Trigonometric Graphs Worked ExamplesIGCSE Maths 0580

(a) The amplitude is the greatest distance from the centre line $y = 0$.

(b) The curve completes one full cycle over $360^{\circ}$.

(c) The single maximum in this interval is at the quarter-cycle point.

(d) The curve crosses the axis at the start, middle and end of the cycle.

(a) Cosine starts at its maximum, unlike sine which starts at zero.

(b) The cosine curve is symmetrical about $x = 180^{\circ}$, so reflecting in that line leaves the value unchanged.

(c) The sine curve is symmetrical about $x = 90^{\circ}$ in its positive arch.

(a) The tangent curve repeats twice over this interval.

(b) The vertical asymptotes in this interval.

(c) Tangent is the ratio of sine to cosine, and division by zero is undefined.

(a) Draw the horizontal line $y = 0.4$ across the grid and read off the $x$-coordinates of every intersection with the curve.

(b) The line $y = 0.4$ lies between $0$ and the maximum of $1$, so it cuts the positive arch of the curve twice.

Sine is positive in the first and second quadrants, so both solutions lie between $0^{\circ}$ and $180^{\circ}$.

They are symmetrical about the maximum at $x = 90^{\circ}$ — the second is $180^{\circ}$ minus the first.

(a) Plot the five key points and join them with one smooth wave — rounded at the top and bottom, not pointed. The curve must reach exactly $1$ and $-1$ on the $y$-axis scale.

(b) Make $\sin x$ the subject. Keep the negative sign.

Use the sketch from (a): the line $y = -0.375$ is below the axis, so it meets the curve twice, both between $180^{\circ}$ and $360^{\circ}$.

The calculator gives a negative angle, which is outside the interval. Its size is the reference angle.

By the symmetry of the sine curve, the two crossings are $22.024\ldots^{\circ}$ after $180^{\circ}$ and $22.024\ldots^{\circ}$ before $360^{\circ}$.

(a)(i) The cosine curve starts at its maximum of $1$ and comes down to cross the $x$-axis at $90^{\circ}$.

(a)(ii) $270^{\circ}$ is where the sine curve reaches its minimum.

(a)(iii) The tangent curve crosses the $x$-axis at $0^{\circ}$, $180^{\circ}$ and $360^{\circ}$.

(b) Make $\sin x$ the subject.

The exact value $\sin 30^{\circ} = \tfrac{1}{2}$ gives the reference angle $30^{\circ}$. The sine curve is below the axis only between $180^{\circ}$ and $360^{\circ}$, and it is symmetrical about its minimum at $270^{\circ}$.

Check: both values are $60^{\circ}$ either side of $270^{\circ}$, as the symmetry of the curve requires. ✓

(a) The first crossing is exactly halfway between the $40^{\circ}$ and $50^{\circ}$ gridlines. Each small square is $0.1$ vertically, so read the height to the nearest small square.

The second crossing is exactly $180^{\circ}$ further on, halfway between $220^{\circ}$ and $230^{\circ}$, at the same distance below the axis.

This agrees with the exact values: $\sin 45^{\circ} = \cos 45^{\circ} = \frac{\sqrt{2}}{2} = 0.707\ldots$, and both are negative at $225^{\circ}$.

(b) $\sin x > \cos x$ wherever the blue curve is above the red curve. That happens between the two crossing points.

(c) A function is negative where its curve is below the $x$-axis. The sine curve is below from $180^{\circ}$ to $360^{\circ}$; the cosine curve is below from $90^{\circ}$ to $270^{\circ}$. Both are below where these overlap.

(a) Draw the horizontal line $y = 0.8$ (two small squares below $1$) and read the $x$-coordinate of each crossing. It meets the curve once on the way down and once on the way back up.

Check with the symmetry of the cosine curve about $x = 180^{\circ}$: the second value should be $360^{\circ}$ minus the first, and $360^{\circ} - 37^{\circ} = 323^{\circ}$. ✓

(b) The graph shows $y = \cos x$, so rearrange the equation to make one side exactly $\cos x$. The other side tells you which line to draw.

Draw $y = -0.6$ (one small square below $-0.5$) and read the two crossings, either side of the minimum at $180^{\circ}$.

(c) Rearranging in the same way gives a value outside the range of the graph. The lowest point of $y = \cos x$ is $-1$, so the line $y = -1.25$ lies entirely below the curve and never meets it.

(a) Read the $x$-coordinate of every point where the two curves meet, including the ends of the interval. Both curves pass through the $x$-axis at $0^{\circ}$, $180^{\circ}$ and $360^{\circ}$, and they cross again on the $60^{\circ}$ and $300^{\circ}$ gridlines.

(b) Substitute $x = 60^{\circ}$ into each side separately, using the exact values $\tan 60^{\circ} = \sqrt{3}$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.

The two sides are equal, so $60^{\circ}$ is an exact solution, not just a close reading. On the graph both curves pass through $y = \sqrt{3} \approx 1.7$ there.

(c) $\tan x > 2\sin x$ where the blue curve is above the red curve. Work along the graph from left to right: from $0^{\circ}$ to $60^{\circ}$ the red curve is higher; from $60^{\circ}$ up to the asymptote at $90^{\circ}$ the blue curve is higher.

Between $90^{\circ}$ and $180^{\circ}$ the blue branch is below the axis while the red curve is above it. Between $180^{\circ}$ and the asymptote at $270^{\circ}$ the blue branch is above the axis and the red curve is below it.

After $270^{\circ}$ the blue branch starts far below the red curve and crosses it at $300^{\circ}$; from there to $360^{\circ}$ it stays above.