Examples on Trigonometric Ratios and Functions Set 6
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
\[\begin{align}&P = \frac{{ - 1}}{2} \times \frac{{ - 1}}{{16}} \times \frac{1}{4} \\ \,\,\,\,\, &\;\;\;= \frac{1}{{128}} \\ \end{align} \]