Solved Questions

How to Use These Questions

The 35 questions below mix every part of this chapter in no fixed order, the way a paper does: a clock angle can follow an identity, and an allied angle can follow a maximum value. Before any working, decide what kind of question it is, because that picks the method: a ratio given (draw the triangle), a condition given (square it, or use the reciprocal partner), an expression to simplify (test at $45^\circ$, then $30^\circ$ or $60^\circ$), or an angle beyond $90^\circ$ (sign first).

The five marked [Tier 2] use triple angles, a product of cosines, $\sin 18^\circ$ and longer eliminations. A Tier 1 question of this kind should take you under a minute; if one takes longer, find its quick route in Shortcuts & Speed Methods and try it again the next day.

Solved Examples

Question 1

The minute hand of a clock is 21 cm long. How far does its tip move in 40 minutes? (Take $\pi = \frac{22}{7}$.)

Answer. In 40 minutes the minute hand turns $40 \times 6^\circ = 240^\circ$, which is $\frac{240}{180}\pi = \frac{4\pi}{3}$ radians. The tip moves along an arc, so $l = r\theta = 21 \times \frac{4}{3} \times \frac{22}{7} = 28 \times \frac{22}{7}$ = 88 cm.

Question 2

In triangle ABC, right-angled at B, AB = 40 cm and BC = 9 cm. Find $\sec A\operatorname{cosec} C - \tan A\cot C$.

Answer. First the hypotenuse: $AC = \sqrt{1600 + 81} = \sqrt{1681} = 41$ cm (the 9-40-41 triplet). At A the opposite side is BC = 9 and the adjacent side is AB = 40; at C it is the other way round. So $\sec A = \frac{41}{40}$, $\operatorname{cosec} C = \frac{41}{40}$, $\tan A = \frac{9}{40}$ and $\cot C = \frac{9}{40}$. The value is $\frac{1681}{1600} - \frac{81}{1600} = \frac{1600}{1600}$ = 1. This is no accident: $A + C = 90^\circ$, so $\operatorname{cosec} C = \sec A$ and $\cot C = \tan A$, and the expression is just $\sec^2 A - \tan^2 A$.

Question 3

Find the value of $\frac{3\operatorname{cosec}^2 60^\circ + 2\cot^2 30^\circ - \sec^2 45^\circ}{4\cos^2 45^\circ + \tan^2 60^\circ}$.

Answer. I write the squared values first: $\operatorname{cosec}^2 60^\circ = \frac{4}{3}$, $\cot^2 30^\circ = 3$, $\sec^2 45^\circ = 2$, $\cos^2 45^\circ = \frac{1}{2}$, $\tan^2 60^\circ = 3$. The top is $4 + 6 - 2 = 8$ and the bottom is $2 + 3 = 5$, so the value is $\frac{8}{5}$.

Question 4

If $\operatorname{cosec}\theta + \cot\theta = 3$, find $\sin\theta + \cos\theta$.

Answer. The partner is the reciprocal: $\operatorname{cosec}\theta - \cot\theta = \frac{1}{3}$. Adding, $2\operatorname{cosec}\theta = \frac{10}{3}$, so $\operatorname{cosec}\theta = \frac{5}{3}$ and $\sin\theta = \frac{3}{5}$. Subtracting, $2\cot\theta = \frac{8}{3}$, so $\cot\theta = \frac{4}{3}$. Then $\cos\theta = \cot\theta \times \sin\theta = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5}$, and $\sin\theta + \cos\theta$ = $\frac{7}{5}$.

Question 5

Find $\sin^2 1^\circ + \sin^2 3^\circ + \sin^2 5^\circ + \cdots + \sin^2 89^\circ$.

Answer. These are the odd angles from $1^\circ$ to $89^\circ$, so there are $\frac{89 - 1}{2} + 1 = 45$ terms. They pair up to $90^\circ$ ($1^\circ$ with $89^\circ$, $3^\circ$ with $87^\circ$, …), giving 22 pairs, and the middle term $\sin^2 45^\circ = \frac{1}{2}$ is left alone. The sum is $22 + \frac{1}{2}$ = $22\frac{1}{2}$.

Question 6

If $\frac{\cos\theta}{1 - \sin\theta} + \frac{\cos\theta}{1 + \sin\theta} = 4$ and $\theta$ is acute, find $\theta$.

Answer. Over a common denominator the left side is $\frac{\cos\theta(1 + \sin\theta + 1 - \sin\theta)}{1 - \sin^2\theta} = \frac{2\cos\theta}{\cos^2\theta} = 2\sec\theta$. So $2\sec\theta = 4$, $\sec\theta = 2$, and $\theta$ = $60^\circ$.

Question 7

Find the value of $\frac{\sin 150^\circ - 5\cos 300^\circ + 7\tan 225^\circ}{\tan 135^\circ + 3\sin 210^\circ}$.

Answer. One angle at a time, sign first:

• $150^\circ$ is in the second quadrant, where sin is positive: $\sin 150^\circ = \sin 30^\circ = \frac{1}{2}$.
• $300^\circ$ is in the fourth, where cos is positive: $\cos 300^\circ = \cos 60^\circ = \frac{1}{2}$.
• $225^\circ$ is in the third, where tan is positive: $\tan 225^\circ = \tan 45^\circ = 1$.
• $\tan 135^\circ = -\tan 45^\circ = -1$ and $\sin 210^\circ = -\sin 30^\circ = -\frac{1}{2}$.

The top is $\frac{1}{2} - \frac{5}{2} + 7 = 5$ and the bottom is $-1 - \frac{3}{2} = -\frac{5}{2}$, so the value is $-2$.

Question 8

Express $47^\circ 30'$ in radians.

Answer. $30'$ is half a degree, so the angle is $47.5^\circ = \frac{95}{2}$ degrees. In radians that is $\frac{95}{2} \times \frac{\pi}{180} = \frac{95\pi}{360}$ = $\frac{19\pi}{72}$.

Question 9

If $15\cos^2\theta + 2\sin\theta = 7$ and $\theta$ is acute, find $\cot\theta$.

Answer. I change $\cos^2\theta$ to $1 - \sin^2\theta$ so that only sin is left: $15 - 15\sin^2\theta + 2\sin\theta = 7$, that is $15\sin^2\theta - 2\sin\theta - 8 = 0$. This factorises as $(5\sin\theta - 4)(3\sin\theta + 2) = 0$. An acute angle has a positive sine, so $\sin\theta = \frac{4}{5}$. Then $\cos\theta = \frac{3}{5}$ and $\cot\theta$ = $\frac{3}{4}$.

Question 10

If $\tan A = \frac{3}{4}$ and $\tan B = \frac{7}{24}$, where A and B are acute, find $\tan(A + B)$ and $\sin(A + B)$.

Answer. $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} = \frac{\frac{3}{4} + \frac{7}{24}}{1 - \frac{7}{32}} = \frac{\frac{25}{24}}{\frac{25}{32}} = \frac{32}{24}$ = $\frac{4}{3}$. It is positive, so $A + B$ is acute, and a tan of $\frac{4}{3}$ belongs to the 3-4-5 triangle. So $\sin(A + B)$ = $\frac{4}{5}$.

Question 11

Simplify $\frac{\cos\theta}{\operatorname{cosec}\theta + 1} + \frac{\cos\theta}{\operatorname{cosec}\theta - 1}$.

Answer. I multiply the top and the bottom of each fraction by $\sin\theta$, which turns the cosecs into 1s: $\frac{\sin\theta\cos\theta}{1 + \sin\theta} + \frac{\sin\theta\cos\theta}{1 - \sin\theta}$. The denominators are a conjugate pair, so the sum is $\sin\theta\cos\theta \times \frac{(1 - \sin\theta) + (1 + \sin\theta)}{1 - \sin^2\theta} = \frac{2\sin\theta\cos\theta}{\cos^2\theta}$ = $2\tan\theta$.

Question 12

Find the greatest and least values of $\sin^2\theta + \cos\theta$.

Answer. One term is squared and the other is not, so the $\sqrt{a^2 + b^2}$ rule does not apply. I write everything in $c = \cos\theta$: $1 - c^2 + c = \frac{5}{4} - \left(c - \frac{1}{2}\right)^2$. The square is 0 when $c = \frac{1}{2}$ ($\theta = 60^\circ$), so the greatest value is $\frac{5}{4}$. The square is largest when c is as far from $\frac{1}{2}$ as it can go, which is $c = -1$ ($\theta = 180^\circ$); then the value is $1 - 1 - 1$ = $-1$.

Question 13 [Tier 2]

Find the value of $\cos 36^\circ\cos 72^\circ$.

Answer. I multiply and divide by $\sin 36^\circ$ and use $2\sin x\cos x = \sin 2x$ twice: $\sin 36^\circ\cos 36^\circ\cos 72^\circ = \frac{1}{2}\sin 72^\circ\cos 72^\circ = \frac{1}{4}\sin 144^\circ$. Since $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$, dividing by $\sin 36^\circ$ leaves $\frac{1}{4}$.

Question 14

In triangle PQR, PQ = PR = 13 cm and QR = 10 cm. Find $\sin Q$ and $\tan\frac{P}{2}$.

Answer. The altitude PD from P cuts QR in half, so QD = 5 cm, and triangle PDQ has hypotenuse 13 and one side 5. That makes PD = 12 cm (5-12-13). So $\sin Q = \frac{PD}{PQ}$ = $\frac{12}{13}$. The altitude also cuts angle P in half, so in triangle PDQ the angle at P is $\frac{P}{2}$, with opposite side QD = 5 and adjacent side PD = 12. So $\tan\frac{P}{2}$ = $\frac{5}{12}$.

Question 15

If $\cos(A + B) = 0$ and $\sin(A - B) = \frac{\sqrt{3}}{2}$, where A and B are acute and $A > B$, find A and B, and the value of $\tan A\tan B$.

Answer. $\cos(A + B) = 0$ gives $A + B = 90^\circ$, and $\sin(A - B) = \frac{\sqrt{3}}{2}$ gives $A - B = 60^\circ$. Adding and halving, $A = 75^\circ$ and $B = 15^\circ$. The two angles add to $90^\circ$, so $\tan B = \cot A$ and $\tan A\tan B$ = 1. With the values, $(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$ agrees.

Question 16

Find the value of $\frac{\cos 38^\circ}{\sin 52^\circ} + \frac{\tan 24^\circ}{\cot 66^\circ} - \frac{3\tan 12^\circ\tan 40^\circ\tan 50^\circ\tan 78^\circ}{\sin^2 33^\circ + \sin^2 57^\circ}$.

Answer. Each pair of angles adds to $90^\circ$. $\cos 38^\circ = \sin 52^\circ$, so the first fraction is 1, and $\tan 24^\circ = \cot 66^\circ$, so the second is 1 too. In the last fraction, $\tan 12^\circ\tan 78^\circ = 1$ and $\tan 40^\circ\tan 50^\circ = 1$, so the top is 3; the bottom is $\sin^2 33^\circ + \cos^2 33^\circ = 1$. The value is $1 + 1 - 3$ = $-1$.

Question 17

If $\tan\theta + \cot\theta = 2\sqrt{2}$, find $\tan^3\theta + \cot^3\theta$.

Answer. With $x = \tan\theta$, I have $x + \frac{1}{x} = 2\sqrt{2}$, and the cube rule from algebra (see Chapter 11) gives $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) = 16\sqrt{2} - 6\sqrt{2}$ = $10\sqrt{2}$.

Question 18

Simplify $\frac{\tan^2\theta}{\sec\theta - 1} + \frac{\cot^2\theta}{\operatorname{cosec}\theta - 1}$, and find its least value for an acute angle $\theta$.

Answer. $\tan^2\theta = \sec^2\theta - 1 = (\sec\theta - 1)(\sec\theta + 1)$, so the first fraction is $\sec\theta + 1$. In the same way $\cot^2\theta = (\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1)$, and the second fraction is $\operatorname{cosec}\theta + 1$. The sum is $\sec\theta + \operatorname{cosec}\theta + 2$. For the least value, AM-GM gives $\sec\theta + \operatorname{cosec}\theta \ge 2\sqrt{\sec\theta\operatorname{cosec}\theta}$, and $\sec\theta\operatorname{cosec}\theta = \frac{1}{\sin\theta\cos\theta}$ is at least 2. Both become equalities at $45^\circ$, where the sum is $\sqrt{2} + \sqrt{2}$, so the least value is $2 + 2\sqrt{2}$.

Question 19

Simplify $\frac{\sin(180^\circ - \theta)\cos(90^\circ + \theta)\tan(270^\circ - \theta)}{\cos(360^\circ - \theta)\sin(90^\circ + \theta)\cot(180^\circ + \theta)}$.

Answer. Taking $\theta$ as a small acute angle, I find each quadrant and sign:

• $\sin(180^\circ - \theta) = \sin\theta$ (second quadrant, sin positive)
• $\cos(90^\circ + \theta) = -\sin\theta$ (second quadrant, cos negative, and $90^\circ$ switches the ratio)
• $\tan(270^\circ - \theta) = \cot\theta$ (third quadrant, tan positive, and $270^\circ$ switches the ratio)
• $\cos(360^\circ - \theta) = \cos\theta$, $\sin(90^\circ + \theta) = \cos\theta$, $\cot(180^\circ + \theta) = \cot\theta$

So the value is $\frac{\sin\theta \times (-\sin\theta) \times \cot\theta}{\cos\theta \times \cos\theta \times \cot\theta}$ = $-\tan^2\theta$.

Question 20

Find the value of $(\cot 11^\circ - 1)(\cot 34^\circ - 1)$.

Answer. $11^\circ + 34^\circ = 45^\circ$, so $\cot(A + B) = 1$, which means $\frac{\cot A\cot B - 1}{\cot A + \cot B} = 1$. Cross-multiplying, $\cot A\cot B - \cot A - \cot B = 1$. Adding 1 to both sides turns the left side into $(\cot A - 1)(\cot B - 1)$, so the value is 2.

Question 21 [Tier 2]

If $\sin\theta = m\sin\phi$ and $\tan\theta = n\tan\phi$, show that $\cos^2\theta = \frac{m^2 - 1}{n^2 - 1}$.

Answer. Dividing the first equation by the second gives $\cos\theta = \frac{m}{n}\cos\phi$. So $\sin\phi = \frac{\sin\theta}{m}$ and $\cos\phi = \frac{n\cos\theta}{m}$. Now $\phi$ has to go, and $\sin^2\phi + \cos^2\phi = 1$ does that: $\frac{\sin^2\theta}{m^2} + \frac{n^2\cos^2\theta}{m^2} = 1$, that is $\sin^2\theta + n^2\cos^2\theta = m^2$. Writing $\sin^2\theta = 1 - \cos^2\theta$ gives $1 + (n^2 - 1)\cos^2\theta = m^2$, so $\cos^2\theta = \frac{m^2 - 1}{n^2 - 1}$.

Question 22

The angles of a triangle are in the ratio 2 : 5 : 8. Find the largest angle in radians.

Answer. The parts add to 15, so the largest angle is $\frac{8}{15} \times 180^\circ = 96^\circ$. Since $180^\circ$ is $\pi$ radians, it is simply $\frac{8}{15}$ of $\pi$: $\frac{8\pi}{15}$.

Question 23

If $\sin\theta - \cos\theta = \frac{7}{13}$ and $\theta$ is acute, find $\sin\theta + \cos\theta$.

Answer. Squaring, $1 - 2\sin\theta\cos\theta = \frac{49}{169}$, so $2\sin\theta\cos\theta = \frac{120}{169}$. Then $(\sin\theta + \cos\theta)^2 = 1 + \frac{120}{169} = \frac{289}{169}$. For an acute angle the sum is positive, so $\sin\theta + \cos\theta$ = $\frac{17}{13}$.

Question 24

Find the greatest and least values of $15\sin\theta - 8\cos\theta$, and of $15\sin^2\theta + 8\cos^2\theta$.

Answer. For the first, $\sqrt{225 + 64} = 17$, so it runs from $-17$ to 17. The second has squares, so I write it as $8 + 7\sin^2\theta$; with $\sin^2\theta$ going from 0 to 1, it runs from 8 to 15. The two look alike but need different methods: the first can go negative, the second never can.

Question 25

In triangle ABC, $\cos\frac{B + C}{2} = \frac{3}{5}$. Find $\tan\frac{A}{2}$.

Answer. The angles add to $180^\circ$, so $\frac{B + C}{2} = 90^\circ - \frac{A}{2}$ and $\cos\frac{B + C}{2} = \sin\frac{A}{2} = \frac{3}{5}$. Half of a triangle's angle is acute, so $\cos\frac{A}{2} = \frac{4}{5}$, and $\tan\frac{A}{2}$ = $\frac{3}{4}$.

Question 26

Simplify $\frac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta}$.

Answer. I use $1 - \cos 2\theta = 2\sin^2\theta$, $1 + \cos 2\theta = 2\cos^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$. The top becomes $2\sin\theta(\sin\theta + \cos\theta)$ and the bottom $2\cos\theta(\cos\theta + \sin\theta)$. The brackets cancel and the answer is $\tan\theta$.

Question 27

Find the value of $\sin 1200^\circ\cos(-1050^\circ)$.

Answer. $1200^\circ - 3 \times 360^\circ = 120^\circ$, in the second quadrant where sin is positive: $\sin 1200^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$. For the cos, the minus sign goes away, and $1050^\circ - 2 \times 360^\circ = 330^\circ$, in the fourth quadrant where cos is positive: $\cos 330^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}$. The product is $\frac{3}{4}$.

Question 28

If $\sqrt{2}\cos 3\theta - 1 = 0$ and $3\theta$ is acute, find $\theta$ and the value of $\sin 2\theta + \tan 3\theta$.

Answer. $\cos 3\theta = \frac{1}{\sqrt{2}}$, so $3\theta = 45^\circ$ and $\theta = 15^\circ$. Then $\sin 30^\circ + \tan 45^\circ = \frac{1}{2} + 1$ = $\frac{3}{2}$.

Question 29

If $x = 5\sec\theta + 2\tan\theta$ and $y = 5\tan\theta + 2\sec\theta$, find $x^2 - y^2$.

Answer. Squaring, $x^2 = 25\sec^2\theta + 20\sec\theta\tan\theta + 4\tan^2\theta$ and $y^2 = 25\tan^2\theta + 20\sec\theta\tan\theta + 4\sec^2\theta$. The middle terms cancel when I subtract: $x^2 - y^2 = 21\sec^2\theta - 21\tan^2\theta = 21(\sec^2\theta - \tan^2\theta)$ = 21.

Question 30 [Tier 2]

Find the value of $\frac{3\sin 40^\circ - 4\sin^3 40^\circ}{4\cos^3 40^\circ - 3\cos 40^\circ}$.

Answer. The top is $\sin 3\theta$ and the bottom is $\cos 3\theta$ with $\theta = 40^\circ$, so the fraction is $\tan 120^\circ$. That is in the second quadrant, where tan is negative: $\tan 120^\circ = -\tan 60^\circ$ = $-\sqrt{3}$.

Question 31

Find the value of $(2\sin\theta + 3\cos\theta)^2 + (3\sin\theta - 2\cos\theta)^2$.

Answer. Expanding, the cross terms are $+12\sin\theta\cos\theta$ and $-12\sin\theta\cos\theta$, so they cancel. What is left is $4\sin^2\theta + 9\cos^2\theta + 9\sin^2\theta + 4\cos^2\theta = 13(\sin^2\theta + \cos^2\theta)$ = 13, the same for every $\theta$.

Question 32

Find the value of $\sin 105^\circ + \cos 105^\circ$.

Answer. $105^\circ = 60^\circ + 45^\circ$. So $\sin 105^\circ = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$, and $\cos 105^\circ = \frac{1}{2} \times \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$, which is negative, as it should be in the second quadrant. Adding, the $\sqrt{3}$ terms cancel: $\frac{2}{2\sqrt{2}}$ = $\frac{1}{\sqrt{2}}$.

Question 33

If $\sec\theta - \tan\theta = \frac{1}{\sqrt{3}}$, find the acute angle $\theta$.

Answer. The partner is the reciprocal: $\sec\theta + \tan\theta = \sqrt{3}$. Adding the two and halving, $\sec\theta = \frac{1}{2}\left(\sqrt{3} + \frac{1}{\sqrt{3}}\right) = \frac{1}{2} \times \frac{4}{\sqrt{3}} = \frac{2}{\sqrt{3}}$. So $\cos\theta = \frac{\sqrt{3}}{2}$ and $\theta$ = $30^\circ$. As a check, $\tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.

Question 34 [Tier 2]

Using $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$, find $\sin^2 54^\circ - \sin^2 18^\circ$.

Answer. First, $\sin 54^\circ = \cos 36^\circ$, since the angles add to $90^\circ$. And $\cos 36^\circ = 1 - 2\sin^2 18^\circ = 1 - 2 \times \frac{6 - 2\sqrt{5}}{16} = \frac{\sqrt{5} + 1}{4}$. So $\sin^2 54^\circ - \sin^2 18^\circ = \frac{(\sqrt{5} + 1)^2 - (\sqrt{5} - 1)^2}{16} = \frac{4\sqrt{5}}{16}$ = $\frac{\sqrt{5}}{4}$.

Question 35 [Tier 2]

If $\frac{\cos^4\alpha}{\cos^2\beta} + \frac{\sin^4\alpha}{\sin^2\beta} = 1$, find $\frac{\cos^4\beta}{\cos^2\alpha} + \frac{\sin^4\beta}{\sin^2\alpha}$.

Answer. I write $a = \cos^2\alpha$ and $b = \cos^2\beta$, so the condition is $\frac{a^2}{b} + \frac{(1 - a)^2}{1 - b} = 1$. Multiplying by $b(1 - b)$: $a^2(1 - b) + (1 - a)^2b = b(1 - b)$. Expanding and cancelling leaves $a^2 - 2ab + b^2 = 0$, so $(a - b)^2 = 0$ and $a = b$. So $\cos^2\alpha = \cos^2\beta$ and $\sin^2\alpha = \sin^2\beta$, and the second expression becomes $\cos^2\beta + \sin^2\beta$ = 1.