# Examples on Trigonometric Ratios and Functions Set 5

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Example-14

A right angle is divided into three positive parts . Prove that for all the possible divisions, $$\tan \alpha + \tan \beta + \tan \gamma > 1 + \tan \alpha \tan \beta \tan \gamma$$

Solution: The trick lies in manipulating the term

$T = \tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma$

to

$T = \frac{1}{{\cos \alpha \;\cos \beta \cos \gamma }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{verify}}} \right)$

and then realizing that T will be minimum when the denominator  $$\cos \alpha \cos \beta \cos \gamma$$  is maximum, which by symmetry occurs at

\begin{align}&\;\;\;\;\; \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha = \beta = \gamma = \frac{\pi }{6}\;\;\;\;\;\; \Rightarrow & {\left( {\cos \alpha \cos \beta \cos \gamma } \right)_{\max }} = \frac{{3\sqrt 3 }}{8} \\ &\qquad\Rightarrow\;\;\;\;\; {T_{\min }} = \frac{8}{{3\sqrt 3 }} > 1 \\ \end{align}

From this, the result follows.

If you are not convinced with using symmetry arguments, here’s a more rigorous approach:

\begin{align}&T = \frac{1}{{\cos \alpha \cos \beta \cos \gamma }} \\ \;\;\; &\;\;= \frac{2}{{\left( {\cos \left( {\beta + \gamma } \right) + \cos \left( {\beta - \gamma } \right)} \right)\cos \alpha }} \\ \;\;\; &\;\;= \frac{2}{{(\sin \alpha + \cos (\beta - \gamma ))\cos \alpha }} \\ \end{align}

Assuming $$\alpha$$ fixed for a while, the minimum value of T for a given occur when $$\beta = \gamma$$ :

${T_{\min }}(\alpha ) = \frac{2}{{(\sin \alpha + 1)\cos \alpha }}$

On the other hand, if we had assumed b fixed, the minimum value of T would occur for $$\alpha = \gamma$$ . This proves that the overall minimum of T occurs when  $$\alpha = \beta = \gamma = \frac{\pi }{6}$$ .

Example-15

If  $$\alpha ,\;\beta ,\gamma \in \left( {0,\frac{\pi }{2}} \right)$$  , prove that

$\sin \alpha + \sin \beta + \sin \gamma > \sin (\alpha + \beta + \gamma )$

Solution: The solution is in two easy steps, if we can recall that
Step-I :

\begin{align}&\sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma ) \\ \,\,\,\,\,\,\,\,\,\,\, &\quad\;\;\;= 4\sin \left( {\frac{{\alpha + \beta }}{2}} \right)\sin \left( {\frac{{\beta + \gamma }}{2}} \right)\sin \left( {\frac{{\gamma + \alpha }}{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{verify}}} \right) \\\end{align}

Step-II : Since $$\alpha ,\beta ,\;\gamma \in \left( {0,\frac{\pi }{2}} \right)$$  , we have

$\frac{{\alpha + \beta }}{2},\;\;\frac{{\beta + \gamma }}{2},\;\;\frac{{\gamma + \alpha }}{2}\;\; \in \;\;\left( {0,\frac{\pi }{2}} \right)$

so their sines are positive.
From these two steps, the result follows.

Example-16

Evaluate $$S = {\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8}$$

Solution:  Since we have $$\sin \theta = \sin (\pi - \theta ),$$

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \frac{{5\pi }}{8} = \sin \frac{{3\pi }}{8}\;\;{\text{and}}\;\sin \frac{{7\pi }}{8} = \sin \frac{\pi }{8} \\ &\Rightarrow\quad S = 2\left( {{{\sin }^4}\frac{\pi }{8} + {{\sin }^4}\frac{{3\pi }}{8}} \right) \\ \end{align}

Now,

\begin{align}&{\sin ^4}\frac{\pi }{8} = {\left( {{{\sin }^4}\frac{\pi }{8}} \right)^2} = {\left( {\frac{{1 - \cos (\pi /4)}}{2}} \right)^2} \\\,\,\,\,\,\,\,\,&\qquad\qquad\qquad\qquad\;= \frac{1}{4}{\left( {1 - \frac{1}{{\sqrt 2 }}} \right)^2} \\ \end{align}

Similarly, $${\sin ^4}\frac{{3\pi }}{8} = \frac{1}{4}{\left( {1 + \frac{1}{{\sqrt 2 }}} \right)^2}$$

$\Rightarrow S = \frac{3}{2}$

Example-17

Evaluate

$S = \left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 + \cos \frac{{5\pi }}{8}} \right)\left( {1 + \cos \frac{{7\pi }}{8}} \right)$

Solution: As in the previous example,

$\cos \frac{{5\pi }}{8} = - \cos \frac{{3\pi }}{8},\;\cos \frac{{7\pi }}{8} = - \cos \frac{\pi }{8}$

\begin{align}&\Rightarrow \qquad S = \left( {1 - {{\cos }^2}\frac{\pi }{8}} \right)\left( {1 - {{\cos }^2}\frac{{3\pi }}{8}} \right) \\ \,\,\,\, &\qquad\qquad= {\sin ^2}\frac{\pi }{8}{\sin ^2}\frac{{3\pi }}{8} \\ \,\,\,\, &\qquad\qquad= \frac{1}{4}\left( {1 - \cos \frac{\pi }{4}} \right)\left( {1 - \cos \frac{{3\pi }}{4}} \right) \\ \,\,\,\, &\qquad\qquad= \frac{1}{4}\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\left( {1 + \frac{1}{{\sqrt 2 }}} \right) = \frac{1}{8} \\ \end{align}

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