Examples on Trigonometric Ratios and Functions Set 1

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Example-1

Let \({\theta _1}\) and \({\theta _2}\) be two different roots of the trigonometric equation 

\[a\cos \theta  + b\sin \theta  = c\]

Find the value of \(\cos ({\theta _1} + {\theta _2})\)
Solution: The equation has two variable quantities \(\sin \theta \;{\text{and}}\;\cos \theta \). Using the following transformations, we can reduce this equation to one containing a single variable quantity:

\[\begin{align}\cos \theta = \frac{1-{{\tan }^{2}}{\begin{align}\frac{\theta }{2}\end{align}}}{1+{{\tan }^{2}}{\begin{align}\frac{\theta }{2}\end{align}}}\;\;\,, \quad \sin \theta =\frac{2\tan {\begin{align}\frac{\theta }{2}\end{align}}}{1+{{\tan }^{2}}{\begin{align}\frac{\theta }{2}\end{align}}}\end{align}\]

Substituting these into the original equation gives:

\[\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\left( \frac{1-{{\tan }^{2}}{\begin{align}\frac{\theta }{2}\end{align}}}{1+{{\tan }^{2}}{\begin{align}\frac{\theta }{2}\end{align}}} \right)+ b\left( \frac{2\tan{\begin{align} \frac{\theta }{2}\end{align}}}{1+{{\tan }^{2}}{\begin{align}\frac{\theta }{2}\end{align}}} \right) = c  \\   \Rightarrow \quad  (a + c){\tan ^2}\frac{\theta }{2}- 2b\tan \frac{\theta }{2} + (c - a) = 0 \\ 
\end{align}\]

This is a quadratic equation in \(\begin{align}\tan \frac{\theta }{2}\end{align}\) and it is given that it has two different roots \({\theta _1}\) and \({\theta _2}\) . 
Therefore,

\[\tan \frac{{{\theta _1}}}{2} + \tan \frac{{{\theta _2}}}{2} = \frac{{2b}}{{a + c}};\;\;\;\tan \frac{{{\theta _1}}}{2}\tan \frac{{{\theta _2}}}{2} = \frac{{c - a}}{{c + a}}\]

Now, we evaluate

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {\frac{{{\theta _1}}}{2} + \frac{{{\theta _2}}}{2}} \right) = \frac{\begin{align}\tan \frac{{{\theta }_{1}}}{2}+\tan \frac{{{\theta }_{2}}}{2}\end{align}}{\begin{align}1-\tan\frac{{{\theta }_{1}}}{2}\tan \frac{{{\theta }_{2}}}{2}\end{align}}= \frac{\begin{align}\frac{2b}{a+c}\end{align}}{\begin{align}1-\frac{c-a}{c+a}\end{align}}\\\   &\quad\Rightarrow \ \tan \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right) = \frac{b}{a}\\ \end{align} \]

Finally,


\[\begin{align} \cos ({{\theta }_{1}}+{{\theta }_{2}})=\frac{1-{{\tan }^{2}}\left({\begin{align} \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \end{align}}\right)}{1+{{\tan }^{2}}\left( \begin{align}\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \end{align}\right)} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\frac{1-\begin{align}\frac{{{b}^{2}}}{{{a}^{2}}}\end{align}}{1+\begin{align}\frac{{{b}^{2}}}{{{a}^{2}}}\end{align}}=\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \end{align}\]

This question should have given you a basic idea on how multiple and sub-multiple angle relations are used to solve trigonometry questions.

Example-2

Evaluate

(a) \(sin\text{ }{{18}^{\circ }}\)                        (b) \(cos\text{ }{{36}^{\circ }}\)

Solution: (a) If we let \(\theta\) = 18º, then 5 \(\theta\) = 90º:

\[\begin{align}&\Rightarrow 2\theta  = 90 - 3\theta\\&\Rightarrow \sin 2\theta  = \sin (90 - 3\theta ) = \cos 3\theta   \\&\Rightarrow 2\sin \theta \cos \theta  = 4{\cos ^3}\theta  - 3\cos \theta  \\   &\Rightarrow  2\sin \theta  - 4{\cos ^2}\theta  + 3 = 0  \\&\Rightarrow  2\sin \theta  - 4(1 - {\sin ^2}\theta ) + 3 = 0  \\&\Rightarrow 4{\sin ^2}\theta  + 2\sin \theta  - 1 = 0   \end{align} \]

Solving this quadratic and picking up the positive root gives us the value of \(sin\text{ }{{18}^{\circ }}\).

\(\begin{align}\sin 18^\circ = \frac{{\sqrt 5 - 1}}{4}\end{align}\)

\(\begin{align}&(b)\qquad \cos 36  = 1 - 2{\sin ^2}18 \\&\qquad\qquad \qquad= 1 - 2{\left( {\frac{{\sqrt 5 - 1}}{4}} \right)^2} \\    &\qquad\qquad \qquad= \frac{{\sqrt 5 + 1}}{4}  \end{align} \)

Example-3

Evaluate

(a) \({{P}_{1}}=\cos 20{}^\text{o}\ \cos 40{}^\text{o}\ \cos 60{}^\text{o}\cos 80{}^\text{o}\)       (b) \({{P}_{2}}=\sin 10{}^\text{o}\ \sin 30{}^\text{o}\ \sin 50{}^\text{o}\ \sin 70{}^\text{o}\)

Solution: (a) The approach in such questions is to combine the appropriate terms, trying to convert the product into a sum:


\[\begin{align}&\cos 20{}^\text{o}\cos 40{}^\text{o}=\frac{1}{2}\left[ \cos (20{}^\text{o}+40{}^\text{o})+\cos (40{}^\text{o}-20{}^\text{o}) \right] \\   \ \ \ \ \ \ \ \ \ \,&\qquad\qquad\quad\;\;\;=\frac{1}{2}\left[ \cos 60{}^\text{o}+\cos 20{}^\text{o} \right]=\frac{1}{2}\left[ \frac{1}{2}+\cos 20{}^\text{o} \right] \\  & \Rightarrow {{P}_{1}}=\frac{1}{4}\left[ \frac{1}{2}+\cos 20{}^\text{o} \right]\cos 80{}^\text{o} \\  & \,\,\,\,\,\,\quad\;\;=\frac{1}{4}\left[ \frac{1}{2}\cos 80{}^\text{o}+\cos 20{}^\text{o}\cos 80{}^\text{o} \right] \\ \end{align}\]
Now,

\[\begin{align}\  & \cos 20{}^\text{o}\cos 80{}^\text{o}=\frac{1}{2}\left[ \cos 100{}^\text{o}+\cos 60{}^\text{o} \right] \\  & \qquad\qquad\qquad=\frac{1}{2}\left[ -\cos 80{}^\text{o}+\cos 60{}^\text{o} \right] \\  & \Rightarrow {{P}_{1}}=\frac{1}{16} \\ \end{align}\]


(b) For  \(\begin{align}{{P}_{2}},\ \sin {{30}^{\circ }}=\frac{1}{2}.\end{align}\)  Now,


\[\sin 10{}^\text{o}\sin 50{}^\text{o}=\frac{1}{2}\left[ \cos 40{}^\text{o}-\cos 60{}^\text{o} \right]\]

\[\begin{align}&\Rightarrow\quad {{P}_{2}}=\frac{1}{4}\left[ (\cos 40{}^\text{o}-\cos 60{}^\text{o})\sin 70{}^\text{o} \right] \\  \,\,\,\,\,\,&\qquad\;\;\quad=\frac{1}{4}\left[ \cos 40{}^\text{o}\sin 70{}^\text{o}-\cos 60{}^\text{o}\sin 70{}^\text{o} \right] \\ \,\,\,\,\,\,&\qquad\;\;\quad=\frac{1}{4}\left[ \frac{1}{2}(\sin 110{}^\text{o}+\sin 30{}^\text{o})-\frac{1}{2}\sin 70{}^\text{o} \right] \\ \end{align}\]

        Since \(\sin 110{}^\text{o}=\sin (180{}^\text{o}-70{}^\text{o})=\sin 70{}^\text{o}\) , we have

\[\Rightarrow \quad  {{P}_{2}}=\frac{1}{16}\]

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