Question

How do you find the amplitude, period and shift for $y = - 5\sin \left( {\dfrac{x}{2}} \right)$ ?

Answer 1

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

Complete step by step solution:
We are given a trigonometric function $y = - 5\sin \left( {\dfrac{x}{2}} \right)$
Comparing this equation with the standard sine function $y = A\sin \left( {Bx + C} \right) + D$ we get
$A = - 5,B = \dfrac{1}{2},C = D = 0$
Amplitude is equal to the modulus of A i.e.
Amplitude $= \left| A \right| = \left| { - 5} \right| = 5$
And period of the function is equal to ratio of $2\pi$ and modulus of $B$
Period = $\dfrac{{2\pi }}{{\left| B \right|}} = \dfrac{{2\pi }}{{\dfrac{1}{2}}} = 4\pi$
Shift of sine function is the ratio of $C$ and $B$ ,
Shift $= \dfrac{C}{B} = \dfrac{0}{{\dfrac{1}{2}}} = 0$
Therefore the amplitude $A$ ,Period and Shift of the sine function $y = - 5\sin \left( {\dfrac{x}{2}} \right)$ is equal to $5,4\pi$ and $0$ respectively.
So, the correct answer is “ $5,4\pi$ and $0$ ”.

Note : 1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
3.Range of sine is in the interval $\left[ { - 1,1} \right]$
4. Domain of sine is in the interval of $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
5. Standard Cosine Function is $y = A\cos \left( {Bx - C} \right) + D$