# Examples on Trigonometric Ratios and Functions Set 6

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Example-18

Find a closed form expression for the product

$P = (2\cos \theta - 1)(2\cos 2\theta - 1)(2\cos {2^2}\theta - 1)....(2\cos {2^{n - 1}}\theta - 1)$

Solution: We have to do some form of ‘collapsing’, a technique typical to such problems. This basically means that we’ve to introduce a term that will lead to a cascade of combinations of successive terms, as shown below, by multiplying and dividing P by  $$(2\cos \theta + 1)$$

$P = \frac{1}{{2\cos \theta + 1}}(4{\cos ^2}\theta - 1)(2\cos 2\theta - 1)......$

Now, since  $$(4{\cos ^2}\theta - 1) = 2(1 + \cos 2\theta ) - 1 = 2\cos 2\theta + 1,$$  this term multiplies with the next in a similar way:

\begin{align} (2\cos 2\theta + 1)(2\cos 2\theta - 1) = 4{\cos ^2}2\theta - 1 \\ \,\,\,\, = 2\cos {2^2}\theta + 1 \\ \end{align}

which multiplies similarly with the next term, and so on.

Eventually, we have

$S = \frac{{2\cos {2^n}\theta + 1}}{{2\cos \theta + 1}}$

Example-19

Let,  $$\tan \alpha = \frac{p}{q}$$  and let $$\beta = \frac{\alpha }{6}$$ . If is acute, find the value of  $$p\;{\text{cosec}}\;2\beta - q\sec 2\beta$$

Solution: We have,

$\sin \alpha = \frac{p}{{\sqrt {{p^2} + {q^2}} }},\;\cos \alpha = \frac{q}{{\sqrt {{p^2} + {q^2}} }}$

Now,

\begin{align}&p\;{\text{cosec}}2\beta - q\sec 2\beta = \frac{p}{{\sin 2\beta }} - \frac{q}{{\cos 2\beta }} \\ \,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;= \sqrt {{p^2} + {q^2}} \left( {\frac{{\frac{p}{{\sqrt {{p^2} + {q^2}} }}}}{{\sin 2\beta }} - \frac{{\frac{q}{{\sqrt {{p^2} + {q^2}} }}}}{{\cos 2\beta }}} \right) \\ &\qquad\qquad\qquad\qquad\;= \sqrt {{p^2} + {q^2}} \left( {\frac{{\sin \alpha }}{{\sin 2\beta }} - \frac{{\cos \alpha }}{{\cos 2\beta }}} \right)\\ &\qquad\qquad\qquad\qquad\;= \sqrt {{p^2} + {q^2}} \frac{{\sin (\alpha - 2\beta )}}{{\frac{1}{2}\sin 4\beta }} \\ \end{align}

Since $$\alpha = 6\beta$$ , this reduces to $$2\sqrt {{p^2} + {q^2}}$$

Example-20

Evaluate a closed form expression for the product.

$P = \cos A\cos 2A\cos {2^2}A\cos {2^3}A.....\cos {2^{n - 1}}A$

Solution: Once again, we have to introduce a term that causes a cascade multiplication of successive terms. That term, as should be evident, is sin A.

\begin{align}&P = \frac{1}{{\sin A}}(\sin A\cos A)\cos 2A\cos {2^2}A...... \\ \,\,\,\,&\;\;= \frac{1}{{2\sin A}}(\sin 2A\cos 2A)\cos {2^2}A...... \\ \,\,\,\, &\;\;= \frac{1}{{{2^2}\sin A}}(\sin {2^2}A\cos {2^2}A)...... \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\vdots \,\,\, \\ \,\,\,\, &\;\;= \frac{{\sin {2^n}A}}{{{2^n}\sin A}} \\ \end{align}.

Example-21

Evaluate

$P = \cos \frac{\pi }{{15}}\cos \frac{{2\pi }}{{15}}\cos \frac{{3\pi }}{{15}}\cos \frac{{4\pi }}{{15}}\cos \frac{{5\pi }}{{15}}\cos \frac{{6\pi }}{{15}}\cos \frac{{7\pi }}{{15}}$

Solution: Using the fact that  $$\cos \frac{{7\pi }}{{15}} = \cos \left( {\pi - \frac{{8\pi }}{{15}}} \right) = - \cos \frac{{8\pi }}{{15}}\;{\text{and}}\;\cos \frac{{5\pi }}{{15}} = \cos \frac{\pi }{3} = \frac{1}{2},\;P\;{\text{can}}$$  be rearranged to

$P = - \frac{1}{2}\left( {\cos \frac{\pi }{{15}}\cos \frac{{2\pi }}{{15}}\cos \frac{{4\pi }}{{15}}\cos \frac{{8\pi }}{{15}}} \right)\;\left( {\cos \frac{{3\pi }}{{15}}\cos \frac{{6\pi }}{{15}}} \right)$

Using the result of the last example, we now have

\begin{align}&P = - \frac{1}{2}\left( {\frac{{\sin \left( {{2^4} \times \frac{\pi }{{15}}} \right)}}{{{2^4} \times \sin \left( {\frac{\pi }{{15}}} \right)}}} \right)\;\;\left( {\frac{{\sin \left( {{2^2} \times \frac{{3\pi }}{{15}}} \right)}}{{{2^2} \times \sin \left( {\frac{{3\pi }}{{15}}} \right)}}} \right) \\ \,\,\,\,\, &\quad= - \frac{1}{2}\frac{{\sin \left( {\frac{{16\pi }}{{15}}} \right)}}{{16\sin \left( {\frac{\pi }{{15}}} \right)}} \times \frac{{\sin \left( {\frac{{12\pi }}{{15}}} \right)}}{{4\sin \left( {\frac{{3\pi }}{{15}}} \right)}} \\ \end{align}

Finally, since $$\sin \left( {\frac{{16\pi }}{{15}}} \right) = - \sin \left( {\frac{\pi }{{15}}} \right)\;{\text{and}}\;\sin \left( {\frac{{12\pi }}{{15}}} \right) = \sin \left( {\frac{{3\pi }}{{15}}} \right)$$  (Why?),
we have

\begin{align}&P = \frac{{ - 1}}{2} \times \frac{{ - 1}}{{16}} \times \frac{1}{4} \\ \,\,\,\,\, &\;\;\;= \frac{1}{{128}} \\ \end{align}