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

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 $