# Trigonometric Elimination

Trigonometric Elimination

## Introduction:

Suppose that the variable $$x$$ and $$y$$ are specified in terms of a variable parameter $$\theta$$, as follows:

$x = a\cos \theta, \;\;y = a\;sin\;\theta$

The parameter $$a$$ is some constant. The question is: how are x and y related? To determine this relation we can eliminate $$\theta$$ and obtain an equation in terms of  $$x$$ and $$y$$ alone:

\begin{align}{x^2} &= {a^2}{\cos ^2}\theta ,{y^2} = {a^2}{\sin ^2}\theta \\ \Rightarrow {x^2} + {y^2} &= {a^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\\ \Rightarrow &\boxed {{x^2} + {y^2} = {a^2}}\end{align}

Thus, we have successfully eliminated the parameter $$\theta$$.

We will now discuss more examples involving the process of trigonometric elimination and we will make the use of basic properties of trigonometric ratios and trigonometric identities for the same.

## Solved Examples on Trigonometric Elimination:

Example 1 :  $$x$$, $$y$$ and $$z$$ are specified in terms of variable parameters  $$\alpha$$ and $$\beta$$ as follows:

\begin{align}&x = p\sin \alpha \cos \beta \\&y = p\sin \alpha \sin \beta \\&z = p\cos \alpha \end{align}

$$p$$ is some constant. Eliminate $$\alpha$$ and $$\beta$$ to obtain a relation in $$x$$, $$y$$ and $$z$$ .

Solution: We have:

$\left\{ {\begin{array}{*{20}{l}} {{x^2} = {p^2}\;{{\sin }^2}\;\alpha \;{{\cos }^2}\;\beta } \\ {{y^2} = {p^2}\;{{\sin }^2}\alpha \;{{\sin }^2}\;\beta } \end{array}} \right.$

\begin{align} \Rightarrow {x^2} + {y^2} &= {p^2}{\sin ^2}\alpha \left( {{{\cos }^2}\beta + {{\sin }^2}\beta } \right) \hfill \\ &= {p^2}{\sin ^2}\alpha \hfill \\ \end{align}

Also, since $${z^2} = = {p^2}{\cos ^2}\alpha$$, we have

\begin{align}{x^2} + {y^2} + {z^2} &= {p^2}{\sin ^2}\alpha + {p^2}{\cos ^2}\alpha \\ &= {p^2}\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)\\ &= {p^2}\end{align}

Thus, the required relation is

$\boxed {{x^2} + {y^2} + {z^2} = {p^2}}$

Example 2 :  If $$a\cos \theta - b\sin \theta = c,$$ find the value of $$a\sin \theta + b\cos \theta$$ .

Solution: Let

\begin{align}&{T_1} = a\cos \theta - b\sin \theta \\&{T_2} = a\sin \theta + b\cos \theta\end{align}

Consider the sum $${T_1}^2 + {T_2}^2$$.

We have:

$\left\{ {\begin{array}{*{20}{l}} {{T_1}^2 = {a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta - 2ab\sin \theta \cos \theta } \\ {{T_2}^2 = {a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta + 2ab\sin \theta \cos \theta } \end{array}} \right.$

\begin{align} \Rightarrow {T_1}^2 + {T_2}^2 &= {a^2} + {b^2} \hfill \\ \Rightarrow {T_2}^2 &= {a^2} + {b^2} - {T_1}^2 \hfill \\ \Rightarrow {T_2}^2 &= {a^2} + {b^2} - {c^2} \hfill \\ \end{align}

Thus,

$\boxed {a\sin \theta + b\cos \theta = \pm \sqrt {{a^2} + {b^2} - {c^2}}}$

Example 3:  Suppose that

\begin{align}&\tan \theta + \sin \theta = m\\&\tan \theta - \sin \theta = n\end{align}

Find the value of  \begin{align}\frac{{{m^2} - {n^2}}}{{\sqrt {mn} }}\end {align}

Solution: We have:

\begin{align}{m^2} - {n^2} &= \left( {m + n} \right)\left( {m - n} \right)\\&= 2\tan \theta \times 2\sin \theta \\&= 4\tan \theta \sin \theta\end{align}

And,

\begin{align}&\sqrt {mn} = \sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } \\ \qquad\;&\qquad\;= \sqrt {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} - {{\sin }^2}\theta } \\ \qquad\;&\qquad\;= \sin \theta \sqrt {\frac{1}{{{{\cos }^2}\theta }} - 1} \\\qquad\; &\qquad\;= \frac{{\sin \theta }}{{\cos \theta }} \times \sqrt {1 - {{\cos }^2}\theta } \\ \qquad\; &\qquad\;= \tan \theta \sin \theta \end{align}

Thus,

\begin{align} \boxed {\frac{{{m^2} - {n^2}}}{{\sqrt {mn} }} = 4} \end {align}

Example 4 : If $$\tan A = n \,\tan B$$ and $$\sin A = m\,\sin B$$, find the value of $${\cos ^2}A$$ in terms of $$m$$ and $$n$$.

Solution: We need to eliminate $$B$$ from these two relations. We have :

$\sin B = \frac{{\sin A}}{m},\tan B = \frac{{\tan A}}{n}$

$\Rightarrow {\cos ^2}B = 1 - \frac{{{{\sin }^2}A}}{{{m^2}}},{\sec ^2}B = 1 + \frac{{{{\tan }^2}A}}{{{n^2}}}$

Taking the product, we have :

\begin{align}&{\cos ^2}B{\sec ^2}B = \left( {1 - \frac{{{{\sin }^2}A}}{{{m^2}}}} \right)\left( {1 + \frac{{{{\tan }^2}A}}{{{n^2}}}} \right)\\ &\Rightarrow 1 = 1 - \frac{{{{\sin }^2}A}}{{{m^2}}} + \frac{{{{\tan }^2}A}}{{{n^2}}} - \frac{{{{\sin }^2}A\;{{\tan }^2}A}}{{{m^2}{n^2}}}\\ &\Rightarrow {n^2}{\sin ^2}A - {m^2}{\tan ^2}A + {\sin ^2}A\;{\tan ^2}A = 0\\ &\Rightarrow {n^2} - {m^2}{\sec ^2}A + {\tan ^2}A = 0\\ &\Rightarrow {n^2} - {m^2}{\sec ^2}A + {\sec ^2}A - 1 = 0\\ &\Rightarrow \left( {{m^2} - 1} \right){\sec ^2}A = {n^2} - 1\\&\Rightarrow \boxed {{\cos ^2}A = \frac{{{m^2} - 1}}{{{n^2} - 1}}}\end{align} Challenge: If $$x = a\sin \theta + b\cos \theta$$ and $$y = a\cos \theta - b\sin \theta$$, prove that $${x^2} + {y^2} = {a^2} + {b^2}$$

⚡Tip: Sqaure L.H.S. and R.H.S. of both the given equations and then, add them.

Trigonometry