# Trigonometry Set-1

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Example - 1

Let \begin{align}\alpha {\rm{, }}\beta {\rm{, }}\gamma \end{align} be acute angles such that

$\cos \alpha = \tan \beta ,\;\cos \beta = \tan \gamma ,\;\cos \gamma = \tan \alpha$

Find the value of \begin{align}\sin \alpha ,\;\sin \beta ,\;\sin \gamma \end{align}

Solution: Let us construct an equation in terms of just one variable, say \begin{align}\alpha \end{align}:

\begin{align} &\qquad\qquad\quad {\cos ^2}\alpha = {\tan ^2}\beta = {\sec ^2}\beta - 1\\ &\qquad\qquad\qquad\quad\;= {\cot ^2}\gamma - 1\\ & \Rightarrow \qquad 1 + {\cos ^2}\alpha = {\cot ^2}\gamma = \frac{{{{\cos }^2}\gamma }}{{{{\sin }^2}\gamma }} = \frac{{{{\cos }^2}\gamma }}{{1 - {{\cos }^2}\gamma }}\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;= \frac{{{{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}\\ &\qquad\qquad\qquad\quad\quad \Rightarrow \qquad 2 - {\sin ^2}\alpha = \frac{{{{\sin }^2}\alpha }}{{1 - 2{{\sin }^2}\alpha }} \end{align}

Solving as a quadratic in \begin{align}{\sin ^2}\alpha \end{align}, we have

\begin{align}\sin \alpha = \frac{{\sqrt 5 - 1}}{2}\end{align}

By symmetry, \begin{align}\sin \beta = \sin \gamma = \frac{{\sqrt 5 - 1}}{2}\end{align}

Example - 2

Find the sum of the series

$S = \sin \theta \sec 3\theta + \sin 3\theta \sec {3^2}\theta + \sin {3^2}\theta \sec {3^3}\theta + .....\;{\rm{to}}\;n\;{\rm{terms}}$

Solution: Our approach should be to try to express the general rth term as a difference:

\begin{align} {T_r} &= \sin \;{3^{r - 1}}\;\theta \sec \,{3^r}\theta \\ & = \left. {\frac{{\sin \;{3^{r - 1}}\theta }}{{\cos {3^r}\theta }} \times \frac{{\cos {3^{r - 1}}\theta }}{{\cos {3^{r - 1}}\theta }}} \right\}\;{\rm{New}}\;{\rm{term}}\;{\rm{introduced}}\\ & = \frac{1}{2}\frac{{\sin 2 \cdot {3^{r - 1}}\theta }}{{\cos \;{3^r}\theta \;\cos {3^{r - 1}}\theta }} = \frac{1}{2}\frac{{\sin \left( {{3^r} - {3^{r - 1}}} \right)\theta }}{{\cos {3^r}\theta \;\cos {3^{r - 1}}\theta }}\\ & = \frac{1}{2}\frac{{\sin {3^r}\theta \;\cos {3^{r - 1}}\theta \; - \cos {3^r}\theta \;\;\sin {3^{r - 1}}\theta }}{{\cos {3^r}\theta \;\;\cos {3^{r - 1}}\theta }}\\ & = \frac{1}{2}\left( {\tan {3^r}\theta \; - \;\tan {3^{r - 1}}\theta } \right) \end{align}

This solves our problem, since now

\begin{align} & S = \frac{1}{2}\left\{ {(\tan 3\theta - \tan \theta ) + (\tan {3^2}\theta - \tan 3\theta ) + ..... + (\tan {3^n}\theta - \tan {3^{n - 1}}\theta )} \right\}\\ & S = \frac{1}{2}(\tan {3^n}\theta - \tan \theta ) \end{align}

Example - 3

If \begin{align}\frac{{\sin (\theta + \alpha )}}{{\cos (\theta - \alpha )}} = \frac{{1 - m}}{{1 + m}},\end{align} find the value of \begin{align}P = \tan \left( {\frac{\pi }{4} - \;\theta } \right)\;\tan \left( {\frac{\pi }{4} - \alpha } \right)\end{align}

Solution: In the given relation, the arguments of the trigonometric functions contain both \begin{align}\theta \;{\rm{and}}\;\alpha \end{align} together, while in the expression we need to evaluate, this is not the case - an indication that we should use combination formulae.

Writing \begin{align}\cos (\theta - \alpha )\;{\rm{as}}\;\sin \left( {\frac{\pi }{2} - (\theta - \alpha )} \right)\end{align} and applying C & D to the given relation, we have:

\begin{align} & \frac{{\sin (\theta + \alpha ) + \sin \left( {\frac{\pi }{2} - (\theta - \alpha )} \right)}}{{\sin (\theta + \alpha ) - \sin \left( {\frac{\pi }{2} - (\theta - \alpha )} \right)}} = - \frac{1}{m}\\ \Rightarrow \qquad & \frac{{\sin \left( {\frac{\pi }{4} + \alpha } \right)\cos \left( { - \frac{\pi }{4} + \theta } \right)}}{{\sin \left( {\frac{{ - \pi }}{4} + \theta } \right)\cos \left( {\frac{\pi }{4} + \alpha } \right)}} = - \frac{1}{m}\\ \Rightarrow \qquad & \tan \left( {\frac{\pi }{4} + \alpha } \right)\cot \left( {\theta - \frac{\pi }{4}} \right) = - \frac{1}{m}\\ \Rightarrow \qquad & \cot \left( {\frac{\pi }{4} - \alpha } \right)\cot \left( {\frac{\pi }{4} - \theta } \right) = \frac{1}{m} \qquad \qquad \qquad \qquad (How?)\\ \Rightarrow \qquad & P = m \end{align}

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