# Examples on Trigonometric Ratios and Functions Set 4

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Example-10

Evaluate a closed-form expression for the following series.

(a) $${{S}_{1}}=\sin \alpha +\sin (\alpha +\beta )+\sin (\alpha +2\beta )+.....+\sin (\alpha +(n-1)\beta )$$

(b) $${{S}_{2}}=\cos \alpha +\cos (\alpha +\beta )+\cos (\alpha +2\beta )+.....+\cos (\alpha +(n-1)\beta )$$

Solution: We will once again use an approach based on complex numbers.

Why? As you’ll soon see, that will modify the series into a G.P.

\begin{align} & {{S}_{1}}=I{{\,}_{m}}\left\{ {{e}^{i\alpha }}+{{e}^{i(\alpha +\beta )}}+{{e}^{i(\alpha +2\beta )}}+......+{{e}^{i(\alpha +(n-1)\beta )}} \right\} \\ & \,\,\,\,\,\,=I{{\,}_{m}}\left\{ {{e}^{i\alpha }}\left( 1+{{e}^{i\beta }}+{{e}^{i2\beta }}+......+{{e}^{i(n-1)\beta }} \right) \right\} \\ & \,\,\,\,\,\,=I{{\,}_{m}}\left\{ {{e}^{i\alpha }}\left( \frac{{{e}^{in\beta }}-1}{{{e}^{i\beta }}-1} \right) \right\} \\ & \,\,\,\,\,\,\,=I{{\ }_{m}}\left\{ (\cos \alpha +i\sin \alpha )\left( \frac{\cos n\beta +i\sin n\beta -1}{\cos \beta +i\sin \beta -1} \right) \right\} \\ & \,\,\,\,\,\,\,=I{{\ }_{m}}\left\{ (\cos \alpha +i\sin \alpha )\left( \frac{-2{{\sin }^{2}}\frac{n\beta }{2}+2i\sin \frac{n\beta }{2}\cos \frac{n\beta }{2}}{-2{{\sin }^{2}}\frac{\beta }{2}+2i\sin \frac{\beta }{2}\cos \frac{\beta }{2}} \right) \right\} \\ & \,\,\,\,\,\,\,=I{{\ }_{m}}\left\{ (\cos \alpha +i\sin \alpha )\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\left( \frac{\cos \frac{n\beta }{2}+i\sin \frac{n\beta }{2}}{\cos \frac{\beta }{2}+i\sin \frac{\beta }{2}} \right) \right\} \\ & \,\,\,\,\,\,=\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\ I{{\,}_{m}}\left\{ \frac{\cos \left( \alpha +\frac{n\beta }{2} \right)+i\sin \left( \alpha +\frac{n\beta }{2} \right)}{\cos \frac{\beta }{2}+i\sin \frac{\beta }{2}} \right\} \\ & \,\,\,\,\,\,=\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\ \ \sin \left( \frac{2\alpha +(n-1)\beta }{2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( verify \right) \\ \end{align}

(b) Using an exactly analogous procedure,

\begin{align} & {{S}_{2}}=\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\ \operatorname{Re}\left\{ \frac{\cos \left( \alpha +\frac{n\beta }{2} \right)+i\sin \left( \alpha +\frac{n\beta }{2} \right)}{\cos \frac{\beta }{2}+i\sin \frac{\beta }{2}} \right\} \\ & \,\,\,\,\,\,\,=\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\ \cos \left( \frac{2\alpha +(n-1)\beta }{2} \right) \\\end{align}

Note the following special cases:

(a)  \begin{align}\sin \alpha +\sin 2\alpha +......+\sin n\alpha =\frac{\sin \left( \frac{n\alpha }{2} \right)\sin \left( \frac{(n+1)\alpha }{2} \right)}{\sin \frac{\alpha }{2}}\end{align}

(b)  \begin{align}\cos \alpha +\cos 2\alpha +......+\cos n\alpha =\frac{\sin \left( \frac{n\alpha }{2} \right)\cos \left( \frac{(n+1)\alpha }{2} \right)}{\sin \frac{\alpha }{2}}\end{align}

Example-11

Sum the series  $$S=\sqrt{1+\cos \alpha }+\sqrt{1+\cos 2\alpha }+\sqrt{1+\cos 3\alpha }+.........\ \text{to}\ n\ \text{terms}$$ .

Solution: $$1+\cos r\alpha =2{{\cos }^{2}}\frac{r\alpha }{2}$$

\begin{align} \Rightarrow\qquad & S = \sqrt 2 \left( {\cos \frac{\alpha }{2} + \cos \alpha + \cos \frac{{3\alpha }}{2} + .......{\text{to}}\;n\;{\text{terms}}} \right) \\ & \,\,\,\,\, = \sqrt 2 \frac{{\sin \frac{{n\alpha }}{4}}}{{\sin \frac{\alpha }{4}}}\cos \left( {(n + 1)\frac{\alpha }{4}} \right)\; \qquad\quad \;\;\;\;\;\left\{ \begin{gathered} {\text{using}}\;{\text{the}}\;{\text{result}}\;{\text{of}}\;{\text{the}}\; \\ {\text{previous}}\;{\text{example}} \\ \end{gathered} \right\}\\ \end{align}

Example-12

Evaluate \begin{align}\cos \frac{\pi }{7} \cdot \;\cos \frac{{2\pi }}{7} \cdot \;\cos \frac{{3\pi }}{7}\end{align}

Solution: If  $$\theta = \frac{\pi }{7},\;{\text{then}}\;7\theta = \pi \;\;\;\;\; \Rightarrow \;\;\;\;\;4\theta = \pi - 3\theta \;\;\;\;\; \Rightarrow \;\;\;\;\;\sin 4\theta = \sin 3\theta$$

Upon expansion, we have

$$\Rightarrow 8{\cos ^3}\theta - 4{\cos ^2}\theta - 4\cos \theta + 1 = 0$$

Note carefully that each of   $$\cos \frac{\pi }{7},\;\;\cos \frac{{3\pi }}{7}\;{\text{and}}\;\cos \frac{{5\pi }}{7}$$ will satisfy this cubic (Why?). These are therefore the three roots of the cubic, and their product is:

$\cos \frac{\pi }{7}\;\;\cos \frac{{3\pi }}{7}\;\,\cos \frac{{5\pi }}{7} = - \frac{1}{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)$

Finally,
$$\cos \frac{{5\pi }}{7} = \;\cos \left( {\pi - \frac{{2\pi }}{7}} \right) = - \cos \frac{{2\pi }}{7}$$ .Replacing this in (1) gives

$\cos \frac{\pi }{7}\;\cos \frac{{2\pi }}{7}\cos \frac{{3\pi }}{7} = \frac{1}{8}$

As an exercise, show that  $$\sin \frac{\pi }{7}\;\sin \frac{{2\pi }}{7}\;\sin \frac{{3\pi }}{7} = \frac{{\sqrt 7 }}{8}$$

Example-13

Let A, B, C be the three angles of a triangle such that  $$A = \frac{\pi }{4}\;{\text{and}}\;\tan B \cdot \tan C = \lambda .$$  Find the possible values of $$lambda$$.

Solution: For A, B, C to represent the angles of a triangle, we must have

$A + B + C = \pi \Rightarrow \;\;\;\;\;B + C = \frac{{3\pi }}{4}$

Now,
$\tan B \cdot \tan C = \tan B \cdot \tan \left( {\frac{{3\pi }}{4} - B} \right) = \lambda$

\begin{align}\Rightarrow & \tan B\left( {\frac{{ - 1 - \tan B}}{{1 - \tan B}}} \right) = \lambda \\ \Rightarrow & {\tan ^2}B + (1 - \lambda )\tan B + \lambda = 0 \\ \end{align}

Since tan B is real, the discriminant of this quadratic must be non-negative:

\begin{align} & {(1 - \lambda )^2} - 4\lambda \geqslant 0 \\ \Rightarrow & \lambda \in \left( { - \infty ,\; - 3 - 2\sqrt 2 } \right] \cup \left[ {3 + 2\sqrt 2 ,\infty } \right) \\ \end{align}