See answer given.

Utilising the formula, \(Period=\frac{360}{b}\);

\(Period=\frac{360}{2}=180 ^{\circ}\)

In this function the amplitude has doubled but the period has not changed from \(f\left(x\right)=\cos\left(x\right)\).

Utilising the formula, \(b=\frac{360}{Period}\);

\(b=\frac{360}{120}=3\)

The point of this question is to show that the \(4\) inside the brackets is not itself the period of the wave, but it does of course affect the period.

See answers given.

In this case for the period, rewrite \(\frac{x}{2}\) as \(\frac{1}{2}x\) and utilise the formula, \(Period=\frac{360}{b}\);

\(Period=\frac{360}{0.5}=720 ^{\circ}\). Other answers are given and more easily understood.

You can use the equations given in video 4, or find these values more intuitively.

The principal axis (or centre line) or the curve is the most logical to identify first. This corresponds to the value \(c\) which is \(-1\).

The amplitude is the height between the principal axis and the maximum (or minimum points. This corresponds to the value \(a\) which is \(2\).

The period of the wave is \(90\) degrees. Intuitively this means that \(4\) waves fit into a \(360\) degree domain which means that the value \(b\) is \(4\). You might also find this using \(b=\frac{360}{Period}\);

\(b=\frac{360}{90}=4\).

Utilising the formulae in this case:

\(a=\frac{\max-\min}{2}=\frac{1.5--1.5}{2}=\frac{3}{2}=1.5\)

To find \(b\), first identify the period of the wave. Half of the wave is shown with the co-ordinates given, so half the period is \(810-270=540\). The period is therefore \(540\cdot2=1080\). Hence, \(b=\frac{360}{1080}=\frac{1}{3}\).

\(c=\frac{\max+\min}{2}=\frac{1.5+-1.5}{2}=\frac{0}{2}=0\)