Question

The period of $\sin \left( {2\pi x + a} \right)\sin \left( {2\pi x + b} \right)$ is:
A.$\dfrac{a}{b}$
B.1
C.2
D.$\pi $

Answer 1

Hint: Here, we are required to determine the period of the given function. For that, we will use the trigonometric identity: sin (A + B) = (sin A cos B + cos A sin B) to solve the given question.

Complete step-by-step answer:
We are given the function: $\sin \left( {2\pi x + a} \right)\sin \left( {2\pi x + b} \right)$
Now, using the trigonometric identity: sin (A + B) = (sin A cos B + cos A sin B) in the given equation of the function whose period needs to be determined, we get
$ \Rightarrow \sin (2\pi x + a)\sin (2\pi x + b) = (\sin 2\pi x\cos a + \cos 2\pi x\sin a) \times (\sin 2\pi x\cos b + \cos 2\pi x\sin b)$
We know that the period of $\sin 2\pi x = 1$ and also the period of $\cos 2\pi x = 1$
Also, cos a, cos b, sin a, sin b are constants in the equation we deduced after using the trigonometric identity.
Hence, the period of the given function: $\sin \left( {2\pi x + a} \right)\sin \left( {2\pi x + b} \right)$ will be 1.
Therefore, option(B) is correct.

Note: In such problems where we are required to find the period of a given function, we simplify it in those simpler terms whose period is known to us. We can define the period of a given function as the distance required to complete one full cycle. Or, in other words, the distance between each repeating wave of the given function.
Also, the sine function is defined as a trigonometric function of an angle, if talking about a right angle triangle, as the ratio of the length of the side that is opposite to that angle to the length of the longest side of the triangle. Or, we can simply say the ratio of the perpendicular to the hypotenuse of a right angle triangle is defined as the sine of that acute angle.