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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