Question

Prove the following results of trigonometric functions of allied angles: -
(1) $\sin \left( \pi -\theta \right)=\sin \theta $, $\cos \left( \pi -\theta \right)=-\cos \theta $, $\tan \left( \pi -\theta \right)=-\tan \theta $
(2) $\sin \left( 2\pi -\theta \right)=-\sin \theta $, $\cos \left( 2\pi -\theta \right)=\cos \theta $, $\tan \left( 2\pi -\theta \right)=-\tan \theta $
(3) $\sin \left( \dfrac{3\pi }{2}-\theta \right)=-\cos \theta $, $\cos \left( \dfrac{3\pi }{2}-\theta \right)=-\sin \theta $, $\tan \left( \dfrac{3\pi }{2}-\theta \right)=\cot \theta $

Answer 1

Hint: To prove the results for the sine function, use the formula $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$ and substitute the values of the angles to evaluate. For the results of the cosine function, use the formula $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b$ and evaluate by substituting proper values of angles. Finally, for the tangent function use the formula $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$ to evaluate the relation. Use the values: - $\sin \pi =0$, $\cos \pi =-1$, $\tan \pi =0$, $\sin 2\pi =0$, $\cos 2\pi =1$, $\tan 2\pi =0$, $\sin \dfrac{3\pi }{2}=-1$, $\cos \dfrac{3\pi }{2}=0$ and $\tan \dfrac{3\pi }{2}=\infty $.

Complete step by step answer:
Here we have been asked to prove certain results of trigonometric functions of allied angles. Let us check them one by one.
(1) Here we need to prove the following three results: - $\sin \left( \pi -\theta \right)=\sin \theta $, $\cos \left( \pi -\theta \right)=-\cos \theta $, $\tan \left( \pi -\theta \right)=-\tan \theta $.
(i) Considering the relation $\sin \left( \pi -\theta \right)$, so using the formula $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$ and substituting the angles a = $\pi $ and b = $\theta $ we get,
$\Rightarrow \sin \left( \pi -\theta \right)=\sin \pi \cos \theta -\cos \pi \sin \theta $
Substituting the values $\sin \pi =0$and $\cos \pi =-1$ we get,
$\begin{align}
  & \Rightarrow \sin \left( \pi -\theta \right)=0\times \cos \theta -\left( -1 \right)\times \sin \theta \\
 & \therefore \sin \left( \pi -\theta \right)=\sin \theta \\
\end{align}$
(ii) Considering the relation $\cos \left( \pi -\theta \right)$, so using the formula $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b$ and substituting the angles a = $\pi $ and b = $\theta $ we get,
\[\begin{align}
  & \Rightarrow \cos \left( \pi -\theta \right)=\cos \pi \cos \theta +\sin \pi \sin \theta \\
 & \Rightarrow \cos \left( \pi -\theta \right)=\left( -1 \right)\times \cos \theta +0\times \sin \theta \\
 & \therefore \cos \left( \pi -\theta \right)=-\cos \theta \\
\end{align}\]
(iii) Considering the relation$\tan \left( \pi -\theta \right)$, so using the formula $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$ and substituting the angles a = $\pi $ and b = $\theta $ we get,
\[\Rightarrow \tan \left( \pi -\theta \right)=\dfrac{\tan \pi -\tan \theta }{1+\tan \pi \tan \theta }\]
Substituting the value $\tan \pi =0$ we get,
\[\begin{align}
  & \Rightarrow \tan \left( \pi -\theta \right)=\dfrac{0-\tan \theta }{1+0\times \tan \theta } \\
 & \therefore \tan \left( \pi -\theta \right)=-\tan \theta \\
\end{align}\]
(2) Here we need to prove the following three results: - $\sin \left( 2\pi -\theta \right)=-\sin \theta $, $\cos \left( 2\pi -\theta \right)=\cos \theta $, $\tan \left( 2\pi -\theta \right)=-\tan \theta $.
(i) Considering the relation $\sin \left( 2\pi -\theta \right)$, so using the formula $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$ and substituting the angles a = $2\pi $ and b = $\theta $ we get,
$\Rightarrow \sin \left( 2\pi -\theta \right)=\sin 2\pi \cos \theta -\cos 2\pi \sin \theta $
Substituting the values $\sin 2\pi =0$ and $\cos 2\pi =1$ we get,
$\begin{align}
  & \Rightarrow \sin \left( 2\pi -\theta \right)=0\times \cos \theta -1\times \sin \theta \\
 & \therefore \sin \left( 2\pi -\theta \right)=-\sin \theta \\
\end{align}$
(ii) Considering the relation $\cos \left( 2\pi -\theta \right)$, so using the formula $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b$ and substituting the angles a = $2\pi $ and b = $\theta $ we get,
\[\begin{align}
  & \Rightarrow \cos \left( 2\pi -\theta \right)=\cos 2\pi \cos \theta +\sin 2\pi \sin \theta \\
 & \Rightarrow \cos \left( 2\pi -\theta \right)=1\times \cos \theta +0\times \sin \theta \\
 & \therefore \cos \left( 2\pi -\theta \right)=\cos \theta \\
\end{align}\]
(iii) Considering the relation $\tan \left( 2\pi -\theta \right)$, so using the formula $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$ and substituting the angles a = $2\pi $ and b = $\theta $ we get,
\[\Rightarrow \tan \left( 2\pi -\theta \right)=\dfrac{\tan 2\pi -\tan \theta }{1+\tan 2\pi \tan \theta }\]
Substituting the value $\tan 2\pi =0$ we get,
\[\begin{align}
  & \Rightarrow \tan \left( 2\pi -\theta \right)=\dfrac{0-\tan \theta }{1+0\times \tan \theta } \\
 & \therefore \tan \left( 2\pi -\theta \right)=-\tan \theta \\
\end{align}\]
(3) Here we need to prove the following three results: - $\sin \left( \dfrac{3\pi }{2}-\theta \right)=-\cos \theta $, $\cos \left( \dfrac{3\pi }{2}-\theta \right)=-\sin \theta $, $\tan \left( \dfrac{3\pi }{2}-\theta \right)=\cot \theta $.
(i) Considering the relation $\sin \left( \dfrac{3\pi }{2}-\theta \right)$, so using the formula $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$ and substituting the angles a = $\dfrac{3\pi }{2}$ and b = $\theta $ we get,
$\Rightarrow \sin \left( \dfrac{3\pi }{2}-\theta \right)=\sin \dfrac{3\pi }{2}\cos \theta -\cos \dfrac{3\pi }{2}\sin \theta $
Substituting the values $\sin \dfrac{3\pi }{2}=-1$and $\cos \dfrac{3\pi }{2}=0$ we get,
$\begin{align}
  & \Rightarrow \sin \left( \dfrac{3\pi }{2}-\theta \right)=\left( -1 \right)\times \cos \theta -0\times \sin \theta \\
 & \therefore \sin \left( \dfrac{3\pi }{2}-\theta \right)=-\cos \theta \\
\end{align}$
(ii) Considering the relation $\cos \left( \dfrac{3\pi }{2}-\theta \right)$, so using the formula $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b$ and substituting the angles a = $\dfrac{3\pi }{2}$ and b = $\theta $ we get,
\[\begin{align}
  & \Rightarrow \cos \left( \dfrac{3\pi }{2}-\theta \right)=\cos \dfrac{3\pi }{2}\cos \theta +\sin \dfrac{3\pi }{2}\sin \theta \\
 & \Rightarrow \cos \left( \dfrac{3\pi }{2}-\theta \right)=0\times \cos \theta +\left( -1 \right)\times \sin \theta \\
 & \therefore \cos \left( \dfrac{3\pi }{2}-\theta \right)=-\sin \theta \\
\end{align}\]
(iii) Considering the relation \[\tan \left( \dfrac{3\pi }{2}-\theta \right)\], so using the formula $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$ and substituting the angles a = $\dfrac{3\pi }{2}$ and b = $\theta $ we get,
\[\Rightarrow \tan \left( \dfrac{3\pi }{2}-\theta \right)=\dfrac{\tan \dfrac{3\pi }{2}-\tan \theta }{1+\tan \dfrac{3\pi }{2}\tan \theta }\]
We know that $\tan \dfrac{3\pi }{2}=\infty $ so we cannot substitute this value directly in the above relation because if we will do so then the expression will become indeterminate form. So dividing the numerator and the denominator with $\tan \dfrac{3\pi }{2}$ and using the fact that any non – zero number divided by infinity is 0, we get,
\[\begin{align}
  & \Rightarrow \tan \left( \dfrac{3\pi }{2}-\theta \right)=\dfrac{1-\left( \dfrac{\tan \theta }{\tan \dfrac{3\pi }{2}} \right)}{\left( \dfrac{1}{\tan \dfrac{3\pi }{2}} \right)+\tan \theta } \\
 & \Rightarrow \tan \left( \dfrac{3\pi }{2}-\theta \right)=\dfrac{1-\left( \dfrac{\tan \theta }{\infty } \right)}{\left( \dfrac{1}{\infty } \right)+\tan \theta } \\
 & \Rightarrow \tan \left( \dfrac{3\pi }{2}-\theta \right)=\dfrac{1}{\tan \theta } \\
\end{align}\]
Using the conversion $\dfrac{1}{\tan \theta }=\cot \theta $ we get,
\[\therefore \tan \left( \dfrac{3\pi }{2}-\theta \right)=\cot \theta \]
Note: Note that you must remember all the above obtained results because they are used as identities and formulas in further topics and chapters of mathematics and therefore they will not be proved everywhere but directly used. Remember the trigonometric identities: - $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$, $\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b$ and $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ in case you have to find the values of trigonometric functions of angles like $\left( \pi +\theta \right)$.