Question

How do you solve $y = - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)x?$ ?

Answer 1

Hint : In order to determine the period and amplitude of the above trigonometric. Compare the cosine function with the standard cosine function i.e. $y = A\cos \left( {Bx - C} \right) + D$ to find the value of $A,B,C,D$ . Amplitude is equal to the modulus of A , And period of the function is equal to ratio of $2\pi$ and modulus of $B$

Complete step-by-step answer :
We are given a trigonometric function $y = - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)x$
Comparing this equation with the standard cosine function $y = A\cos \left( {Bx - C} \right) + D$ we get
$A = - \dfrac{1}{2},B = \dfrac{\pi }{3},C = D = 0$
Amplitude is equal to the modulus of A i.e.
Amplitude $= \left| A \right| = \left| { - \dfrac{1}{2}} \right| = \dfrac{1}{2}$
And period of the function is equal to ratio of $2\pi$ and modulus of $B$
Period = $\dfrac{{2\pi }}{{\left| B \right|}} = \dfrac{{2\pi }}{{\dfrac{\pi }{3}}} = 6$
Therefore the amplitude $A$ and period of the cosine function
$y = - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)x$ is equal to $\dfrac{1}{2}$ and $6$ respectively.
So, the correct answer is “ $y = - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)x$ is equal to $\dfrac{1}{2}$ and $6$ ”.

Note : 1. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta$ for all values of $\theta$ and n $\in$ N.
2. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta$
Therefore, $\sin \theta$ and $\tan \theta$ and their reciprocals, $\cos ec\theta$ and $\cot \theta$ are odd functions whereas $\cos \theta$ and its reciprocal $\sec \theta$ are even functions.