# Trigonometry Set-2

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Example - 4

Find the sum of

\begin{align}S = {\cos ^4}\theta + {\cos ^4}\left( {\theta + \frac{{2\pi }}{n}} \right) + {\cos ^4}\left( {\theta + \frac{{4\pi }}{n}} \right) + ......\;{\rm{to}}\;n\;{\rm{terms}}\end{align}

Solution: First, let us express \begin{align}{\cos ^4}x\end{align} as a linear term:

\begin{align} {\cos ^4}x & = {\left( {\frac{{1 + \cos 2x}}{2}} \right)^2} = \frac{1}{4}(1 + 2\cos 2x + {\cos ^2}2x)\\ & = \frac{1}{4}\left( {1 + 2\cos 2x + \left( {\frac{{1 + \cos 4x}}{2}} \right)} \right)\\ & = \frac{1}{8}\left( {3 + 4\cos 2x + \cos 4x} \right) \end{align}

Thus,

\begin{align}S = \sum\limits_{r = 1}^n {{{\cos }^4}} \left( {\theta + (r - 1)\frac{{2\pi }}{n}} \right) = \frac{1}{8}\;\sum\limits_{r = 1}^n {} \left( {3 + 4\cos \left( {2\theta + (r - 1)\frac{{4\pi }}{n}} \right) + \cos \left( {4\theta + (r - 1)\frac{{8\pi }}{n}} \right)} \right)\end{align}

Now we sum the two cos series using the general relation for the sum of a cos series. Verify that both these series sum to zero. Therefore,

\begin{align}S = \frac{{3n}}{8}\end{align}

Example - 5

If \begin{align}x + y + z = xyz,\end{align} show that

\begin{align}\frac{{2x}}{{1 - {x^2}}} + \frac{{2y}}{{1 - {y^2}}} + \frac{{2z}}{{1 - {z^2}}} = \frac{{8xyz}}{{(1 - {x^2})(1 - {y^2})(1 - {z^2})}}\end{align}

Solution: The solution consists of two steps, and is motivated by the fact that terms on the LHS \begin{align}\left( {{\rm{eg}}{\rm{.}}\;\frac{{2x}}{{1 - {x^2}}}} \right)\end{align} resemble the expansion formula of \begin{align}\tan 2\theta \end{align}. Therefore, we let

$x = \tan A,\;y = \tan B,\;z = \tan C$

(1) From the given relation, we have

\begin{align} & \tan A + \tan B + \tan C = \tan A\;\tan B\;\tan C\\ \Rightarrow \qquad & \tan (A + B + C) = 0 & & & (How?)\\ \Rightarrow \qquad & A + B + C = n\pi \end{align}

(2) \begin{align}2A + 2B + 2C = 2n\pi \end{align}

\begin{align} \Rightarrow \qquad & \tan 2A + \tan 2B + \tan 2C = \tan 2A\tan 2B\tan 2C & & (How?)\\ \Rightarrow \qquad & \frac{{2x}}{{1 - {x^2}}} + \frac{{2y}}{{1 - {y^2}}} + \frac{{2z}}{{1 - {z^2}}} = \frac{{8xyz}}{{(1 - {x^2})(1 - {y^2})(1 - {z^2})}} \end{align}

Example - 6

Eliminate \begin{align}\theta \end{align} from the equations.

\begin{align}\frac{{\cos (\alpha - 3\theta )}}{{{{\cos }^3}\theta }} = \frac{{\sin (\alpha - 3\theta )}}{{{{\sin }^3}\theta }} = m\end{align}

Solution:

\begin{align} \cos \alpha \cos 3\theta + \sin \alpha \;\sin 3\theta = \frac{m}{4}(3\cos \theta + \cos 3\theta ) \qquad \qquad \qquad ...(1)\\ - \cos \alpha \;\sin 3\theta + \sin \alpha \cos 3\theta = \frac{m}{4}(3\sin \theta - \sin 3\theta ) \qquad \qquad \qquad ...(2) \end{align}

Our approach is motivated by an attempt to generate \begin{align}\cos 4\theta \end{align} and \begin{align}\sin 4\theta \end{align} terms from (1) and (2), from which $$\theta$$ can then be eliminated.

\begin{align} (1) \times \cos 3\theta - (2) \times \sin 3\theta \;{\rm{gives}}\\ \cos \alpha & = \frac{m}{4}(3\cos \theta \cos 3\theta - 3\sin \theta \sin 3\theta + 1)\\ & = \frac{m}{4}(3\cos 4\theta + 1)\\ \Rightarrow \qquad 3\cos 4\theta & = \frac{{4\cos \alpha }}{m} - 1 & & & & ...(3)\\ (1)\sin 3\theta + (2) \times \cos 3\theta \;{\text{similarly}}\;{\text{yields}}\\ 3\sin 4\theta & = \frac{{4\sin \alpha }}{m} & & & & ...(4)\\ {(3)^2} + {(4)^2}\;{\text{eliminating}}\;\theta :\\ {\left( {\frac{{4\cos \alpha }}{m} - 1} \right)^2} & + {\left( {\frac{{4\sin \alpha }}{m}} \right)^2} = 9\\ \Rightarrow \qquad {m^2} &+ m\cos \alpha - 2 = 0 \end{align}

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