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.

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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

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

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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,

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

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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,

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

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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 :

\[\begin{align}&{\cos ^2}B{\sec ^2}B = \left( {1 - \frac{{{{\sin }^2}A}}{{{m^2}}}} \right)\left( {1 + \frac{{{{\tan }^2}A}}{{{n^2}}}} \right)\\ &\Rightarrow 1 = 1 - \frac{{{{\sin }^2}A}}{{{m^2}}} + \frac{{{{\tan }^2}A}}{{{n^2}}} - \frac{{{{\sin }^2}A\;{{\tan }^2}A}}{{{m^2}{n^2}}}\\ &\Rightarrow {n^2}{\sin ^2}A - {m^2}{\tan ^2}A + {\sin ^2}A\;{\tan ^2}A = 0\\ &\Rightarrow {n^2} - {m^2}{\sec ^2}A + {\tan ^2}A = 0\\ &\Rightarrow {n^2} - {m^2}{\sec ^2}A + {\sec ^2}A - 1 = 0\\ &\Rightarrow \left( {{m^2} - 1} \right){\sec ^2}A = {n^2} - 1\\&\Rightarrow \boxed {{\cos ^2}A = \frac{{{m^2} - 1}}{{{n^2} - 1}}}\end{align}\]

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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.

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