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0606 P11 - Nov 2020 - Q2 - 5 marks
8164
(a) Write down the amplitude of \(1+4\cos\frac{x}{3}\).
(b) Write down the period of \(1+4\cos\frac{x}{3}\).
(c) Sketch the graph of \(y=1+4\cos\frac{x}{3}\) for \(-180^\circ\leq x\leq180^\circ\).
Solution
Answer: amplitude \(4\); period \(1080^\circ\); key points include \((0,5)\) and \((\pm180^\circ,3)\).
For \(y=1+4\cos\frac{x}{3}\), the amplitude is the coefficient of the cosine term, so the amplitude is \(4\).
The basic cosine period is \(360^\circ\). Since the angle is \(\frac{x}{3}\), the period is
\(360^\circ\times3=1080^\circ.\)
For the sketch,
\(y(0)=1+4\cos0=5.\)
Also
\(y(180^\circ)=1+4\cos60^\circ=3\)
and
\(y(-180^\circ)=1+4\cos(-60^\circ)=3.\)
The curve is symmetrical about the \(y\)-axis because cosine is an even function. It should therefore be a smooth cosine arc with maximum \((0,5)\) and endpoints \((-180^\circ,3)\) and \((180^\circ,3)\).
