesChallenge 3: Prove \(\begin{align}\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\end{align}\)

⚡Tip: Use Property, \(tan(A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A{\text{tanB}}}}\)


Example 3: Prove the following half angle formulae:

  1. \(\begin{align}\sin \frac{A}{2} = \sqrt {\frac{{1 - \cos A}}{2}}\end{align} \)
  2. \(\begin{align}\cos \frac{A}{2} = \sqrt {\frac{{1 + \cos A}}{2}} \end{align}\)
  3. \(\begin{align}tan \frac{A}{2} = \sqrt {\frac{{1 - \cos A}}{{1 + \cos A}}} = \frac{{1 - \cos A}}{{\sin\; A}}\\ = \frac{{{\mathop{\rm sin\;A}\nolimits} }}{{1 + \cos A}}\end{align}\)

Solution: We have (using the result of the previous example)

\[\begin{align}\cos A = \cos \left( {2 \times \frac{A}{2}} \right) = 1 - 2{\sin ^2}\frac{A}{2}\\ \qquad= 2{\cos ^2}\frac{A}{2} - 1\end{align}\]

Thus, 

\[\begin{gathered}
  \cos A = 1 - 2{\sin ^2}\frac{A}{2} \Rightarrow \boxed{\sin \frac{A}{2} = \sqrt {\frac{{1 - \cos A}}{2}} } \hfill \\
  \cos A = 2{\cos ^2}\frac{A}{2} - 1 \Rightarrow \boxed{\cos \frac{A}{2} = \sqrt {\frac{{1 + \cos A}}{2}} } \hfill \\ 
\end{gathered} \]
Dividing these two gives,

\[\frac{{\sin \frac{A}{2}}}{{\cos \frac{A}{2}}} = \boxed{\tan \frac{A}{2} = \sqrt {\frac{{1 - \cos A}}{{1 + \cos A}}} }\]
If we multiply the numerator and the denominator by \(\sqrt {1 - \cos A}\) , we have

\[\boxed{\tan \frac{A}{2} = \frac{{1 - \cos A}}{{\sqrt {1 - {{\cos }^2}A} }} = \frac{{1 - {\text{cosA}}}}{{\sin A}}}\]
Similarly, if we had multiplied by \(\sqrt {1 + \cos A} \) we would have obtained

\[\boxed{\tan \frac{A}{2} = \frac{{\sqrt {1 - {{\cos }^2}A} }}{{1 + \cos A}} = \frac{{{\text{sin}}\;{\text{A}}}}{{1 + \cos A}}}\]


Example 4: Find the values of \(\sin 7{\frac{1}{2}^\circ }\) and \(\cos 7{\frac{1}{2}^\circ }\).

Solution: Using the results of the previous example, we have 

\[\begin{align}&\sin 7{\frac{1}{2}^\circ } = \sqrt {\frac{{1 - \cos {{15}^\circ}}}{2}} \\&\cos 7{\frac{1}{2}^\circ } = \sqrt {\frac{{1 + \cos {{15}^\circ }}}{2}} \end{align}\]

We can calculate \(\cos {15^\circ}\) as follows:

\[\begin{align}&\cos {15^\circ } = \cos \left( {{{45}^\circ } - {{30}^\circ }} \right)\\ &\qquad\;\;\;= \cos {45^\circ }\cos {30^\circ } + \sin {45^\circ}\sin {30^\circ }\\ &\qquad\;\;\;= \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} \times \frac{1}{2}\\ &\qquad\;\;\;= \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\end{align}\]

Thus,

\[\begin{align}& \sin 7{\frac{1}{2}^\circ } = \sqrt {\frac{{1 - \left( {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)}}{2}} \\&\qquad\quad= \sqrt {\frac{{2\sqrt 2 - \sqrt 3 - 1}}{{4\sqrt 2 }}} \\&\qquad\quad= \sqrt {\frac{{8 - 2\sqrt 6 - 2\sqrt 2 }}{{16}}} \;{\rm{  }}\\ \end{align}\]
\[ \Rightarrow \boxed{\sin 7{{\frac{1}{2}}^\circ } = \frac{{\sqrt {2(4 - \sqrt 6  - \sqrt 2 } )}}{4}}\]

 Similarly,

\[\begin{align}&\cos 7{\frac{1}{2}^\circ } = \sqrt {\frac{{1 + \cos {{15}^\circ }}}{2}} \\ &\qquad\quad= \sqrt {\frac{{1 + \left( {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)}}{2}} \\ \end{align}\]

\[ \Rightarrow \boxed{\cos 7{{\frac{1}{2}}^\circ } = \frac{{\sqrt {2\left( {4 + \sqrt 6  + \sqrt 2 } \right)} }}{4}}\]

yesChallenge 4: Prove the following.

\[{\sin ^2}7{\frac{1}{2}^\circ} + {\cos ^2}7{\frac{1}{2}^\circ } = 1\]

⚡Tip: Use results of example 4.


Example 5: Determine expressions for \(\sin 3\theta ,\) and \(\tan 3\theta \).

Solution: We have:

\[\begin{align}
  \sin 3\theta  &= \sin \left( {2\theta  + \theta } \right) \hfill \\
  \qquad \;\; &= \sin 2\theta \cos \theta  + \cos 2\theta \sin \theta  \hfill \\
  \qquad \;\; &= \left( {2\sin \theta \cos \theta } \right)\cos \theta  + \left( {1 - 2{{\sin }^2}\theta } \right)\sin \theta  \hfill \\
  \qquad \;\; &= 2\sin \theta co{s^2}\theta  + \sin \theta  - 2{\sin ^3}\theta  \hfill \\
  \qquad \;\; &= 2\sin \theta (1 - {\sin ^2}\theta ) + \sin \theta  - 2{\sin ^3}\theta  \hfill \\
   &\Rightarrow \boxed{\sin 3\theta \; = 3\sin \theta  - 4{{\sin }^3}\theta } \hfill \\ 
\end{align} \]

Now, we have

\[\eqalign{
  & \tan 3\theta  = \frac{{\sin 3\theta }}{{\cos 3\theta }}  \cr 
  & \qquad \;\; = \frac{{3\sin \theta  - 4{{\sin }^3}\theta }}{{4{{\cos }^3}\theta  - 3\cos \theta }}  \cr 
  & \qquad \;\; = \frac{{\sin \theta \;(3 - 4{{\sin }^2}\theta )}}{{\cos \theta \;(4{{\cos }^2}\theta  - 3)}}  \cr 
  & \qquad \;\; = \frac{{\tan \theta \;(3{{\sec }^2}\theta  - 4{{\tan }^2}\theta )}}{{4 - 3\;(1 + {{\sec }^2}\theta )}}  \cr 
  & \qquad \;\; = \frac{{\tan \theta \;{\text{ }}(3\;(1 + {{\tan }^2}\theta ) - 4\;{{\tan }^2}\theta )}}{{4 - 3(1 + {{\tan }^2}\theta )}}  \cr 
  & \qquad \;\; = \frac{{\tan \theta \;(3 - {{\tan }^2}\theta )}}{{1 - 3{\mkern 1mu} {{\tan }^2}\theta }} \cr} \]

Thus, we have

\[\boxed{\tan 3\theta  = \frac{{3\tan \theta  - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}}\]

yesChallenge 5: Determine expressions for \(\cos 3\theta \).


Example 6: Evaluate \(sin {18^\circ }\).

Solution: (a) Let \(\theta = {18^\circ}\).Thus, \(5\theta = {90^\circ}\),and so:

\[\begin{align}&2\theta = {90^\circ } - 3\theta \\ &\Rightarrow \quad \sin 2\theta = \sin \left( {{{90}^\circ } - 3\theta } \right)\end{align}\]

Using the fact \(\sin (90^\circ  - \theta ) = cos\theta \), we have:

\[\begin{align}  &\sin 2\theta  = \cos 3\theta  \hfill \\
  &\Rightarrow \quad 2\sin \theta \cos \theta  = 4co{s^3}\theta  - 3\cos \theta  \hfill \\
  &\Rightarrow \quad 2\sin \theta  = 4{\cos ^2}\theta  - 3 \hfill \\
  &\Rightarrow \quad2\sin \theta  = 1 - 4{\sin ^2}\theta (\because {\cos ^2}\theta  = 1 - {\sin ^2}\theta ) \hfill \\  &\Rightarrow \quad 4{\sin ^2}\theta  - 2\sin \theta  - 1 = 0 \hfill \\
\end{align} \]

This is a quadratic in \(\sin \theta \). Solving this and taking the positive root (why?), We have:

\[\sin \theta  = \boxed{\sin 18^\circ  = \frac{{\sqrt 5 {\text{ }} - 1}}{4}}\]

yesChallenge 6: Evaluate \(cos {36^\circ }\).

⚡Tip: Use property \(\cos 2\theta  = 1 - 2{\sin ^2}\theta \)


 

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