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,

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

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