Trigonometric Graphs and Ratios of Any Angle

Work out \cos 0^\circ and \cos 180^\circ, and remember that the graph of y=\cos x is a smooth curve.

1. \cos 0^\circ=1, so the graph starts at (0,\ 1): this rules out A and B
2. Graph A starts at (0,\ 0): it is y=\sin x
3. Graph B starts at (0,\ -1): it is the cos graph upside down
4. Graph D is made of straight lines with a sharp point at (180,\ -1); the cos graph is a smooth curve through (45,\ 0.71), not (45,\ 0.5)
5. Answer: Graph C

\tan x=\dfrac{\sin x}{\cos x}, so look for the angles where you would be dividing by zero.

1. \tan x=\dfrac{\sin x}{\cos x} cannot be worked out when \cos x=0
2. \cos x=0 at x=90^\circ and x=270^\circ
3. These are the vertical asymptotes of the graph of y=\tan x
4. Answer: 90^\circ, 270^\circ

The graph of y=\sin x for 0^\circ\leqslant x\leqslant 180^\circ is symmetrical about the line x=90^\circ.

1. The sine graph is symmetrical about x=90^\circ
2. 48^\circ is 42^\circ before 90^\circ, so the other angle is 42^\circ after 90^\circ
3. \sin x=\sin(180^\circ-48^\circ)
4. Answer: x=132^\circ

Sketch y=\cos x for 0^\circ\leqslant x\leqslant 360^\circ and mark the point at x=40^\circ: the graph is symmetrical about x=180^\circ and about x=90^\circ (with the sign changing).

1. On the cos graph, \cos(180^\circ-40^\circ)=-\cos 40^\circ, so \cos 140^\circ=-k
2. The graph is symmetrical about x=180^\circ, so \cos 220^\circ=\cos 140^\circ=-k
3. \cos 320^\circ=\cos 40^\circ=k (positive, not negative)
4. \cos 130^\circ and \cos 50^\circ come from 50^\circ, not 40^\circ, so they are not \pm k
5. Answer: \cos 220^\circ

Decide whether \tan 160^\circ is positive or negative, then use the symmetry of the tan graph about (180^\circ,\ 0).

1. 160^\circ is 20^\circ before 180^\circ
2. The tan graph has rotational symmetry about (180^\circ,\ 0), so \tan(180^\circ-20^\circ)=-\tan 20^\circ
3. \tan x is negative between 90^\circ and 180^\circ
4. Answer: \tan 160^\circ=-k

In a right-angled triangle, the sine of one acute angle is the cosine of the other acute angle.

1. The graph of y=\cos x is the graph of y=\sin x moved 90^\circ to the left
2. In a right-angled triangle with angles 50^\circ and 40^\circ, the side opposite 50^\circ is adjacent to 40^\circ
3. So \cos 40^\circ=\sin 50^\circ
4. Answer: x=40^\circ

Picture the graphs of y=\sin x and y=\tan x and find where each one is below or above the x-axis.

1. \sin\theta<0 for 180^\circ<\theta<360^\circ
2. \tan\theta>0 for 0^\circ<\theta<90^\circ and 180^\circ<\theta<270^\circ
3. Both are true only for 180^\circ<\theta<270^\circ
4. Answer: 180^\circ<\theta<270^\circ

The graph of y=\cos x for 0^\circ\leqslant x\leqslant 360^\circ is symmetrical about the line x=180^\circ.

1. The cos graph is symmetrical about x=180^\circ
2. a is a degrees after 0^\circ, so the matching angle is a degrees before 360^\circ
3. Check with a=60^\circ: \cos 300^\circ=\cos 60^\circ=0.5
4. Answer: x=360^\circ-a

Picture the cos graph: where does it cross the x-axis between 0^\circ and 360^\circ, and how often does the pattern repeat?

1. For 0^\circ\leqslant x\leqslant 360^\circ the cos graph crosses the x-axis at 90^\circ and 270^\circ
2. The graph repeats every 360^\circ, so add 360^\circ to each: 450^\circ and 630^\circ
3. Answer: 90^\circ, 270^\circ, 450^\circ, 630^\circ

\cos x=-k is the reflection of \cos x=k in the x-axis: find the matching points in each part of the graph that is below the axis.

1. \cos 70^\circ=k, and 70^\circ is 20^\circ before 90^\circ
2. The graph is negative from 90^\circ to 270^\circ; 20^\circ after 90^\circ gives \cos 110^\circ=-k
3. By symmetry about x=180^\circ: 360^\circ-110^\circ=250^\circ
4. The graph repeats every 360^\circ: 110^\circ+360^\circ=470^\circ (250^\circ+360^\circ=610^\circ is too big)
5. Answer: 110^\circ, 250^\circ, 470^\circ

Use the graphs to write \sin 120^\circ and \cos 150^\circ in terms of \sin 60^\circ and \cos 30^\circ, taking care with the signs.

1. \sin 120^\circ=\sin 60^\circ=\dfrac{\sqrt3}{2} (symmetry about x=90^\circ)
2. \cos 150^\circ=-\cos 30^\circ=-\dfrac{\sqrt3}{2} (cos is negative between 90^\circ and 270^\circ)
3. \dfrac{\sqrt3}{2}\times\left(-\dfrac{\sqrt3}{2}\right)=-\dfrac{3}{4}
4. Answer: -\dfrac34

Count the solutions in each 360^\circ cycle of the sine graph, and look carefully at the last part, from 720^\circ to 1000^\circ.

1. \sin x is negative between 180^\circ and 360^\circ, so there are 2 solutions from 0^\circ to 360^\circ: 197.5^\circ and 342.5^\circ
2. From 360^\circ to 720^\circ there are 2 more: 557.5^\circ and 702.5^\circ
3. From 720^\circ to 1000^\circ: 917.5^\circ is a solution, but the next one, 1062.5^\circ, is too big
4. 2+2+1=5
5. Answer: 5

Use the graph to decide in which two parts of 0^\circ to 360^\circ \tan x is negative.

1. \tan x is negative for 90^\circ<x<180^\circ and 270^\circ<x<360^\circ
2. The tan graph has rotational symmetry about (180^\circ,\ 0), so \tan(180^\circ-a)=-\tan a=-k
3. The graph repeats every 180^\circ, so \tan(360^\circ-a)=-k too
4. 180^\circ+a gives +k, and 360^\circ+a is outside the range
5. Answer: 180^\circ-a and 360^\circ-a

Find the two angles where \cos x=\frac12, then use the graph to see whether \cos x is below \frac12 between them or outside them.

1. \cos 60^\circ=\frac12, and by symmetry about x=180^\circ, \cos 300^\circ=\frac12
2. The cos graph starts at 1, falls to -1 at 180^\circ and rises back to 1 at 360^\circ
3. So \cos x is \frac12 or less from 60^\circ to 300^\circ
4. Answer: 60^\circ\leqslant x\leqslant 300^\circ

Your calculator gives a negative angle for \tan^{-1}(-1.5): use the fact that the tan graph repeats every 180^\circ to move it into the range.

1. \tan^{-1}(-1.5)=-56.309\ldots^\circ, which is not in the range
2. The tan graph repeats every 180^\circ
3. A: -56.309\ldots^\circ+180^\circ=123.690\ldots^\circ
4. B: 123.690\ldots^\circ+180^\circ=303.690\ldots^\circ
5. C: 303.690\ldots^\circ+180^\circ=483.690\ldots^\circ
6. Answer: x=483.7^\circ

Mark a point at x=a, between 90^\circ and 180^\circ, on the graph, and use the symmetry of the graph to find the other points at the same height or the opposite height.

Part (a)

1. The sine graph is symmetrical about x=90^\circ
2. a is between 90^\circ and 180^\circ, so the matching angle is between 0^\circ and 90^\circ
3. Answer: x=180^\circ-a

Part (b)

1. The graph from 180^\circ to 360^\circ is the graph from 0^\circ to 180^\circ reflected in the x-axis and moved 180^\circ to the right
2. So \sin(180^\circ+a)=-\sin a=-k
3. a is between 90^\circ and 180^\circ, so 180^\circ+a is between 270^\circ and 360^\circ
4. (The other solution, 360^\circ-a, is between 180^\circ and 270^\circ)
5. Answer: x=180^\circ+a

Part (c)

1. The graph of y=\cos x is the graph of y=\sin x moved 90^\circ to the left
2. So \cos(a-90^\circ)=\sin a=k
3. a is between 90^\circ and 180^\circ, so a-90^\circ is between 0^\circ and 90^\circ
4. Check with a=120^\circ: \cos 30^\circ=\sin 120^\circ=\dfrac{\sqrt3}{2}
5. Answer: x=a-90^\circ

Where the graphs meet, \sin x=\cos x: divide both sides by \cos x.

Part (a)

1. At P, \sin x=\cos x, so \dfrac{\sin x}{\cos x}=1, which means \tan x=1
2. \tan 45^\circ=1, so x=45^\circ
3. y=\sin 45^\circ=\dfrac{\sqrt2}{2} (from the 1:1:\sqrt2 triangle)
4. Answer: P\left(45,\ \dfrac{\sqrt2}{2}\right)

Part (b)

1. \tan x=1 again; the tan graph repeats every 180^\circ
2. x=45^\circ+180^\circ=225^\circ
3. \sin 225^\circ=-\sin 45^\circ=-\dfrac{\sqrt2}{2} (sine is negative between 180^\circ and 360^\circ)
4. Answer: Q\left(225,\ -\dfrac{\sqrt2}{2}\right)

Part (c)

1. The sine graph is above the cos graph between the two points where they meet
2. Answer: 45^\circ<x<225^\circ

Picture the graph of y=\cos x for 0^\circ<x<360^\circ and a horizontal line y=c: count how many times the line meets the curve as c changes, with the end points not included.

Part (a)

1. k=4: \cos x=\dfrac{4-3}{4}=\dfrac14
2. x=\cos^{-1}\left(\dfrac14\right)=75.522\ldots^\circ
3. The cos graph is symmetrical about x=180^\circ: 360^\circ-75.522\ldots^\circ=284.477\ldots^\circ
4. Answer: 75.5^\circ, 284.5^\circ

Part (b)

1. For 0^\circ<x<360^\circ, a horizontal line y=c meets the cos graph twice when -1<c<1
2. (c=1 gives only x=0^\circ and 360^\circ, which are not allowed; c=-1 gives only x=180^\circ)
3. So -1<\dfrac{k-3}{4}<1
4. Multiply by 4: -4<k-3<4
5. Add 3: -1<k<7
6. Answer: -1<k<7

Part (c)

1. The line y=c meets the cos graph once, at x=180^\circ, only when c=-1
2. \dfrac{k-3}{4}=-1, so k-3=-4
3. Answer: k=-1

The largest and smallest values of \sin x are 1 and -1: work out what \mathrm{f}(x) is at each.

Part (a)

1. \mathrm{f}(x) is greatest when 4\sin x is as small as possible, that is when \sin x=-1
2. \sin x=-1 at the lowest point of the sine graph
3. Answer: x=270^\circ

Part (b)

1. \sin x takes every value from -1 to 1
2. \sin x=1: \mathrm{f}(x)=5-4=1
3. \sin x=-1: \mathrm{f}(x)=5+4=9
4. Answer: 1\leqslant\mathrm{f}(x)\leqslant 9

Part (c)

1. 5-4\sin x=7, so -4\sin x=2
2. \sin x=-\dfrac12
3. \sin 30^\circ=\dfrac12, and sine is negative between 180^\circ and 360^\circ
4. x=180^\circ+30^\circ or x=360^\circ-30^\circ
5. Answer: x=210^\circ, 330^\circ

Decide which quadrant \theta is in from the signs of \sin\theta and \cos\theta before you use your calculator.

Part (a)

1. \sin^2\theta+\cos^2\theta=1, so \cos^2\theta=1-\dfrac{400}{841}=\dfrac{441}{841}
2. \cos\theta>0, so \cos\theta=\dfrac{21}{29}
3. \tan\theta=\dfrac{\sin\theta}{\cos\theta}=-\dfrac{20}{29}\div\dfrac{21}{29}
4. Answer: \tan\theta=-\dfrac{20}{21}

Part (b)

1. \sin\theta<0 and \cos\theta>0, so \theta is between 270^\circ and 360^\circ
2. The acute angle with \sin=\dfrac{20}{29} is \sin^{-1}\!\left(\dfrac{20}{29}\right)=43.60\ldots^\circ
3. Using the symmetry of the sine graph: \theta=360^\circ-43.60\ldots^\circ
4. Answer: \theta=316.4^\circ (1 d.p.)

Part (c)

1. -\tan\theta=\dfrac{20}{21}, which is positive
2. \tan^{-1}\!\left(\dfrac{20}{21}\right)=43.60\ldots^\circ
3. The tan graph repeats every 180^\circ, so the value between 180^\circ and 360^\circ is 180^\circ+43.60\ldots^\circ
4. Answer: x=223.6^\circ (1 d.p.)