Question

Simplify: $\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta $.
(A) $1$
(B) $ - 1$
(C) $\sin \theta $
(D) $ - \sin \theta $

Answer 1

Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\sin \left( {\pi + x} \right) = - \sin x$ and $\sin \left( {\pi - x} \right) = \sin x$ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We will use the above mentioned trigonometric identities to simplify the expression and then open the brackets to get to the required answer.

Complete answer:
In the given problem, we have to simplify the product $\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta $.
So, $\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta $
We know the trigonometric formula $\sin \left( {\pi + x} \right) = - \sin x$. So, we get,
$ = $ $ - \sin \theta \sin \left( {\pi - \theta } \right)\cos e{c^2}\theta $
We also know the trigonometric formula $\sin \left( {\pi - x} \right) = \sin x$. So, we get,
$ = $ $ - \sin \theta \sin \theta \cos e{c^2}\theta $
Now, expressing the product of $\sin \theta $ with itself as ${\sin ^2}\theta $, we get,
$ = $ $ - {\sin ^2}\theta \cos e{c^2}\theta $
Now, we know that cosecant and sine are reciprocal trigonometric functions of each other. So, we get the expression as,
$ = $ \[ - \dfrac{{{{\sin }^2}\theta }}{{\;{{\sin }^2}\theta }}\]
Now, cancelling common factors in numerator and denominator, we get,
$ = $ \[ - 1\]
Hence, the product $\sin \left( {\pi + \theta } \right)\sin \left( {\pi - \theta } \right)\cos e{c^2}\theta $ can be simplified as $( - 1)$ by the use of basic algebraic rules and simple trigonometric formulae.
So, option (B) is the correct answer.

Additional information: Trigonometric functions are also called Circular functions. Trigonometric functions are the functions that relate an angle of a right angled triangle to the ratio of two side lengths. There are $6$ trigonometric functions, namely: $\sin (x)$,$\cos (x)$,$\tan (x)$,$\cos ec(x)$,$\sec (x)$ and \[\cot \left( x \right)\] . Also, $\cos ec(x)$ ,$\sec (x)$and \[\cot \left( x \right)\] are the reciprocals of $\sin (x)$,$\cos (x)$ and $\tan (x)$ respectively.

Note:
The problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: $\sin \left( {\pi + x} \right) = - \sin x$ and $\sin \left( {\pi - x} \right) = \sin x$ . Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. We can also simplify the given expression using the compound angle formulae for sine. Questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers.