Standard Angles and Complementary Angles

The 45° Values: Half a Square

Cut a square of side 1 along a diagonal. Each half is a right triangle with two equal sides of length 1, so its two acute angles are equal: 45° each. The diagonal is $\sqrt{1^2 + 1^2} = \sqrt{2}$.

At either 45° angle the opposite and adjacent sides are both 1 and the hypotenuse is $\sqrt{2}$, so

$$\sin 45^\circ = \frac{1}{\sqrt{2}} \qquad \cos 45^\circ = \frac{1}{\sqrt{2}} \qquad \tan 45^\circ = 1$$

$$\operatorname{cosec} 45^\circ = \sqrt{2} \qquad \sec 45^\circ = \sqrt{2} \qquad \cot 45^\circ = 1$$

Rationalised, $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$, which is about 0.707. Options may print either form, so treat them as the same number.

45° is the only acute angle where the sine equals the cosine, because only there are the opposite and adjacent sides equal. For the same reason it is the only acute angle with $\tan\theta = 1$.

Sides of any 45-45-90 triangle. They are in the ratio $1 : 1 : \sqrt{2}$. A leg of 7 cm gives a hypotenuse of $7\sqrt{2}$ cm. A hypotenuse of 10 cm gives legs of $\frac{10}{\sqrt{2}} = 5\sqrt{2}$ cm, and the area of that triangle is $\frac{1}{2} \times 5\sqrt{2} \times 5\sqrt{2} = 25$ cm².

Squares of the values. $\sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$ and $\sec^2 45^\circ = (\sqrt{2})^2 = 2$. Squaring $\frac{1}{\sqrt{2}}$ squares the root as well, so the answer is $\frac{1}{2}$; writing $\frac{1}{4}$ (squaring as if the root were a 2) is a slip to avoid. As a check, $\sin^2 45^\circ + \cos^2 45^\circ = \frac{1}{2} + \frac{1}{2} = 1$.

The 30° and 60° Values: Half an Equilateral Triangle

Take an equilateral triangle ABC of side 2. All its angles are 60°. The altitude AD meets BC at its midpoint and cuts the top angle in half, so triangle ABD has 30° at A, 60° at B and 90° at D, with AB = 2, BD = 1 and $AD = \sqrt{4 - 1} = \sqrt{3}$.

At 30° (angle A): the opposite side is BD = 1, the adjacent side is $AD = \sqrt{3}$ and the hypotenuse is 2.

$$\sin 30^\circ = \frac{1}{2} \qquad \cos 30^\circ = \frac{\sqrt{3}}{2} \qquad \tan 30^\circ = \frac{1}{\sqrt{3}}$$

At 60° (angle B): the opposite side is $AD = \sqrt{3}$ and the adjacent side is BD = 1.

$$\sin 60^\circ = \frac{\sqrt{3}}{2} \qquad \cos 60^\circ = \frac{1}{2} \qquad \tan 60^\circ = \sqrt{3}$$

The reciprocals follow: $\operatorname{cosec} 30^\circ = 2$, $\sec 30^\circ = \frac{2}{\sqrt{3}}$, $\cot 30^\circ = \sqrt{3}$, and $\operatorname{cosec} 60^\circ = \frac{2}{\sqrt{3}}$, $\sec 60^\circ = 2$, $\cot 60^\circ = \frac{1}{\sqrt{3}}$.

Look at the pairs: $\sin 30^\circ = \cos 60^\circ$ and $\tan 30^\circ = \cot 60^\circ$. That is because 30° and 60° add up to 90°; the notes on complementary angles further down explain it.

Sides of any 30-60-90 triangle. They are in the ratio $1 : \sqrt{3} : 2$, facing 30°, 60° and 90°. The side facing 30° is half the hypotenuse. A hypotenuse of 14 cm gives sides of 7 cm and $7\sqrt{3}$ cm. If the side facing 60° is 9 cm, the shortest side is $\frac{9}{\sqrt{3}} = 3\sqrt{3}$ cm and the hypotenuse is $6\sqrt{3}$ cm.

The values $\frac{1}{\sqrt{3}}$ and $\sqrt{3}$ are easy to swap in a hurry. Picture the triangle: 60° is the steeper angle, so it has the bigger tan, $\sqrt{3} \approx 1.73$, while $\tan 30^\circ \approx 0.58$.

The Table of Standard Values, and How to Remember It

For 0° and 90°, use a circle of radius 1. A point on it at angle $\theta$ sits at $(\cos\theta, \sin\theta)$: its distance across is the cosine and its height is the sine, as shown in Angles, Radians and Trigonometric Ratios. At 0° the point is (1, 0), so $\cos 0^\circ = 1$ and $\sin 0^\circ = 0$. At 90° it is (0, 1), so $\cos 90^\circ = 0$ and $\sin 90^\circ = 1$.

All the values together:

How to remember the sine row. Write 0, 1, 2, 3, 4. Take the square root of each and divide by 2:

$$\frac{\sqrt{0}}{2}, \ \frac{\sqrt{1}}{2}, \ \frac{\sqrt{2}}{2}, \ \frac{\sqrt{3}}{2}, \ \frac{\sqrt{4}}{2} \quad\text{that is}\quad 0, \ \frac{1}{2}, \ \frac{1}{\sqrt{2}}, \ \frac{\sqrt{3}}{2}, \ 1$$

The cosine row is the same five numbers read backwards. The tan row is sine divided by cosine: 0, $\frac{1}{\sqrt{3}}$, 1, $\sqrt{3}$, and then $\frac{1}{0}$ at 90°. The last three rows are the first three turned upside down.

Values that are not defined. Division by zero has no answer, so $\tan 90^\circ = \frac{1}{0}$ and $\sec 90^\circ = \frac{1}{\cos 90^\circ}$ are not defined; nor are $\cot 0^\circ$ and $\operatorname{cosec} 0^\circ$. Some books write these as infinity, and an option saying ‘not defined’, ‘undefined’ or ‘infinite’ means the same thing. Zero itself is a proper value: $\tan 0^\circ = 0$ and $\cot 90^\circ = 0$.

[SSC Tip] When one option says ‘not defined’, look for a term that divides by $\cos 90^\circ$ or $\sin 0^\circ$. One such term makes the whole expression undefined, however neatly the other terms add up.

Working Out Expressions with Standard Angles

Replace each ratio by its value from the table, then simplify. Two habits help: square the whole value, root and all, and clear roots from the bottom at the end.

$$4\sin^2 60^\circ + 3\tan^2 30^\circ - 8\sin 45^\circ \cos 45^\circ = 4 \times \frac{3}{4} + 3 \times \frac{1}{3} - 8 \times \frac{1}{2} = 3 + 1 - 4 = 0$$

A sum times a difference. Spot the pattern $(a + b)(a - b) = a^2 - b^2$ and the root disappears:

$$(\operatorname{cosec} 30^\circ + \cot 45^\circ + \sin 45^\circ)(\operatorname{cosec} 30^\circ + \cot 45^\circ - \sin 45^\circ) = \left(3 + \frac{1}{\sqrt{2}}\right)\left(3 - \frac{1}{\sqrt{2}}\right) = 9 - \frac{1}{2} = \frac{17}{2}$$

A fraction with roots. Simplify the top and the bottom separately, then divide:

$$\frac{\tan 60^\circ + \cot 60^\circ}{\sec 30^\circ} = \frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{\frac{2}{\sqrt{3}}} = \frac{\frac{4}{\sqrt{3}}}{\frac{2}{\sqrt{3}}} = 2$$

Squaring. $\cos^2 30^\circ = \frac{3}{4}$, $\operatorname{cosec}^2 45^\circ = 2$, $\sec^2 30^\circ = \frac{4}{3}$ and $\tan^2 60^\circ = 3$. One slip is writing $\cos^2 30^\circ$ as $\frac{\sqrt{3}}{4}$, squaring the 2 but not the root.

Rationalising. Multiply top and bottom by the root: $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$, $\frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$ and $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. The options may show either form.

With 0° and 90°. A term multiplied by $\cos 90^\circ$, $\sin 0^\circ$ or $\tan 0^\circ$ is 0. A term that is 1 is easy to drop by accident: in $\sin 90^\circ + \cos 0^\circ + \tan 45^\circ$ every term is 1, and the value is 3.

Same angle or not. $\sin^2\theta + \cos^2\theta = 1$ needs the same angle in both terms. $\sin^2 30^\circ + \cos^2 30^\circ = 1$, but $\sin^2 30^\circ + \cos^2 60^\circ = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

A Ratio of a Sum Is Not a Sum of Ratios

The standard values show at once that a trigonometric ratio does not split over a plus sign or a multiple:

• $\sin(30^\circ + 60^\circ) = \sin 90^\circ = 1$, but $\sin 30^\circ + \sin 60^\circ = \frac{1 + \sqrt{3}}{2}$, about 1.37.
• $\cos(60^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}$, but $\cos 60^\circ - \cos 30^\circ = \frac{1 - \sqrt{3}}{2}$, which is negative.
• $\sin(2 \times 30^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$, but $2\sin 30^\circ = 1$.
• $\tan(2 \times 30^\circ) = \sqrt{3}$, but $2\tan 30^\circ = \frac{2}{\sqrt{3}}$.

So $\sin(A + B) \neq \sin A + \sin B$ and $\sin 2A \neq 2\sin A$. The correct formulas for $\sin(A + B)$, $\cos 2A$ and the rest are in Allied, Compound and Double Angles. If an answer came from opening a bracket this way, it is wrong.

Testing a formula at a standard angle. Put $\theta = 30^\circ$ into a claimed formula and see whether both sides agree.

• $\cos 2\theta = 1 - 2\sin^2\theta$: the left side is $\cos 60^\circ = \frac{1}{2}$ and the right side is $1 - 2 \times \frac{1}{4} = \frac{1}{2}$. They agree.
• $\frac{2\tan\theta}{1 - \tan^2\theta}$ at $\theta = 30^\circ$ is $\frac{2/\sqrt{3}}{2/3} = \sqrt{3}$, which is $\tan 60^\circ$. So this expression gives the tan of the double angle.
• $\sin 2\theta = 2\sin\theta$ fails at once: $\frac{\sqrt{3}}{2}$ on the left, 1 on the right.

One angle that agrees does not prove a formula, but one angle that disagrees is enough to throw it out. Identities and Simplification uses this test to check the options of a simplification question, and shows when two options can agree at the same angle.

Finding the Angle from Its Value

Between 0° and 90° the sine keeps rising, so each value of the sine belongs to exactly one angle there; the same holds for cosine and tan. Read the table backwards: $\sin\theta = \frac{\sqrt{3}}{2}$ means $\theta = 60^\circ$, and $\tan\theta = \frac{1}{\sqrt{3}}$ means $\theta = 30^\circ$.

A multiple of the angle inside. Find the whole bracket first, then $\theta$.

• $2\cos 2\theta = 1$: $\cos 2\theta = \frac{1}{2}$, so $2\theta = 60^\circ$ and $\theta = 30^\circ$.
• $\tan(\theta - 10^\circ) = \frac{1}{\sqrt{3}}$: $\theta - 10^\circ = 30^\circ$, so $\theta = 40^\circ$.
• $\sqrt{3}\sin(2\theta + 10^\circ) = \frac{3}{2}$: $\sin(2\theta + 10^\circ) = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$, so $2\theta + 10^\circ = 60^\circ$ and $\theta = 25^\circ$.

Stopping at $2\theta = 60^\circ$ and writing 60° as the answer is the slip to avoid.

A square. $4\cos^2\theta = 3$ gives $\cos\theta = \frac{\sqrt{3}}{2}$ (the positive root, since the angle is acute), so $\theta = 30^\circ$.

Sine and cosine in one equation. Divide by $\cos\theta$ to get tan.

• $\sin\theta = \cos\theta$ gives $\tan\theta = 1$, so $\theta = 45^\circ$.
• $\sqrt{3}\sin\theta = \cos\theta$ gives $\tan\theta = \frac{1}{\sqrt{3}}$, so $\theta = 30^\circ$.

A quadratic in sin or cos. $2\sin^2\theta - 3\sin\theta + 1 = 0$ factorises as $(2\sin\theta - 1)(\sin\theta - 1) = 0$, so $\sin\theta = \frac{1}{2}$ or 1, which means $\theta = 30^\circ$ or 90°. If the question says $\theta$ is acute, only 30° is left. A root such as $\sin\theta = 2$ is thrown out, since a sine is never more than 1.

Then use the angle. Once $\theta$ is known, the rest is the table. If $2\cos 2\theta = 1$, then $\theta = 30^\circ$ and

$$\sin 3\theta + \cos 3\theta = \sin 90^\circ + \cos 90^\circ = 1 + 0 = 1$$

Two Equations in A + B and A − B

Some questions give two equations, one in A + B and one in A − B. Turn each one into an angle with the table, then add and subtract.

If $\tan(A + B) = 1$ and $\sin(A - B) = \frac{1}{2}$, with both brackets acute, then

$$A + B = 45^\circ, \quad A - B = 30^\circ \quad\Rightarrow\quad 2A = 75^\circ, \quad 2B = 15^\circ$$

So A = 37.5° = 37°30′ and B = 7.5° = 7°30′. An answer in degrees and minutes is quite normal here; write half a degree as 30 minutes, not 50.

A second one: $\cos(A + B) = 0$ and $\tan(A - B) = \sqrt{3}$ give A + B = 90° and A − B = 60°, so A = 75° and B = 15°. The ratios of 75° and 15° themselves are found in Allied, Compound and Double Angles.

With a multiple of A. $\sin(2A - B) = 1$ and $\tan(A + B) = \sqrt{3}$ give

$$2A - B = 90^\circ, \quad A + B = 60^\circ$$

Adding the two removes B: 3A = 150°, so A = 50° and B = 10°.

Checks before you answer.

• With A and B both positive, A + B is larger than A − B. If your table reading makes A − B the bigger one, B would come out negative, so re-read which equation holds which bracket.
• Check that each bracket is in the range the question allows. In these questions the brackets are between 0° and 90°, which is why each value gives one angle only.
• Put your A and B back into both equations. It takes a few seconds and catches a reading of $\frac{\sqrt{3}}{2}$ as $\cos 60^\circ$ instead of $\cos 30^\circ$.

Complementary Angles

Two angles are complementary when they add up to 90°. The two acute angles of a right triangle are complementary: if one is $\theta$, the other is $90^\circ - \theta$.

In the figure the side a is opposite $\theta$ but next to the angle $90^\circ - \theta$ at C. So the one fraction $\frac{a}{h}$ is both the sine of $\theta$ and the cosine of $90^\circ - \theta$. The same happens with every ratio:

$$\sin(90^\circ - \theta) = \cos\theta \qquad \cos(90^\circ - \theta) = \sin\theta$$

$$\tan(90^\circ - \theta) = \cot\theta \qquad \cot(90^\circ - \theta) = \tan\theta$$

$$\sec(90^\circ - \theta) = \operatorname{cosec}\theta \qquad \operatorname{cosec}(90^\circ - \theta) = \sec\theta$$

The ‘co’ in cosine, cotangent and cosecant stands for complement: the cosine of an angle is the sine of its complement. Each ratio turns into its ‘co’ partner, and the partner turns back. The table agrees: $\sin 30^\circ = \cos 60^\circ = \frac{1}{2}$ and $\tan 60^\circ = \cot 30^\circ = \sqrt{3}$.

Using it. Look for two angles that add up to 90°, rewrite one of them, and things cancel.

• $\frac{\sin 67^\circ}{\cos 23^\circ} = \frac{\cos 23^\circ}{\cos 23^\circ} = 1$
• $\cos 48^\circ - \sin 42^\circ = \cos 48^\circ - \cos 48^\circ = 0$
• $\sec 70^\circ \sin 20^\circ = \sec 70^\circ \cos 70^\circ = 1$
• $\tan 26^\circ \tan 64^\circ = \tan 26^\circ \cot 26^\circ = 1$

With squares. $\sin^2 32^\circ + \sin^2 58^\circ = \sin^2 32^\circ + \cos^2 32^\circ = 1$, and $\sec^2 40^\circ - \cot^2 50^\circ = \sec^2 40^\circ - \tan^2 40^\circ = 1$. Identities and Simplification has these two identities.

[SSC Tip] The rule needs angles that add up to 90°, not the same angle. $\sin 50^\circ$ and $\cos 50^\circ$ are different numbers, while $\sin 50^\circ$ and $\cos 40^\circ$ are equal. Check the sum of the two angles before cancelling anything.

Finding an Angle from a Complementary Pair

If A and B are acute and $\sin A = \cos B$, then $\sin A = \sin(90^\circ - B)$. Between 0° and 90° each value of the sine belongs to one angle only, so

$$A = 90^\circ - B, \quad\text{that is}\quad A + B = 90^\circ$$

The same is true when $\tan A = \cot B$ or $\sec A = \operatorname{cosec} B$.

Solving for the unknown. Add the two angles and set the sum equal to 90°.

• $\sin 4\theta = \cos(\theta + 15^\circ)$: $4\theta + \theta + 15^\circ = 90^\circ$, so $5\theta = 75^\circ$ and $\theta = 15^\circ$. Check: $\sin 60^\circ = \cos 30^\circ$.
• $\sec 4\theta = \operatorname{cosec}(\theta - 20^\circ)$: $5\theta - 20^\circ = 90^\circ$, so $\theta = 22^\circ$. The two angles are 88° and 2°, both acute, as they must be.

The tempting mistake is to set the two angles equal, as in $4\theta = \theta + 15^\circ$, which gives 5°. Equal angles go with equal ratios of the same name ($\sin A = \sin B$); a sine equal to a cosine needs angles that add up to 90°.

When A + B = 90° is given. Then

• $\sin A = \cos B$ and $\tan A = \cot B$, so $\sin A \sec B = 1$ and $\tan A \tan B = 1$;
• $\sin^2 A + \sin^2 B = \sin^2 A + \cos^2 A = 1$.

If also $\sin A = \frac{7}{25}$, then $\cos B = \frac{7}{25}$ as well; the 7, 24, 25 triangle then gives $\tan B = \frac{24}{7}$.

An angle and a relation. If A + B = 90° and A = 3B + 10°, then 4B + 10° = 90°, so B = 20° and A = 70°. Questions of this kind may hide the 90° behind a ratio: ‘$\sin A = \cos B$ and A = 3B + 10°’ is the same question.

Long Products: Pair Each Angle with Its Complement

Products of tan, or of cot. Since $\tan(90^\circ - \theta) = \cot\theta$,

$$\tan\theta \tan(90^\circ - \theta) = \tan\theta \cot\theta = 1$$

So in $\tan 1^\circ \tan 2^\circ \tan 3^\circ \cdots \tan 89^\circ$ the factors pair off: $\tan 1^\circ$ with $\tan 89^\circ$, $\tan 2^\circ$ with $\tan 88^\circ$, and so on up to $\tan 44^\circ$ with $\tan 46^\circ$. Each pair is 1, and the factor left in the middle is $\tan 45^\circ = 1$, so the whole product is 1. A product of cotangents behaves the same way.

A shorter product works the same, as long as you watch for a factor with no partner. In $\tan 15^\circ \tan 30^\circ \tan 40^\circ \tan 50^\circ \tan 75^\circ$, 15° pairs with 75° and 40° with 50°, and $\tan 30^\circ = \frac{1}{\sqrt{3}}$ is left over. That is the value of the product.

A product with a zero factor. If one factor is 0, the whole product is 0, however long it is.

• $\cos 1^\circ \cos 2^\circ \cos 3^\circ \cdots \cos 100^\circ$ contains $\cos 90^\circ = 0$, so it equals 0. You never need the cosines above 90°.
• $(2\sin 1^\circ - 1)(2\sin 2^\circ - 1) \cdots (2\sin 89^\circ - 1)$ contains $2\sin 30^\circ - 1 = 0$, so it is 0 too.

Before multiplying anything, scan the factors for one that vanishes at a standard angle.

Which pairs give 1. A ratio times the reciprocal of its partner: tan with tan, cot with cot, sin with sec, cos with cosec. $\sin 20^\circ \sec 70^\circ = \frac{\sin 20^\circ}{\cos 70^\circ} = 1$, but $\sin 20^\circ \sin 70^\circ = \sin 20^\circ \cos 20^\circ$, which is not 1.

Logarithms. The log of a product is the sum of the logs, so

$$\log\tan 1^\circ + \log\tan 2^\circ + \cdots + \log\tan 89^\circ = \log(\tan 1^\circ \tan 2^\circ \cdots \tan 89^\circ) = \log 1 = 0$$

The answer is 0, not 1: the 1 is the product, and the question asks for its log.

Long Sums of Squares

Pairing works for sums of squares as well, because

$$\sin^2\theta + \sin^2(90^\circ - \theta) = \sin^2\theta + \cos^2\theta = 1$$

and in the same way $\cos^2\theta + \cos^2(90^\circ - \theta) = 1$.

Take $\sin^2 6^\circ + \sin^2 12^\circ + \sin^2 18^\circ + \cdots + \sin^2 84^\circ$.

1. Count the terms. From 6° to 84° in steps of 6° there are $\frac{84 - 6}{6} + 1 = 14$ terms.
2. Pair them. 6° with 84°, 12° with 78°, and so on inwards: 7 pairs, each worth 1.
3. Add. The sum is 7.

When 45° is in the list. It has no partner, and $\sin^2 45^\circ = \cos^2 45^\circ = \frac{1}{2}$. In $\cos^2 3^\circ + \cos^2 6^\circ + \cdots + \cos^2 87^\circ$ there are 29 terms: 14 pairs and $\cos^2 45^\circ$ in the middle, so the sum is $14 + \frac{1}{2} = 14\frac{1}{2}$.

Both answers are half the number of terms. That shortcut holds whenever the first and last angles add up to 90°.

When the list runs to 90°. Then the ends no longer pair up, so count with care. In $\sin^2 15^\circ + \sin^2 30^\circ + \sin^2 45^\circ + \sin^2 60^\circ + \sin^2 75^\circ + \sin^2 90^\circ$, the pairs (15°, 75°) and (30°, 60°) make 2, $\sin^2 45^\circ = \frac{1}{2}$ and $\sin^2 90^\circ = 1$, so the sum is $3\frac{1}{2}$, not half of 6. If 0° and 90° are both present, they pair like any other two: $\cos^2 0^\circ + \cos^2 90^\circ = 1 + 0 = 1$.

Sums of plain sines do not pair at all: $\sin 10^\circ + \sin 80^\circ$ is not 1. The squares are what make the pairs work.

[SSC Tip] Count the terms with (last − first) ÷ step + 1 instead of by eye, and check whether 45° or 90° is in the list. Each of them changes the answer by a half or a whole.

Angles of a Triangle

The three angles of any triangle add up to 180°. Halve this: $\frac{A + B + C}{2} = 90^\circ$, so

$$\frac{B + C}{2} = 90^\circ - \frac{A}{2}$$

The half-angles $\frac{B + C}{2}$ and $\frac{A}{2}$ are complementary, and every complementary rule applies:

$$\sin\frac{B + C}{2} = \cos\frac{A}{2} \qquad \cos\frac{B + C}{2} = \sin\frac{A}{2} \qquad \tan\frac{A + B}{2} = \cot\frac{C}{2}$$

So $\tan\frac{B + C}{2} \tan\frac{A}{2} = 1$ in every triangle. If $A = 60^\circ$, then $\frac{B + C}{2} = 60^\circ$ and $\sec\frac{B + C}{2} = 2$; the rule gives the same as $\operatorname{cosec}\frac{A}{2} = \operatorname{cosec} 30^\circ = 2$.

Which half goes with which. The angle outside the bracket is the one missing from it: $\frac{A + B}{2}$ pairs with $\frac{C}{2}$, $\frac{A + C}{2}$ with $\frac{B}{2}$. Matching $\frac{A + B}{2}$ with $\frac{A}{2}$ is the error to avoid.

A right-angled triangle. If $C = 90^\circ$, then A + B = 90° with no halving at all: $\sin A = \cos B$, $\tan A \tan B = 1$ and $\sin^2 A + \sin^2 B = 1$.

Angles given in a ratio. Angles in the ratio 1 : 1 : 2 are 45°, 45° and 90°, so for that triangle $\sin A \cos B + \sin C = \frac{1}{2} + 1 = \frac{3}{2}$.

Without halving, the angles are not complementary. B + C = 180° − A is the supplement of A, not its complement. Then $\sin(B + C) = \sin A$, but $\cos(B + C) = -\cos A$, with a minus sign. These need angles above 90° and are done in Allied, Compound and Double Angles.

Solved Examples

Question 1

What is the value of $3\tan^2 30^\circ + \sec^2 45^\circ - 4\cos^2 60^\circ$? (a) 3 (b) 2 (c) 1 (d) 4

Answer. From the table, $\tan^2 30^\circ = \frac{1}{3}$, $\sec^2 45^\circ = (\sqrt{2})^2 = 2$ and $\cos^2 60^\circ = \frac{1}{4}$. So the value is $3 \times \frac{1}{3} + 2 - 4 \times \frac{1}{4} = 1 + 2 - 1 =$ (b) 2.

Question 2

What is the value of $\frac{\sin 60^\circ + \cos 30^\circ}{\tan 45^\circ + \sec 60^\circ}$? (a) $\sqrt{3}$ (b) $\frac{\sqrt{3}}{2}$ (c) $\frac{2}{\sqrt{3}}$ (d) $\frac{1}{\sqrt{3}}$

Answer. The top is $\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}$. The bottom is 1 + 2 = 3, since $\sec 60^\circ = 2$. So the value is $\frac{\sqrt{3}}{3}$, which is (d) $\frac{1}{\sqrt{3}}$ after cancelling one $\sqrt{3}$.

Question 3

Which of these is not defined? (a) $\sin 90^\circ$ (b) $\cot 90^\circ$ (c) $\sec 90^\circ$ (d) $\operatorname{cosec} 90^\circ$

Answer. $\sec 90^\circ = \frac{1}{\cos 90^\circ} = \frac{1}{0}$, which has no value. The others are fine: $\sin 90^\circ = 1$, $\cot 90^\circ = \frac{\cos 90^\circ}{\sin 90^\circ} = 0$ and $\operatorname{cosec} 90^\circ = 1$. So the answer is (c) $\sec 90^\circ$.

Question 4

If $\theta = 30^\circ$, what is the value of $\frac{1 - \tan^2\theta}{1 + \tan^2\theta}$? (a) $\frac{1}{2}$ (b) $\frac{\sqrt{3}}{2}$ (c) $\frac{1}{\sqrt{3}}$ (d) 2

Answer. $\tan^2 30^\circ = \frac{1}{3}$. So the top is $\frac{2}{3}$, the bottom is $\frac{4}{3}$, and the value is $\frac{2}{3} \div \frac{4}{3} =$ (a) $\frac{1}{2}$. This is $\cos 60^\circ$, the cosine of $2\theta$; Allied, Compound and Double Angles shows why.

Question 5

What is the value of $\sin(45^\circ + 45^\circ) - (\sin 45^\circ + \sin 45^\circ)$? (a) 0 (b) $1 - \sqrt{2}$ (c) $\sqrt{2} - 1$ (d) 1

Answer. The first part is $\sin 90^\circ = 1$. The second is $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. So the value is (b) $1 - \sqrt{2}$, a negative number. Getting 0 means $\sin(45^\circ + 45^\circ)$ was opened up as $\sin 45^\circ + \sin 45^\circ$, which is not allowed.

Question 6

In triangle ABC, right-angled at B, $\angle A = 30^\circ$ and AC = 16 cm. What is AB + BC? (a) $8\sqrt{3}$ cm (b) $16(\sqrt{3} + 1)$ cm (c) $8(\sqrt{3} - 1)$ cm (d) $8(\sqrt{3} + 1)$ cm

Answer. AC is the hypotenuse. BC faces the 30° angle, so $BC = AC \sin 30^\circ = 16 \times \frac{1}{2} = 8$ cm. AB is next to the 30° angle, so $AB = AC \cos 30^\circ = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3}$ cm. Together they make $8\sqrt{3} + 8 =$ (d) $8(\sqrt{3} + 1)$ cm.

Question 7

If $\sqrt{3}\sec 2\theta = 2$ and $2\theta$ is acute, what is $\theta$? (a) $15^\circ$ (b) $30^\circ$ (c) $60^\circ$ (d) $7.5^\circ$

Answer. Dividing by $\sqrt{3}$: $\sec 2\theta = \frac{2}{\sqrt{3}}$, so $\cos 2\theta = \frac{\sqrt{3}}{2}$. That makes $2\theta = 30^\circ$, and $\theta =$ (a) $15^\circ$. 30° is the value of $2\theta$, not of $\theta$.

Question 8

If $\cos(2\theta - 30^\circ) = \frac{1}{2}$ and $2\theta - 30^\circ$ is acute, what is $\tan\theta$? (a) $\sqrt{3}$ (b) $\frac{1}{\sqrt{3}}$ (c) 1 (d) 0

Answer. $\cos 60^\circ = \frac{1}{2}$, so $2\theta - 30^\circ = 60^\circ$, which gives $2\theta = 90^\circ$ and $\theta = 45^\circ$. Then $\tan\theta = \tan 45^\circ =$ (c) 1.

Question 9

If $2\cos^2\theta + \cos\theta - 1 = 0$ and $0^\circ < \theta < 90^\circ$, what is $\theta$? (a) $30^\circ$ (b) $60^\circ$ (c) $45^\circ$ (d) $180^\circ$

Answer. This is a quadratic in $\cos\theta$. It factorises as $(2\cos\theta - 1)(\cos\theta + 1) = 0$, so $\cos\theta = \frac{1}{2}$ or $\cos\theta = -1$. Between 0° and 90° the cosine is positive, so $\cos\theta = \frac{1}{2}$ and $\theta =$ (b) $60^\circ$.

Question 10

If $\sin(A + B) = 1$ and $\cos(A - B) = \frac{\sqrt{3}}{2}$, where A > B and both brackets lie between 0° and 90°, what are A and B? (a) A = 60°, B = 30° (b) A = 45°, B = 45° (c) A = 30°, B = 60° (d) A = 75°, B = 15°

Answer. $\sin(A + B) = 1$ gives A + B = 90°, and $\cos(A - B) = \frac{\sqrt{3}}{2}$ gives A − B = 30°. Adding, 2A = 120°, so A = 60° and then B = 30°. The answer is (a) A = 60°, B = 30°. Reading $\frac{\sqrt{3}}{2}$ as the cosine of 60° would lead to 75° and 15°.

Question 11

If $\cos(A + B) = \frac{1}{2}$ and $\tan(A - B) = \frac{1}{\sqrt{3}}$, where A > B and A + B is acute, what is B? (a) $30^\circ$ (b) $45^\circ$ (c) $22.5^\circ$ (d) $15^\circ$

Answer. From the table, A + B = 60° and A − B = 30°. Subtracting the second from the first, 2B = 30°, so B = (d) $15^\circ$. A is 45°.

Question 12

The angles A, B and C of a triangle are in the ratio 1 : 2 : 3. What is $\sin A + \cos B + \sin C$? (a) 1 (b) $\frac{3}{2}$ (c) 2 (d) $1 + \frac{\sqrt{3}}{2}$

Answer. 1 + 2 + 3 = 6 parts share 180°, so the angles are 30°, 60° and 90°. Then $\sin 30^\circ + \cos 60^\circ + \sin 90^\circ = \frac{1}{2} + \frac{1}{2} + 1 =$ (c) 2.

Question 13

What is the value of $\frac{\sin 37^\circ}{\cos 53^\circ} + 3\tan 28^\circ \tan 62^\circ - \cos^2 25^\circ - \cos^2 65^\circ$? (a) 3 (b) 4 (c) 2 (d) 5

Answer. Each pair of angles adds up to 90°. $\cos 53^\circ = \sin 37^\circ$, so the first term is 1. $\tan 62^\circ = \cot 28^\circ$, so $3\tan 28^\circ \tan 62^\circ = 3$. And $\cos 65^\circ = \sin 25^\circ$, so the last two terms are $-(\cos^2 25^\circ + \sin^2 25^\circ) = -1$. Total: 1 + 3 − 1 = (a) 3.

Question 14

If $\sin(3\theta - 12^\circ) = \cos(\theta + 22^\circ)$, where both angles are acute, what is $\theta$? (a) $17^\circ$ (b) $24^\circ$ (c) $20^\circ$ (d) $26^\circ$

Answer. A sine equal to a cosine means the two angles are complementary: $(3\theta - 12^\circ) + (\theta + 22^\circ) = 90^\circ$. So $4\theta + 10^\circ = 90^\circ$ and $\theta =$ (c) $20^\circ$. Check: the angles are 48° and 42°, which add up to 90°. Setting the two angles equal instead gives 17°, and that is wrong.

Question 15

If $\tan 2A = \cot(A - 18^\circ)$, where 2A is acute, what is A? (a) $18^\circ$ (b) $36^\circ$ (c) $24^\circ$ (d) $54^\circ$

Answer. The tan of one angle equals the cot of the other, so the angles are complementary: $2A + (A - 18^\circ) = 90^\circ$, which gives 3A = 108° and A = (b) $36^\circ$. Check: 2A = 72° and A − 18° = 18°, which add up to 90°.

Question 16

A and B are complementary angles and $\tan A = \frac{91}{60}$. What is $\sec B$? (a) $\frac{109}{60}$ (b) $\frac{60}{109}$ (c) $\frac{91}{109}$ (d) $\frac{109}{91}$

Answer. Since B = 90° − A, $\sec B = \sec(90^\circ - A) = \operatorname{cosec} A$. From $\tan A = \frac{91}{60}$ the hypotenuse is $\sqrt{8281 + 3600} = \sqrt{11881} = 109$, so $\operatorname{cosec} A = \frac{109}{91}$, and $\sec B =$ (d) $\frac{109}{91}$. Working out $\sec A$ by mistake gives $\frac{109}{60}$.

Question 17

What is the value of $\tan 9^\circ \tan 32^\circ \tan 58^\circ \tan 81^\circ \sin 60^\circ$? (a) $\frac{\sqrt{3}}{2}$ (b) 1 (c) $\sqrt{3}$ (d) $\frac{1}{2}$

Answer. 9° pairs with 81° and 32° with 58°. $\tan 9^\circ \tan 81^\circ = \tan 9^\circ \cot 9^\circ = 1$, and in the same way $\tan 32^\circ \tan 58^\circ = 1$. Only $\sin 60^\circ$ is left, so the value is (a) $\frac{\sqrt{3}}{2}$.

Question 18

What is the value of $(\sin 1^\circ - \cos 1^\circ)(\sin 2^\circ - \cos 2^\circ) \cdots (\sin 89^\circ - \cos 89^\circ)$? (a) 1 (b) $-1$ (c) 0 (d) $\frac{1}{\sqrt{2}}$

Answer. One of the brackets is $\sin 45^\circ - \cos 45^\circ = 0$. A product with a zero factor is 0, so the value is (c) 0. There is no need to look at the other 88 brackets.

Question 19

What is the value of $\sin^2 2^\circ + \sin^2 4^\circ + \sin^2 6^\circ + \cdots + \sin^2 88^\circ$? (a) $\frac{45}{2}$ (b) 22 (c) 44 (d) 21

Answer. From 2° to 88° in steps of 2° there are $\frac{88 - 2}{2} + 1 = 44$ terms. The first and last add up to 90°, and so does every pair working inwards: 2° with 88°, 4° with 86°, up to 44° with 46°. 45 is odd, so 45° is not in the list. That makes 22 pairs, each equal to 1, and the sum is (b) 22.

Question 20

What is the value of $\sin^2 10^\circ + \sin^2 20^\circ + \sin^2 30^\circ + \cdots + \sin^2 90^\circ$? (a) 4 (b) $\frac{9}{2}$ (c) 9 (d) 5

Answer. There are 9 terms. Pairing 10° with 80°, 20° with 70°, 30° with 60° and 40° with 50° gives 4 pairs, worth 4. The last term, $\sin^2 90^\circ$, has no partner and equals 1. So the sum is (d) 5. Half the number of terms would give $\frac{9}{2}$, but that shortcut needs the first and last angles to add up to 90°.

Question 21

What is the value of $\log\tan 1^\circ + \log\tan 2^\circ + \log\tan 3^\circ + \cdots + \log\tan 89^\circ$? (a) 0 (b) 1 (c) $-1$ (d) $\frac{89}{2}$

Answer. Adding logs is the same as taking the log of the product, so the sum is $\log(\tan 1^\circ \tan 2^\circ \cdots \tan 89^\circ)$. The product is 1, because the angles pair off as $\tan\theta \cot\theta = 1$ and the middle factor is $\tan 45^\circ = 1$. And $\log 1 =$ (a) 0.

Question 22

If A, B and C are the angles of a triangle, then $\sin\frac{A + B}{2}$ is equal to (a) $\sin\frac{C}{2}$ (b) $\cos\frac{C}{2}$ (c) $\cos C$ (d) $\sin C$

Answer. A + B + C = 180°, so $\frac{A + B}{2} = 90^\circ - \frac{C}{2}$. Then $\sin\frac{A + B}{2} = \sin\left(90^\circ - \frac{C}{2}\right) =$ (b) $\cos\frac{C}{2}$.

Question 23

In triangle ABC, $\angle A = 60^\circ$. What is $\cos\frac{B + C}{2}$? (a) $\frac{\sqrt{3}}{2}$ (b) 0 (c) 1 (d) $\frac{1}{2}$

Answer. B + C = 180° − 60° = 120°, so $\frac{B + C}{2} = 60^\circ$ and the value is $\cos 60^\circ =$ (d) $\frac{1}{2}$. The rule $\cos\frac{B + C}{2} = \sin\frac{A}{2} = \sin 30^\circ$ gives the same.

Question 24

In triangle ABC, right-angled at C, $\tan A = \sqrt{3}$. What is $\sin A \cos B + \cos A \sin B$? (a) $\frac{3}{4}$ (b) $\frac{\sqrt{3}}{2}$ (c) 1 (d) $\frac{1}{2}$

Answer. $\tan A = \sqrt{3}$ means A = 60°, and since C = 90°, B = 30°. So the value is $\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = \frac{3}{4} + \frac{1}{4} =$ (c) 1.