Examples on Trigonometric Ratios and Functions Set 2

Example-4

If  \(a\cos \theta -b\sin \theta =c,\)  find the value of \(a\sin \theta +b\cos \theta \) :

Solution:  \({{(a\cos \theta -b\sin \theta )}^{2}}+{{(a\sin \theta +b\cos \theta )}^{2}}={{a}^{2}}+{{b}^{2}}\)

\(\Rightarrow \quad  a\sin \theta +b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}\)

Example-5

Evaluate to n terms the series

\[S=\frac{1}{\sin \theta \sin 2\theta }+\frac{1}{\sin 2\theta \sin 3\theta }+\frac{1}{\sin 3\theta \sin 4\theta }+..........\]

Solution: The trick is to convert this series of reciprocals into a sum-series where alternate terms cancel out, as shown below:

\[S=\frac{1}{\sin \theta }\left\{ \frac{\sin \theta }{\sin \theta \sin 2\theta }+\frac{\sin \theta }{\sin 2\theta \sin 3\theta }+\frac{\sin \theta }{\sin 3\theta \sin 4\theta }+........ \right\}\]

Now, the r^th term in this series \({{T}_{r}}\) can be manipulated as shown below:

\[\begin{align} & {{T}_{r}}=\frac{\sin \theta }{\sin r\theta \sin (r+1)\theta }=\frac{\sin \left( (r+1)\theta -r\theta \right)}{\sin r\theta \ \sin (r+1)\theta } \\ & \qquad\qquad\qquad \qquad\qquad=\frac{\sin (r+1)\theta \cos r\theta -\cos (r+1)\theta \sin r\theta }{\sin r\theta \ \sin (r+1)\theta } \\& \qquad\qquad\qquad \qquad\qquad=\cot r\theta -\cot (r+1)\theta \end{align}\]

So we now have,

\[\begin{align}  & S=\frac{1}{\sin \theta }\left\{ (\cot \theta -\cot 2\theta )+(\cot 2\theta -\cot 3\theta )+(\cot 3\theta -\cot 4\theta )+........ \right\} \\  & \,\,\,\,\,=\frac{\cot \theta -\cot (n+1)\theta }{\sin \theta } \\ 
\end{align}\]

Example-6

If  \(\begin{align}\frac{{{\sin }^{4}}\alpha }{a}+\frac{{{\cos }^{4}}\alpha }{b}=\frac{1}{a+b},\ \ \text{evaluate}\ \ \frac{{{\sin }^{8}}\alpha }{{{a}^{3}}}+\frac{{{\cos }^{8}}\alpha }{{{b}^{3}}}\end{align}\)

Solution:  Rearranging the given relation gives us

\[\begin{align}&\quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1+\frac{b}{a} \right){{\sin }^{4}}\alpha +\left( 1+\frac{a}{b} \right){{\cos }^{4}}\alpha =1 \\  & \Rightarrow\qquad {{\sin }^{4}}\alpha +{{\cos }^{4}}\alpha +\frac{b}{a}{{\sin }^{4}}\alpha +\frac{a}{b}{{\cos }^{4}}\alpha =1 \\\end{align}\]

Since ,\({{\sin }^{4}}\alpha +{{\cos }^{4}}\alpha ={{\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha  \right)}^{2}}-2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha =1-2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \)
we have,

\[\begin{align} & \Rightarrow \quad\frac{b}{a}{{\sin }^{4}}\alpha +\frac{a}{b}{{\cos }^{4}}\alpha -2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha =0 \\ & \Rightarrow \quad {{\left( \sqrt{\frac{b}{a}}{{\sin }^{2}}\alpha -\sqrt{\frac{a}{b}}{{\cos }^{2}}\alpha \right)}^{2}}=0 \\ & \Rightarrow \quad \frac{{{\sin }^{2}}\alpha }{a}=\frac{{{\cos }^{2}}\alpha }{b}=\frac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }{a+b}=\frac{1}{a+b} \\ & \Rightarrow \quad {{\sin }^{2}}\alpha =\frac{a}{a+b},\ \ {{\cos }^{2}}\alpha =\frac{b}{a+b} \\ & \Rightarrow\quad \frac{{{\sin }^{8}}\alpha }{{{a}^{3}}}=\frac{{{\cos }^{8}}\alpha }{{{b}^{3}}}=\frac{a}{{{(a+b)}^{4}}}+\frac{b}{{{(a+b)}^{4}}} \\ & \qquad\qquad\qquad\qquad\quad\;=\frac{1}{{{(a+b)}^{3}}} \\ \end{align}\]

Example-7

Evaluate \(\cot 7\frac{1}{2}{}^\text{o}\)
Solution:

\[S=\cot 7\frac{1}{2}{}^\text{o}=\frac{\cos 7\frac{1}{2}{}^\text{o}}{\sin 7\frac{1}{2}{}^\text{o}}\]

                Multiplying both the numerator and denominator by \(2\cos 7\frac{1}{2}{}^\text{o}\) , we have 

\[S=\frac{2{{\cos }^{2}}7\frac{1}{2}{}^\text{o}}{2\sin 7\frac{1}{2}{}^\text{o}\cos 7\frac{1}{2}{}^\text{o}}=\frac{1+\cos 15{}^\text{o}}{\sin 15{}^\text{o}}\]

               Now,
\[\begin{align}  & \cos 15{}^\text{o}=\cos (45{}^\text{o}-30{}^\text{o})=\cos 45{}^\text{o}\cos 30{}^\text{o}+\sin 45{}^\text{o}\sin 30{}^\text{o} \\ & \qquad\qquad\qquad\qquad\qquad\;\;=\frac{\sqrt{3}+1}{2\sqrt{2}} \\\end{align}\]

\[\begin{align}& \sin 15{}^\text{o}=\sin (45{}^\text{o}-30{}^\text{o})=\sin 45{}^\text{o}\cos 30{}^\text{o}-\cos 45{}^\text{o}\sin 30{}^\text{o} \\  &\qquad\qquad\qquad\qquad\qquad =\frac{\sqrt{3}-1}{2\sqrt{2}} \\ \end{align}\]

\(\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\begin{align}\Rightarrow S=\frac{1+\begin{align}\frac{\sqrt{3}+1}{2\sqrt{2}}\end{align}}{\begin{align}\frac{\sqrt{3}-1}{2\sqrt{2}}\end{align}}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\end{align}\) upon simplification