# Examples on Trigonometric Ratios and Functions Set 7

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Example-22

Let A and B be angles such that

\begin{align}&\sqrt 2 \cos A = {\cos ^3}B + \cos \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right) \\ & \sqrt 2 \sin A = - {\sin ^3}B + \sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right) \\ \end{align}

Evaluate  $$\sin (A - B)$$.

Solution:

\begin{align}&\sin (A - B) = \sin A\cos B - \cos A\sin B \\ \,\,\,\,\,\,\, &\qquad\qquad\;\;\;= \frac{1}{{\sqrt 2 }}( - {\sin ^3}B + \sin B)\cos B - \frac{1}{{\sqrt 2 }}({\cos ^3}B + \cos B)\sin B \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\;\;\;= \frac{{ - 1}}{{2\sqrt 2 }}\sin 2B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {\text{3}} \right){\text{ }}\left( {{\text{Verify}}} \right) \\ \end{align}

By (1)2 + (2)2, we have

\begin{align}2&= \left\{ {{{\cos }^6}B + {{\sin }^6}B} \right\} + {\cos ^2}B + {\sin ^2}B + \left\{ {2({{\cos }^4}B - {{\sin }^4}B)} \right\} \\&= \left\{ {{{({{\cos }^2}B + {{\sin }^2}B)}^3} - 3{{\cos }^2}B{{\sin }^2}B({{\cos }^2}B + {{\sin }^2}B)} \right\} + 1 \\ &\qquad\qquad\qquad\qquad\qquad+ 2\left\{ {{{\cos }^2}B - {{\sin }^2}B} \right\} \\&= 1 - 3{\cos ^2}B{\sin ^2}B + 1 + 2\cos 2B \\ \Rightarrow \qquad \qquad \quad 0 &= - \frac{3}{4}{\sin ^2}2B + 2\cos 2B \\ \Rightarrow \quad \quad 8\cos 2B &= 3{\sin ^2}2B = 3(1 - {\cos ^2}2B) \\ \Rightarrow \quad \quad \;\; \cos 2B &= \frac{1}{3} \qquad\qquad\qquad\qquad\qquad({\text{Verify}}) \\ \Rightarrow \!\qquad \quad \sin 2B &= \pm \frac{{2\sqrt 2 }}{3} \\ \Rightarrow \quad \sin (A - B) &= \pm \frac{1}{3} \qquad\qquad\qquad\qquad\qquad{\text{from}}\;(3) \\ \end{align}

Example-23

Find the range of (a) $$f(x) = \frac{{\tan 3x}}{{\tan x}}$$    (b)  $$f(x) = \frac{{\tan 3x}}{{\tan 2x}}$$

Solution:

(a)  \begin{align}&\quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = f(x) = \frac{{\left( {\frac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}} \right)}}{{\tan x}} = \frac{{3 - {{\tan }^2}x}}{{1 - 3{{\tan }^2}x}} \\ &\Rightarrow \qquad {\tan ^2}x = \frac{{y - 3}}{{3y - 1}}\\ \end{align}

Since $${\tan ^2}x$$  is necessarily non-negative, are have

$\frac{{y - 3}}{{3y - 1}} \geqslant 0\;\;\;\;\; \Rightarrow \;\;\;y \notin \left[ {\frac{1}{3},3} \right)$

Also, note that  $$y = 3 \Rightarrow \tan x = 0$$  for which the ratio  $$\frac{{\tan 3x}}{{\tan x}}$$  is non-defined. So technically speaking, it is incorrect to include 3 in the range, even though

$\mathop {\lim }\limits_{x \to 0} \;\frac{{\tan 3x}}{{\tan x}} = 3$

Therefore, the range is  $$\mathbb{R}\backslash \left[ {\frac{1}{3},3} \right]$$

(b) \begin{align}y = f(x) = \frac{{\tan 3x}}{{\tan 2x}} = \frac{{\left( {\frac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}} \right)}}{{\left( {\frac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}}\end{align}

Rearranging this yields

${\tan ^4}x - 2(2 - 3y){\tan ^2}x + 3 - 2y = 0$

This is a quadratic in  $${\tan ^2}x,$$  which means that the equation

${z^2} - 2(2 - 3y)z + 3 - 2y = 0,\;\;\;z = {\tan ^2}x$

should have at least one non-negative root

This implies that

(I)  $$D \geqslant 0 \Rightarrow y \leqslant - 1\;\;{\text{or}}\;\;y \geqslant 1$$
and
(II) Either there’s just one positive root  $$\Rightarrow f(0) < 0 \Rightarrow y > \frac{3}{2}$$

OR

Both roots are positive

\begin{align}&\Rightarrow \qquad f(0) > 0 \\ &\qquad\;\;\;\;\;\;\ {\text{and}} \\ &\qquad\;\;\;\;\;\;\frac{{ - b}}{{2a}} > 0 \;\;\;\;\; \Rightarrow \;\;\;\;y < \frac{2}{3} \,\,\,\,\,\, \Rightarrow y \leqslant - 1 \\ \end{align}

$\Rightarrow \qquad y > \frac{3}{2}\;\;{\text{or}}\;\;y \leqslant - 1$

Example-24

Evaluate
(a)  \begin{align}\tan \frac{\pi }{7}\tan \frac{{2\pi }}{7}\tan \frac{{3\pi }}{7}\end{align}

(b)   \begin{align}\left( {{{\tan }^2}\frac{\pi }{7} + {{\tan }^2}\frac{{2\pi }}{7} + {{\tan }^2}\frac{{3\pi }}{7}} \right)\;\left( {{{\cot }^2}\frac{\pi }{7} + {{\cot }^2}\frac{{2\pi }}{7} + {{\cot }^2}\frac{{3\pi }}{7}} \right)\end{align}

Solution: (a) If  \begin{align}&\theta = \frac{\pi }{7},\;\;{\text{then}}\;7\theta = \pi \quad \Rightarrow\quad 4\theta = \pi - 3\theta \\ &\qquad\qquad\qquad\qquad\qquad\Rightarrow\quad \tan 4\theta = - \tan 3\theta \\ &\Rightarrow \frac{{4\tan \theta - 4{{\tan }^3}\theta }}{{1 - 6{{\tan }^2}\theta + {{\tan }^4}\theta }} = \frac{{{{\tan }^3}\theta - 3\tan \theta }}{{1 - 3{{\tan }^2}\theta }} \\ \end{align}

Simplifying this to a polynomial equation in we have

${y^6} - 21{y^4} + 35{y^2} - 7 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)$

which has the roots   $$\tan \frac{\pi }{7},\;\tan \frac{{2\pi }}{7}.....,\;\tan \frac{{6\pi }}{7}$$       (Why?)

Since  $$\tan \frac{{6\pi }}{7} = - \;\tan \frac{\pi }{7},\;\tan \frac{{5\pi }}{7} = - \tan \frac{{2\pi }}{7},\;\tan \frac{{4\pi }}{7} = - \tan \frac{{3\pi }}{7}$$  , the product of roots P becomes

\begin{align}&\qquad \;\;P = {\tan ^2}\frac{\pi }{7}\;\;{\tan ^2}\frac{{2\pi }}{7}\;\;{\tan ^2}\frac{{3\pi }}{7} = 7\\ &\Rightarrow\quad \tan \frac{\pi }{7}\;\tan \frac{{2\pi }}{7}\;\tan \frac{{3\pi }}{7} = \sqrt 7 \\ \end{align}

(b) Note that the 6th degree polynomial equation in (1) can be reduced to the 3rd degree polynomial equation

${z^3} - 21{z^2} + 35z - 7 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)$

which has the roots  $${\tan ^2}\frac{\pi }{7},\;\;{\tan ^2}\frac{{2\pi }}{7},\;{\tan ^2}\frac{{3\pi }}{7}$$

If we let $$z = \frac{1}{u},$$  (2) changes to the cubic

$7{u^3} - 35u + 21u - 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right)$

with roots  $${\cot ^2}\frac{\pi }{7},\;{\cot ^2}\frac{{2\pi }}{7},\;{\cot ^2}\frac{{3\pi }}{7}$$

From (2),

${\tan ^2}\frac{\pi }{7} + {\tan ^2}\frac{{2\pi }}{7} + {\tan ^2}\frac{{3\pi }}{7} = 21$

and from (3),

${\cot ^2}\frac{\pi }{7} + {\cot ^2}\frac{{2\pi }}{7} + {\cot ^2}\frac{{3\pi }}{7} = 5$

$$\Rightarrow$$ The required product is 105.

Example-25

Find a closed form expression for the series

$S = \tan \alpha + 2\tan 2\alpha + {2^2}\tan {2^2}\alpha + .... + {2^{n - 1}}\tan {2^{n - 1}}\alpha$

Solution: Once again, we have to produce a cascade. Note that

$\tan \alpha - \cot \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} - \frac{{\cos \alpha }}{{\sin \alpha }} = - 2\cot 2\alpha$

Thus,

\begin{align}&S - \cot \alpha = (\tan \alpha - \cot \alpha ) + 2\tan 2\alpha + {2^2}\tan {2^2}\alpha + .... \\ \,\,\,\, &\qquad\qquad= (2\tan 2\alpha - 2\cot 2\alpha ) + {2^2}\tan {2^2}\alpha + ..... \\ \,\,\,\, &\qquad\qquad= ({2^2}\tan {2^2}\alpha - {2^2}\cot {2^2}\alpha ) + ..... \\ \,\,\,\, &\qquad\qquad= - {2^n}\cot {2^n}\alpha \\ \Rightarrow &\quad\qquad S = \cot \alpha - {2^n}\cot {2^n}\alpha \\ \end{align}