Question

Find the value of $\sin (\pi + \theta )\sin (\pi - \theta )\cos e{c^2}\theta =$.
A. $1$
B. $- 1$
C. $\sin \theta$
D. $- \sin \theta$

Answer 1

Hint: The trigonometric functions are circular functions with the function of an angle of a triangle. To solve such questions the properties of the trigonometric ratios and the All STC rule can be applied. After applying the correct rules and properties the expression can be further simplified to get the final answer.

Complete step by step solution:
By using the All STC rule, we can say that the $\sin$ function is positive in the first and the second quadrant, and the function is negative in the third and the fourth quadrant. This can be represented in the combination of odd and even angles.
For odd functions of the angles $\sin$ can be represented as
$\Rightarrow \sin (\pi + \theta ) = \sin (3\pi + \theta ) = \sin [(2n + 1)\pi + \theta ] = - \sin \theta$ $.............(i)$
Similarly, the even functions of the angles $\sin$ can be represented as
$\Rightarrow \sin (2\pi + \theta ) = \sin (4\pi + \theta ) = \sin [2n\pi + \theta ] = \sin \theta$ $.................(ii)$
Also, it is clear that $\sin (\pi - \theta )$ will be the same as $\sin \theta$ because then the function $\sin (\pi - \theta )$ is in the second quadrant, and the value of $\sin$ is always positive in the second quadrant.
It is given to simplify the expression $\sin (\pi + \theta )\sin (\pi - \theta )\cos e{c^2}\theta$
To simplify it we will first substitute the value of $\sin (\pi + \theta )$in the given expression which is $- \sin \theta$ and the value of $\sin (\pi - \theta )$ which is $\sin \theta$ from the equation $(i)$ and $(ii)$ respectively, to get
$\sin (\pi + \theta )\sin (\pi - \theta )\cos e{c^2}\theta = - \sin \theta \times \sin \theta \times \cos e{c^2}\theta$
Simplifying the above expression we get
$\Rightarrow - {\sin ^2}\theta \times \cos e{c^2}\theta$ $.................(iii)$
From basic trigonometric identities, we also know that $\sin \theta = \dfrac{1}{{\cos ec\theta }}$ , so substituting this formula in the equation $(iii)$ we get
$\Rightarrow - \dfrac{1}{{\cos e{c^2}\theta }} \times \cos e{c^2}\theta$
Canceling the like terms in the above expression and simplifying it we get
$= - 1$
Hence on simplifying the given expression $\sin (\pi + \theta )\sin (\pi - \theta )\cos e{c^2}\theta$ we get the final answer as $- 1$ .

Therefore, the correct answer for this will be option B.

Note: Remember the ALL STC rule while solving such questions. It is also known as the ASTC rule in trigonometry. The rule states that all the trigonometric ratios in the first quadrant ( ${0^\circ }$ to ${90^\circ }$ ) are positive. In the second quadrant ( ${90^\circ }$ to ${180^\circ }$ ) the ratios $\sin$ and $\cos ec$ are positive. The trigonometric ratios $\tan$ and $\cot$ are positive in the third quadrant ( ${180^\circ }$ to ${270^\circ }$ ) and the ratios $\cos$ and $\sec$ are positive in the fourth quadrant ( ${270^\circ }$ to ${360^\circ }$ )