Exercise 11.2 Solutions — Trigonometric Functions and their Graphs

(i) \(y = -\sin 2x, \quad x \in [-2\pi, 2\pi]\)

Period \(= \frac{2\pi}{2} = \pi\). Because of the minus sign, the wave starts by going downwards. Key values for one period starting at \(x = 0\):

Plot these points for \(0 \le x \le \pi\), then repeat the same wave on \([\pi, 2\pi]\) and, using the odd symmetry of the function, mirror it onto \([-2\pi, 0]\). The graph oscillates between \(-1\) and \(1\).

(ii) \(y = 2\cos 2x, \quad x \in [-2\pi, 2\pi]\)

Period \(= \pi\), amplitude \(= 2\). The cosine curve starts at its maximum:

Plot for \(0 \le x \le \pi\), repeat on \([\pi, 2\pi]\), and mirror evenly (cosine is even) onto \([-2\pi, 0]\). The wave runs between \(y = -2\) and \(y = 2\).

(iii) \(y = \tan 2x, \quad x \in [-\pi, \pi]\)

Period \(= \frac{\pi}{2}\). Vertical asymptotes occur where \(\cos 2x = 0\), i.e. \(2x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}\), so at \(x = \pm\frac{\pi}{4}, \pm\frac{3\pi}{4}\). Draw dashed vertical lines at these asymptotes, then in each strip draw the tangent branch passing through \((0, 0)\):

(iv) \(y = \tan \frac{x}{2}, \quad x \in [-2\pi, 2\pi]\)

Period \(= \frac{\pi}{1/2} = 2\pi\). Asymptotes where \(\cos\frac{x}{2} = 0\), i.e. at \(x = \pm\pi\) (inside the interval). One full branch of tangent occupies the whole interval \((-2\pi, 2\pi)\), passing through \((0, 0)\) with, for example, \(\tan\frac{x}{2} = \pm 1\) at \(x = \pm\frac{\pi}{2}\).

(v) \(y = \sin \frac{\pi}{2}x, \quad x \in [0, 2\pi]\)

Here \(b = \frac{\pi}{2}\), so period \(= \frac{2\pi}{\pi/2} = 4\). Choose \(x\)-values that make the angle \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\), namely \(x = 0, 1, 2, 3, 4\):

One complete sine wave fits in \([0, 4]\); the remaining interval \([4, 2\pi]\) (i.e. up to \(\approx 6.28\)) shows the start of the next wave, reaching \(1\) again at \(x = 5\).

(vi) \(y = \cos \frac{\pi}{2}x, \quad x \in [-\pi, \pi]\)

Period \(= 4\) again. With \(x = -2, -1, 0, 1, 2\) the angle becomes \(-\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi\):

Plot from \(x = -\pi\) (a little before \(-2\)) to \(x = \pi\) (a little after \(2\)): a cosine hump centred at the origin, even about the \(y\)-axis.

Common mistake: Using the ordinary sine/cosine shape without rescaling the \(x\)-axis — always compute the period first and mark the quarter-period points; otherwise the graph will not fit the given interval.

(i) \(y = \sin x\) and \(y = \sin 2x\)

Period of \(\sin x\) is \(2\pi\); period of \(\sin 2x\) is \(\pi\). Draw both on \([0, 2\pi]\):

The graph of \(y = \sin 2x\) completes two full waves while \(y = \sin x\) completes one — doubling the frequency compresses the wave horizontally by a factor of \(2\).

(ii) \(y = \cos x\) and \(y = \cos 2x\)

Periods are \(2\pi\) and \(\pi\) respectively. On \([0, 2\pi]\):

Again \(\cos 2x\) oscillates twice as fast as \(\cos x\); both start at their maximum \((0, 1)\) on the \(y\)-axis.

(i) \(\sin x = \cos x, \quad x \in [0, \pi]\)

Draw \(y = \sin x\) (starts at \(0\), peaks at \(\frac{\pi}{2}\)) and \(y = \cos x\) (starts at \(1\), crosses zero at \(\frac{\pi}{2}\)) on the same axes for \(0 \le x \le \pi\). The two curves cross exactly once, at the point where both equal \(\frac{1}{\sqrt{2}}\):

(ii) \(\sin x = x, \quad x \in [0, \pi]\)

Draw \(y = \sin x\) and the straight line \(y = x\) through the origin. Near the origin the line rises faster (slope \(1\)) than the sine curve (slope \(\cos 0 = 1\) at the origin, but decreasing immediately after), and for \(x > 0\) we always have \(\sin x < x\). Hence the two graphs meet only at the origin: