Question

Does $\sin \left( {\pi - x} \right) = \sin \left( x \right){\text{ ?}}$

Answer 1

Hint: We need to know some rules of conversion to answer this question.
As the angle $\left( \theta \right)$ of a trigonometric function changes, then we have to take care of two things in the resultant function :
$\left( 1 \right)$ Conversion to another function:
Rules of conversion: Conversion of one trigonometric function to another as the angle varies from quadrant to quadrant. The important point to note here is that the conversion will only be done when the reference angle involves the vertical axis or y axis, if the angle of reference is with respect to horizontal axis or x axis there will be no conversion and function will remain the same.
 Conversion rules are:
$\left( {\text{i}} \right)\sin \leftrightarrow \cos $
$\left( {{\text{ii}}} \right)\sec \leftrightarrow \cos ec$
$\left( {{\text{iii}}} \right)\tan \leftrightarrow \cot $


Complete step-by-step answer:
The given trigonometric equality is ;
$ \Rightarrow \sin \left( {\pi - x} \right) = \sin \left( x \right)$
L.H.S. $ = \sin \left( {\pi - x} \right)$ and R.H.S. $ = \sin \left( x \right)$
We will try to simplify the L.H.S. by using the properties mentioned above and try to prove it equal to the R.H.S. ;
L.H.S. $ = \sin \left( {\pi - x} \right){\text{ }}......\left( 1 \right)$
By the property $2$ for sine function, we know that;
$ \Rightarrow \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B{\text{ }}......\left( 2 \right)$
Comparing equation $1$ with equation $2$ , let ;
$ \Rightarrow A = \pi {\text{ and }}B = x$
Therefore, we can expand equation $1$ by the formula mentioned in equation $2$ as ;
$L.H.S. \Rightarrow \sin \left( {\pi - x} \right) = \sin \pi \cos x - \cos \pi \sin x{\text{ }}......\left( 3 \right)$
By the standard values for sine and cosine functions, we know that;
$\because \sin \pi = 0$ and $\because \cos \pi = - 1$
Putting the above values of $\sin \pi {\text{ and }}\cos \pi $ in equation $3$ , we get;
$L.H.S. \Rightarrow \sin \left( {\pi - x} \right) = 0 \times \left\{ {\cos \left( x \right)} \right\} - \left\{ {\left( { - 1} \right) \times \sin \left( x \right)} \right\}$
On further simplification, we get;
$L.H.S. \Rightarrow \sin \left( {\pi - x} \right) = 0 - \left\{ { - \sin \left( x \right)} \right\}$
$L.H.S. \Rightarrow \sin \left( {\pi - x} \right) = \sin \left( x \right)$
By the above trigonometric equality, we can say that;
L.H.S.=R.H.S.
Therefore, the value of $\sin \left( {\pi - x} \right) = \sin \left( x \right)$ .
Hence proved.

Note: We can verify our answer for the given question directly, if the above rules are kept in mind. The given function is $\sin \left( {\pi - x} \right)$ means the angle is less than $\pi {\text{ or 18}}{{\text{0}}^0}$ and our function lies in the second quadrant. Recall that in the second quadrant sine function is positive means our resultant function will be positive. Secondly, talking about the conversion, as the reference angle is $\pi $ which involves x-axis means no need of conversion, hence the resultant function will be sine only. Therefore, $\sin \left( {\pi - x} \right) = \sin x$ .