# Examples on Trigonometric Ratios and Functions Set 1

Go back to  'Trigonometry'

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

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