Allied, Compound and Double Angles

Angles Beyond 90° and the Four Quadrants

In a right triangle the other two angles are acute, so the ratios met so far stop at 90°. SSC papers also ask for values such as $\sin 120^\circ$, $\cos 210^\circ$ or $\tan 315^\circ$. For these the angle is drawn on a pair of axes.

Start on the positive x-axis and turn anticlockwise through an angle $\theta$. Take a point P on the turning arm at a distance r from the origin O, with coordinates (x, y). Then

$$\sin\theta = \frac{y}{r} \qquad \cos\theta = \frac{x}{r} \qquad \tan\theta = \frac{y}{x}$$

and r is always positive. For an acute angle, x is the adjacent side, y the opposite side and r the hypotenuse, so these are the old definitions. With r = 1 the point is simply $P(\cos\theta, \sin\theta)$.

P stays on a circle of radius 1, so neither coordinate can go above 1 or below −1. That is why $\sin\theta$ and $\cos\theta$ lie between −1 and 1 for every angle, not only for acute ones.

Signs in the four quadrants. The axes cut the plane into four quadrants, numbered anticlockwise. As P goes round, x and y change sign, and the ratios change with them.

Quadrant	Angle	x	y	Positive ratios
I	0° to 90°	+	+	all six
II	90° to 180°	−	+	$\sin$ and $\operatorname{cosec}$ only
III	180° to 270°	−	−	$\tan$ and $\cot$ only
IV	270° to 360°	+	−	$\cos$ and $\sec$ only

Read the positive ratios anticlockwise from the first quadrant: All, Sin, Tan, Cos. The usual memory line is Add Sugar To Coffee. A reciprocal always has the same sign as its partner, so cosec follows sin, sec follows cos and cot follows tan.

• 160° is in the second quadrant: $\sin 160^\circ$ is positive, while $\cos 160^\circ$ and $\tan 160^\circ$ are negative.
• 250° is in the third quadrant: only $\tan 250^\circ$ and $\cot 250^\circ$ are positive.
• 340° is in the fourth quadrant: $\cos 340^\circ$ is positive, and $\sin 340^\circ$ and $\tan 340^\circ$ are negative.

Angles on the axes. At 0°, 90°, 180°, 270° and 360° the point P sits on an axis, at (1, 0), (0, 1), (−1, 0), (0, −1) and back at (1, 0). The values come straight from these coordinates:

$\theta$	0°	90°	180°	270°	360°
$\sin\theta$	0	1	0	−1	0
$\cos\theta$	1	0	−1	0	1
$\tan\theta$	0	not defined	0	not defined	0

So $\cos 180^\circ + \sin 270^\circ = -1 + (-1) = -2$. The tangent has no value at 90° or 270°, because x = 0 there.

Allied Angles with 180° and 360°

Angles such as $180^\circ - \theta$, $180^\circ + \theta$, $360^\circ - \theta$ and $90^\circ + \theta$ are called allied angles of $\theta$. Their ratios are the ratios of $\theta$ itself, with a sign in front.

In the figure P makes an acute angle $\theta$ and has coordinates (a, b) on a circle of radius 1. The point for $180^\circ - \theta$ is the mirror image of P in the y-axis, so it is (−a, b). The point for $180^\circ + \theta$ is straight across O, at (−a, −b), and the point for $360^\circ - \theta$ is the mirror image in the x-axis, at (a, −b).

Reading the coordinates gives

$$\sin(180^\circ - \theta) = \sin\theta \qquad \cos(180^\circ - \theta) = -\cos\theta \qquad \tan(180^\circ - \theta) = -\tan\theta$$

$$\sin(180^\circ + \theta) = -\sin\theta \qquad \cos(180^\circ + \theta) = -\cos\theta \qquad \tan(180^\circ + \theta) = \tan\theta$$

$$\sin(360^\circ - \theta) = -\sin\theta \qquad \cos(360^\circ - \theta) = \cos\theta \qquad \tan(360^\circ - \theta) = -\tan\theta$$

You don't have to learn nine lines. One rule covers all of them: with 180° or 360°, the ratio keeps its name and the sign comes from the quadrant. Treat $\theta$ as acute, see which quadrant the whole angle falls in, and take the sign from Add Sugar To Coffee. For $\cos(180^\circ + \theta)$ the angle is in the third quadrant, where cos is negative, so $\cos(180^\circ + \theta) = -\cos\theta$.

Values to know. Write the angle as 180° or 360° plus or minus 30°, 45° or 60°. Values at 120°, 135° and 150° are asked in Tier 1 as well as Tier 2.

Angle	Written as	$\sin$	$\cos$	$\tan$
120°	180° − 60°	$\frac{\sqrt{3}}{2}$	$-\frac{1}{2}$	$-\sqrt{3}$
135°	180° − 45°	$\frac{1}{\sqrt{2}}$	$-\frac{1}{\sqrt{2}}$	$-1$
150°	180° − 30°	$\frac{1}{2}$	$-\frac{\sqrt{3}}{2}$	$-\frac{1}{\sqrt{3}}$
210°	180° + 30°	$-\frac{1}{2}$	$-\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
225°	180° + 45°	$-\frac{1}{\sqrt{2}}$	$-\frac{1}{\sqrt{2}}$	1
240°	180° + 60°	$-\frac{\sqrt{3}}{2}$	$-\frac{1}{2}$	$\sqrt{3}$
300°	360° − 60°	$-\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$-\sqrt{3}$
315°	360° − 45°	$-\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	$-1$
330°	360° − 30°	$-\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$-\frac{1}{\sqrt{3}}$

The numbers are always those of 30°, 45° or 60°; only the signs change. So $\sin 150^\circ + \cos 300^\circ = \frac{1}{2} + \frac{1}{2} = 1$, while $\tan 135^\circ \times \tan 240^\circ = (-1) \times \sqrt{3} = -\sqrt{3}$.

An obtuse angle from its ratio. In a triangle one angle may be more than 90°, and sine is positive in the second quadrant as well. So $\sin A = \frac{1}{2}$ allows A = 30° or A = 150°, while $\cos A = -\frac{1}{2}$ fixes A = 120°. A cosine or a tangent settles which angle it is; a sine alone does not.

Angles of a triangle. A + B + C = 180°, so A + B = 180° − C, and

$$\sin(A + B) = \sin C \qquad \cos(A + B) = -\cos C \qquad \tan(A + B) = -\tan C$$

The minus sign on cos and tan is the part that gets dropped.

Allied Angles with 90° and 270°

Now turn OP through a further 90°. In the figure P(a, b) moves to Q(−b, a). The two shaded triangles are the same triangle turned on its side, so a and b swap places and one of them picks up a minus sign.

Q is the point for $90^\circ + \theta$, so

$$\sin(90^\circ + \theta) = a = \cos\theta \qquad \cos(90^\circ + \theta) = -b = -\sin\theta \qquad \tan(90^\circ + \theta) = \frac{a}{-b} = -\cot\theta$$

The rule: with 90° or 270°, the ratio changes to its partner (sin to cos, tan to cot, sec to cosec, and back), and the sign still comes from the quadrant of the whole angle.

	$90^\circ - \theta$	$90^\circ + \theta$	$270^\circ - \theta$	$270^\circ + \theta$
$\sin$	$\cos\theta$	$\cos\theta$	$-\cos\theta$	$-\cos\theta$
$\cos$	$\sin\theta$	$-\sin\theta$	$-\sin\theta$	$\sin\theta$
$\tan$	$\cot\theta$	$-\cot\theta$	$\cot\theta$	$-\cot\theta$

The first column is the complementary-angle rule from Standard Angles and Complementary Angles. Take $\cos(270^\circ - \theta)$: the ratio changes to sin, and $270^\circ - \theta$ is in the third quadrant, where cos is negative, so the answer is $-\sin\theta$.

Two ways to split an angle. 120° is 180° − 60° and also 90° + 30°. The first split gives $\sin 120^\circ = \sin 60^\circ$, the second gives $\sin 120^\circ = \cos 30^\circ$, and both are $\frac{\sqrt{3}}{2}$. Use whichever split you see first.

Simplifying a mixed expression. Change each factor on its own, write its sign in front, and only then cancel:

$$\frac{\sin(180^\circ + \theta)\cos(90^\circ + \theta)}{\cos(180^\circ - \theta)\sin(90^\circ + \theta)} = \frac{(-\sin\theta)(-\sin\theta)}{(-\cos\theta)(\cos\theta)} = -\tan^2\theta$$

[SSC Tip] In a long product, write every factor with its sign first and count the minus signs at the end: an even number gives a plus, an odd number a minus. Carrying the signs through each line of working is where most of the mistakes happen.

Negative Angles and Angles Larger than 360°

A negative angle is turned clockwise. The arm for $-\theta$ is the mirror image of the arm for $\theta$ in the x-axis, so (a, b) becomes (a, −b):

$$\sin(-\theta) = -\sin\theta \qquad \cos(-\theta) = \cos\theta \qquad \tan(-\theta) = -\tan\theta$$

Only cos and its reciprocal sec ignore the minus sign. So $\cos(-60^\circ) = \frac{1}{2}$, $\sin(-30^\circ) = -\frac{1}{2}$ and $\tan(-45^\circ) = -1$.

Large angles. One full turn of 360° brings the arm back to where it started, so adding or removing whole turns changes nothing:

$$\sin(n \times 360^\circ + \theta) = \sin\theta$$

and the same holds for every ratio. Divide the angle by 360, keep the remainder, and use the allied-angle rules on the remainder.

• 1020 = 2 × 360 + 300, so $\sin 1020^\circ = \sin 300^\circ = -\frac{\sqrt{3}}{2}$.
• 765 = 2 × 360 + 45, so $\cos 765^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$.
• $\sin(-1050^\circ) = -\sin 1050^\circ$, and 1050 = 2 × 360 + 330, so the value is $-\sin 330^\circ = -\left(-\frac{1}{2}\right) = \frac{1}{2}$.

The graph shows why this works. $\sin\theta$ rises from 0 to 1 at 90°, comes back to 0 at 180°, goes down to −1 at 270° and returns to 0 at 360°; after that the same wave repeats. The red dots mark $\sin 30^\circ = \sin 150^\circ = \frac{1}{2}$ and $\sin 210^\circ = \sin 330^\circ = -\frac{1}{2}$: each height is reached twice in one turn. The dashed curve for $\cos\theta$ is the same wave moved 90° to the left.

Tangent repeats every 180°. Since $\tan(180^\circ + \theta) = \tan\theta$, you may remove 180°s from a tangent or a cotangent. 930 = 5 × 180 + 30, so $\tan 930^\circ = \tan 30^\circ = \frac{1}{\sqrt{3}}$.

[SSC Tip] Remove 180°s only for tan and cot. For sin and cos a half turn flips the sign: $\sin 210^\circ = -\frac{1}{2}$ although $\sin 30^\circ = \frac{1}{2}$. With sin and cos, take away full turns of 360° only.

Compound Angles: Sine and Cosine of A + B and A − B

For the sum or the difference of two angles:

$$\sin(A + B) = \sin A\cos B + \cos A\sin B$$

$$\sin(A - B) = \sin A\cos B - \cos A\sin B$$

$$\cos(A + B) = \cos A\cos B - \sin A\sin B$$

$$\cos(A - B) = \cos A\cos B + \sin A\sin B$$

To keep them apart: the sine formula mixes the ratios (sin with cos) and keeps the sign of A ± B; the cosine formula keeps like with like (cos with cos, sin with sin) and flips the sign.

A ratio does not split over a sum. $\cos(30^\circ + 60^\circ) = \cos 90^\circ = 0$, but $\cos 30^\circ + \cos 60^\circ = \frac{\sqrt{3} + 1}{2}$. Every right-hand side above is made of products, never a plain sum of two ratios.

A + B from given ratios. Let $\sin A = \frac{4}{5}$ and $\cos B = \frac{5}{13}$, with A and B acute. The triangles give $\cos A = \frac{3}{5}$ and $\sin B = \frac{12}{13}$. Then

$$\sin(A + B) = \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13} = \frac{20 + 36}{65} = \frac{56}{65}$$

$$\cos(A + B) = \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13} = \frac{15 - 48}{65} = -\frac{33}{65}$$

The cosine is negative, so A + B is more than 90°. That is possible, since each angle is above 50°. A quick check: 56² + 33² = 3136 + 1089 = 4225 = 65².

Folding a formula back up. When a question shows the right-hand side, spot the pattern and write it as one ratio:

• $\sin 48^\circ\cos 18^\circ - \cos 48^\circ\sin 18^\circ = \sin 30^\circ = \frac{1}{2}$
• $\cos 70^\circ\cos 25^\circ + \sin 70^\circ\sin 25^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$

Sum times difference. Multiply the two sine formulas and write $\cos^2 A = 1 - \sin^2 A$, $\cos^2 B = 1 - \sin^2 B$. This gives two short results:

$$\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B \qquad \cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B$$

So $\sin^2 75^\circ - \sin^2 15^\circ = \sin 90^\circ\sin 60^\circ = \frac{\sqrt{3}}{2}$, and the values of 15° and 75° are never needed.

Compound Angles: Tangent, and the Values of 15° and 75°

$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} \qquad \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$$

The sign in the denominator is always the opposite of the sign on top.

Finding A + B. If $\tan A = \frac{1}{4}$ and $\tan B = \frac{3}{5}$, with A and B acute, then

$$\tan(A + B) = \frac{\frac{1}{4} + \frac{3}{5}}{1 - \frac{3}{20}} = \frac{\frac{17}{20}}{\frac{17}{20}} = 1$$

so A + B = 45°, though neither angle is a standard one.

15° and 75°. Write 15° = 45° − 30°:

$$\sin 15^\circ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$$

$$\cos 15^\circ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$$

$$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{2} = 2 - \sqrt{3}$$

Since 75° = 90° − 15°, the values of 75° are the same numbers swapped:

	15°	75°
$\sin$	$\frac{\sqrt{3} - 1}{2\sqrt{2}}$	$\frac{\sqrt{3} + 1}{2\sqrt{2}}$
$\cos$	$\frac{\sqrt{3} + 1}{2\sqrt{2}}$	$\frac{\sqrt{3} - 1}{2\sqrt{2}}$
$\tan$	$2 - \sqrt{3}$	$2 + \sqrt{3}$

Some options clear the root from the bottom: $\frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$. As decimals $\sin 15^\circ \approx 0.26$ and $\cos 15^\circ \approx 0.97$, which tells you at once which form belongs to which.

A picture of the tangent. Take the 30°-60°-90° triangle ABC with AC = 1, BC = $\sqrt{3}$ and AB = 2, and extend CB to D so that BD = BA = 2. Triangle ABD is isosceles with $\angle ABD = 150^\circ$, so each of its base angles is 15°. In the right triangle ADC,

$$\tan 15^\circ = \frac{AC}{DC} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3} \qquad \tan 75^\circ = \frac{DC}{AC} = 2 + \sqrt{3}$$

Three facts follow: $\tan 15^\circ + \tan 75^\circ = 4$, $\tan 75^\circ - \tan 15^\circ = 2\sqrt{3}$ and $\tan 15^\circ\tan 75^\circ = 1$.

105°. Combine with the allied rules: 105° = 180° − 75°, so $\sin 105^\circ = \sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}$ and $\cos 105^\circ = -\cos 75^\circ = \frac{1 - \sqrt{3}}{2\sqrt{2}}$. Expanding $\cos(60^\circ + 45^\circ)$ gives the same negative value.

Tangent Forms with 45°

Put A = 45° in the tangent formulas. Since $\tan 45^\circ = 1$,

$$\tan(45^\circ + \theta) = \frac{1 + \tan\theta}{1 - \tan\theta} \qquad \tan(45^\circ - \theta) = \frac{1 - \tan\theta}{1 + \tan\theta}$$

Multiply the top and the bottom of the first one by $\cos\theta$ and it reads

$$\frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \tan(45^\circ + \theta)$$

So $\frac{1 + \tan 15^\circ}{1 - \tan 15^\circ} = \tan 60^\circ = \sqrt{3}$, and $\frac{\cos 10^\circ + \sin 10^\circ}{\cos 10^\circ - \sin 10^\circ} = \tan 55^\circ$. When the plus sign is at the bottom, the angle is $45^\circ - \theta$ instead.

Two angles adding to 45°. If A + B = 45°, then $\tan(A + B) = 1$, so

$$\tan A + \tan B = 1 - \tan A\tan B \Rightarrow 1 + \tan A + \tan B + \tan A\tan B = 2 \Rightarrow (1 + \tan A)(1 + \tan B) = 2$$

So $(1 + \tan 20^\circ)(1 + \tan 25^\circ) = 2$ and $(1 + \tan 5^\circ)(1 + \tan 40^\circ) = 2$, though none of these tangents is a neat number. Dividing the first line by $\tan A\tan B$ gives the cotangent version, $(\cot A - 1)(\cot B - 1) = 2$; for instance $(\cot 10^\circ - 1)(\cot 35^\circ - 1) = 2$.

A long product breaks into such pairs. In $(1 + \tan 1^\circ)(1 + \tan 2^\circ) \cdots (1 + \tan 44^\circ)$, pair 1° with 44°, 2° with 43°, and so on. That makes 22 pairs, each worth 2, so the product is $2^{22}$.

Two angles adding to 90°. Now $\tan A\tan B = 1$, so the denominator in $\tan(A - B)$ is 1 + 1 = 2:

$$\tan(A - B) = \frac{\tan A - \tan B}{2} \Rightarrow \tan A - \tan B = 2\tan(A - B)$$

With A = 70° and B = 20° this gives $\tan 70^\circ - \tan 20^\circ = 2\tan 50^\circ$.

Double Angles

Put B = A in the compound-angle formulas and call the angle $\theta$:

$$\sin 2\theta = 2\sin\theta\cos\theta$$

$$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$$

$$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$$

The three forms of $\cos 2\theta$ come from replacing $\sin^2\theta$ by $1 - \cos^2\theta$, or $\cos^2\theta$ by $1 - \sin^2\theta$. Pick the form that uses the ratio the question gives you. Dividing by $\cos^2\theta + \sin^2\theta$ (which is just 1) puts sine and cosine in terms of $\tan\theta$ alone:

$$\sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta} \qquad \cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}$$

Where they come from. In the figure AB is a diameter of a circle of radius 1 and $\angle PAB = \theta$. The angle at the centre is twice the angle at the circumference, so $\angle POB = 2\theta$, which makes $PM = \sin 2\theta$ and $OM = \cos 2\theta$. The angle APB in the semicircle is 90°, so $AP = AB\cos\theta = 2\cos\theta$. Now $PM = AP\sin\theta = 2\sin\theta\cos\theta$. Also $AM = AP\cos\theta = 2\cos^2\theta$, and $AM = 1 + OM$, which gives $\cos 2\theta = 2\cos^2\theta - 1$.

From one ratio to the double angle. If $\cos\theta = \frac{12}{13}$ with $\theta$ acute, then $\sin\theta = \frac{5}{13}$ and

• $\sin 2\theta = 2 \times \frac{5}{13} \times \frac{12}{13} = \frac{120}{169}$
• $\cos 2\theta = 2 \times \frac{144}{169} - 1 = \frac{119}{169}$
• $\tan 2\theta = \frac{120}{119}$, the sine divided by the cosine.

If only $\tan\theta$ is given, use the tan forms. With $\tan\theta = \frac{1}{2}$: $\sin 2\theta = \frac{1}{1 + \frac{1}{4}} = \frac{4}{5}$, $\cos 2\theta = \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} = \frac{3}{5}$ and $\tan 2\theta = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}$.

Reading the formulas backwards. Spot the double-angle shape and write it as one ratio of the doubled angle:

• $2\sin 15^\circ\cos 15^\circ = \sin 30^\circ = \frac{1}{2}$, so $\sin 15^\circ\cos 15^\circ = \frac{1}{4}$
• $\cos^2 15^\circ - \sin^2 15^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}$
• $2\cos^2 22.5^\circ - 1 = \cos 45^\circ = \frac{1}{\sqrt{2}}$
• $\frac{2\tan 15^\circ}{1 - \tan^2 15^\circ} = \tan 30^\circ = \frac{1}{\sqrt{3}}$

[SSC Tip] When the doubled angle goes past 90°, the sign changes. $\cos^2 60^\circ - \sin^2 60^\circ = \cos 120^\circ = -\frac{1}{2}$ (check: $\frac{1}{4} - \frac{3}{4}$). An expression built from squares can still have a negative value.

Removing the 1: Half-Angle Forms

From the last two forms of $\cos 2\theta$:

$$1 - \cos 2\theta = 2\sin^2\theta \qquad 1 + \cos 2\theta = 2\cos^2\theta$$

Each one gets rid of the 1 and halves the angle. Comparing them with $\sin 2\theta = 2\sin\theta\cos\theta$, or with each other:

$$\frac{1 - \cos 2\theta}{\sin 2\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \tan\theta \qquad \frac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta \qquad \frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2\theta$$

The same idea works with $\sin 2\theta$, because $\sin^2\theta + \cos^2\theta = 1$:

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

The value of 22.5°. Put $\theta = 22.5^\circ$ in the first result:

$$\tan 22.5^\circ = \frac{1 - \cos 45^\circ}{\sin 45^\circ} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1$$

and so $\cot 22.5^\circ = \tan 67.5^\circ = \sqrt{2} + 1$.

Two handy results.

$$\cot\theta - \tan\theta = \frac{\cos^2\theta - \sin^2\theta}{\sin\theta\cos\theta} = 2\cot 2\theta \qquad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta} = 2\operatorname{cosec} 2\theta$$

Check the first with 22.5°: $(\sqrt{2} + 1) - (\sqrt{2} - 1) = 2$, and $2\cot 45^\circ = 2$.

Simplifying. Turn every $1 + \cos 2\theta$ or $1 - \cos 2\theta$ into a square first, then take out the common factor:

$$\frac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} = \frac{\sin\theta(1 + 2\cos\theta)}{\cos\theta + 2\cos^2\theta} = \frac{\sin\theta(1 + 2\cos\theta)}{\cos\theta(1 + 2\cos\theta)} = \tan\theta$$

Square roots open the same way, from the inside out. For $0^\circ < \theta < 45^\circ$:

$$\sqrt{2 + \sqrt{2 + 2\cos 4\theta}} = \sqrt{2 + \sqrt{4\cos^2 2\theta}} = \sqrt{2 + 2\cos 2\theta} = \sqrt{4\cos^2\theta} = 2\cos\theta$$

[Tier 2] Triple Angles, and the Angles 18° and 36°

$$\sin 3\theta = 3\sin\theta - 4\sin^3\theta \qquad \cos 3\theta = 4\cos^3\theta - 3\cos\theta$$

$$\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$$

They come from writing $3\theta = 2\theta + \theta$ and using the double-angle results. The sine formula starts with 3 sin, the cosine formula with 4 cos cubed. If you are unsure which way round, test at $\theta = 30^\circ$: $3 \times \frac{1}{2} - 4 \times \frac{1}{8} = 1 = \sin 90^\circ$.

Read them backwards as well:

• $3\sin 10^\circ - 4\sin^3 10^\circ = \sin 30^\circ = \frac{1}{2}$
• $4\cos^3 20^\circ - 3\cos 20^\circ = \cos 60^\circ = \frac{1}{2}$
• If $\sin\theta = \frac{1}{4}$, then $\sin 3\theta = \frac{3}{4} - 4 \times \frac{1}{64} = \frac{11}{16}$.

The value of 18°. Let $\theta = 18^\circ$. Then $5\theta = 90^\circ$, so $2\theta = 90^\circ - 3\theta$ and $\sin 2\theta = \cos 3\theta$:

$$2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$$

Divide by $\cos\theta$ and write $\cos^2\theta = 1 - \sin^2\theta$: $2\sin\theta = 1 - 4\sin^2\theta$, that is $4\sin^2\theta + 2\sin\theta - 1 = 0$. The positive root of this quadratic is

$$\sin 18^\circ = \frac{\sqrt{5} - 1}{4} \approx 0.309$$

which is also $\cos 72^\circ$. From it,

$$\cos 36^\circ = 1 - 2\sin^2 18^\circ = 1 - \frac{6 - 2\sqrt{5}}{8} = \frac{\sqrt{5} + 1}{4} \approx 0.809$$

which is also $\sin 54^\circ$. Multiplying the two, $\sin 18^\circ\cos 36^\circ = \frac{5 - 1}{16} = \frac{1}{4}$.

A paper may give $\sin 18^\circ$ and ask for $\cos 18^\circ$: $\cos^2 18^\circ = 1 - \frac{6 - 2\sqrt{5}}{16} = \frac{10 + 2\sqrt{5}}{16}$, so $\cos 18^\circ = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$.

[Tier 2] Products and Sums of Sines and Cosines

Adding and subtracting the compound-angle formulas turns a product into a sum:

$$2\sin A\cos B = \sin(A + B) + \sin(A - B)$$

$$2\cos A\cos B = \cos(A - B) + \cos(A + B)$$

$$2\sin A\sin B = \cos(A - B) - \cos(A + B)$$

Read backwards, with C = A + B and D = A − B, they turn a sum into a product:

$$\sin C + \sin D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2} \qquad \sin C - \sin D = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2}$$

$$\cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2} \qquad \cos C - \cos D = 2\sin\frac{C + D}{2}\sin\frac{D - C}{2}$$

Watch the last one: the half-difference is taken as D − C.

A typical use is a fraction in which the same factor appears on top and at the bottom:

$$\frac{\sin 75^\circ - \sin 15^\circ}{\cos 75^\circ + \cos 15^\circ} = \frac{2\cos 45^\circ\sin 30^\circ}{2\cos 45^\circ\cos 30^\circ} = \tan 30^\circ = \frac{1}{\sqrt{3}}$$

Products of three ratios. Using the sum-times-difference result, $\sin(60^\circ - \theta)\sin(60^\circ + \theta) = \sin^2 60^\circ - \sin^2\theta = \frac{3}{4} - \sin^2\theta$. Multiply by $\sin\theta$:

$$\sin\theta\sin(60^\circ - \theta)\sin(60^\circ + \theta) = \frac{3\sin\theta - 4\sin^3\theta}{4} = \frac{1}{4}\sin 3\theta$$

In the same way

$$\cos\theta\cos(60^\circ - \theta)\cos(60^\circ + \theta) = \frac{1}{4}\cos 3\theta \qquad \tan\theta\tan(60^\circ - \theta)\tan(60^\circ + \theta) = \tan 3\theta$$

With $\theta = 20^\circ$ the three angles are 20°, 40° and 80°:

• $\sin 20^\circ\sin 40^\circ\sin 80^\circ = \frac{1}{4}\sin 60^\circ = \frac{\sqrt{3}}{8}$
• $\cos 20^\circ\cos 40^\circ\cos 80^\circ = \frac{1}{4}\cos 60^\circ = \frac{1}{8}$
• $\tan 20^\circ\tan 40^\circ\tan 80^\circ = \tan 60^\circ = \sqrt{3}$

If a factor of 60° is thrown in as well, multiply by its value: $\sin 20^\circ\sin 40^\circ\sin 60^\circ\sin 80^\circ = \frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2} = \frac{3}{16}$.

Solved Examples

Question 1

Which of the following is negative? (a) $\sin 160^\circ$ (b) $\tan 230^\circ$ (c) $\cos 310^\circ$ (d) $\cos 130^\circ$

Answer. I only need the quadrant of each angle. 160° is in the second quadrant, where sine is positive. 230° is in the third, where tan is positive. 310° is in the fourth, where cos is positive. 130° is in the second quadrant, and cos is negative there, so the answer is (d) $\cos 130^\circ$.

Question 2

What is the value of $\sin 120^\circ + \cos 150^\circ + \tan 135^\circ$? (a) $-1$ (b) 1 (c) $\sqrt{3} - 1$ (d) 0

Answer. I write each angle as 180° minus a standard angle. $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$. Cos is negative in the second quadrant, so $\cos 150^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$. Tan is negative there too: $\tan 135^\circ = -\tan 45^\circ = -1$. The first two cancel, which leaves (a) $-1$. Keeping a plus sign on $\tan 135^\circ$ would give 1.

Question 3

If $\sin\theta = \frac{20}{29}$ and $\theta$ lies between 90° and 180°, what is $\tan\theta$? (a) $\frac{20}{21}$ (b) $-\frac{21}{29}$ (c) $-\frac{20}{21}$ (d) $-\frac{21}{20}$

Answer. First the size, then the sign. $\cos^2\theta = 1 - \frac{400}{841} = \frac{441}{841}$, so the size of $\cos\theta$ is $\frac{21}{29}$ (20, 21, 29 is a Pythagorean triplet). $\theta$ is in the second quadrant, where cos is negative, so $\cos\theta = -\frac{21}{29}$. Then $\tan\theta = \frac{20}{29} \div \left(-\frac{21}{29}\right)$ = (c) $-\frac{20}{21}$.

Question 4

What is the value of $\sec 1200^\circ$? (a) 2 (b) $-2$ (c) $-\frac{2}{\sqrt{3}}$ (d) $\frac{2}{\sqrt{3}}$

Answer. I take out the full turns: 1200 = 3 × 360 + 120, so $\sec 1200^\circ = \sec 120^\circ$. Sec has the same sign as cos, and 120° is in the second quadrant, where cos is negative. $\cos 120^\circ = -\cos 60^\circ = -\frac{1}{2}$, so $\sec 120^\circ$ = (b) $-2$.

Question 5

What is the value of $\tan(-1035^\circ)$? (a) 1 (b) $-1$ (c) $\sqrt{3}$ (d) $-\frac{1}{\sqrt{3}}$

Answer. Tan takes the minus sign outside: $\tan(-1035^\circ) = -\tan 1035^\circ$. Now 1035 = 2 × 360 + 315, and $\tan 315^\circ = -\tan 45^\circ = -1$ (fourth quadrant). So the value is $-(-1)$ = (a) 1. A shorter way: adding three full turns (1080°) to −1035° gives 45°, and $\tan 45^\circ = 1$ at once.

Question 6

Simplify $\frac{\sin(180^\circ - \theta)\cos(90^\circ + \theta)\tan(270^\circ - \theta)}{\cos(360^\circ - \theta)\sin(-\theta)\cot(180^\circ + \theta)}$. (a) $-\tan\theta$ (b) $\cot\theta$ (c) 1 (d) $\tan\theta$

Answer. I change one factor at a time. On top: $\sin(180^\circ - \theta) = \sin\theta$, $\cos(90^\circ + \theta) = -\sin\theta$, and $\tan(270^\circ - \theta) = \cot\theta$ (270° turns tan into cot, and tan is positive in the third quadrant). At the bottom: $\cos(360^\circ - \theta) = \cos\theta$, $\sin(-\theta) = -\sin\theta$ and $\cot(180^\circ + \theta) = \cot\theta$. There is one minus sign on top and one below, so they cancel. $\sin\theta$ and $\cot\theta$ also cancel, and what is left is $\frac{\sin\theta}{\cos\theta}$ = (d) $\tan\theta$.

Question 7

In a triangle ABC (C is not a right angle), what is the value of $\tan(A + B) + \tan C$? (a) $2\tan C$ (b) $-2\tan C$ (c) 0 (d) 1

Answer. The three angles add up to 180°, so A + B = 180° − C. Tan is negative in the second quadrant, so $\tan(180^\circ - C) = -\tan C$. Adding $\tan C$ to it gives (c) 0. The condition on C is there because $\tan 90^\circ$ has no value.

Question 8

What is the value of $\cos 165^\circ$? (a) $\frac{\sqrt{3} + 1}{2\sqrt{2}}$ (b) $-\frac{\sqrt{3} + 1}{2\sqrt{2}}$ (c) $-\frac{\sqrt{3} - 1}{2\sqrt{2}}$ (d) $\frac{\sqrt{3} - 1}{2\sqrt{2}}$

Answer. 165° = 180° − 15°, and cos is negative in the second quadrant, so $\cos 165^\circ = -\cos 15^\circ$. Now $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$. So the answer is (b) $-\frac{\sqrt{3} + 1}{2\sqrt{2}}$. The size makes sense: 165° is close to 180°, so the cosine should be close to −1, and $\frac{2.73}{2.83} \approx 0.97$.

Question 9

If $\sin A = \frac{8}{17}$ and $\cos B = \frac{4}{5}$, where A and B are acute, what is $\sin(A + B)$? (a) $\frac{77}{85}$ (b) $\frac{36}{85}$ (c) $\frac{13}{85}$ (d) $\frac{84}{85}$

Answer. I need all four ratios first. 8, 15, 17 is a triplet, so $\cos A = \frac{15}{17}$, and from 3, 4, 5, $\sin B = \frac{3}{5}$. Then $\sin(A + B) = \sin A\cos B + \cos A\sin B = \frac{8}{17} \times \frac{4}{5} + \frac{15}{17} \times \frac{3}{5} = \frac{32 + 45}{85}$ = (a) $\frac{77}{85}$.

Question 10

What is the value of $\sin 25^\circ\cos 35^\circ + \cos 25^\circ\sin 35^\circ$? (a) $\frac{1}{2}$ (b) $\frac{1}{\sqrt{2}}$ (c) $\sin 10^\circ$ (d) $\frac{\sqrt{3}}{2}$

Answer. This is the right-hand side of $\sin(A + B)$ with A = 25° and B = 35°. So it equals $\sin 60^\circ$ = (d) $\frac{\sqrt{3}}{2}$. With a minus sign in the middle it would have been $\sin(25^\circ - 35^\circ) = -\sin 10^\circ$.

Question 11

If $\tan A = 2$ and $\tan B = 3$, where A and B are acute, what is A + B? (a) 45° (b) 135° (c) 120° (d) 150°

Answer. $\tan(A + B) = \frac{2 + 3}{1 - 2 \times 3} = \frac{5}{-5} = -1$. Each tangent is more than 1, so each angle is more than 45°, and their sum lies between 90° and 180°. In that range the tangent is −1 only at 135°, so A + B = (b) 135°. Reading the −1 as 1 leads to 45°, which is too small for two angles that are each above 45°.

Question 12

What is the value of $\tan 75^\circ + \cot 75^\circ$? (a) $2\sqrt{3}$ (b) 2 (c) 4 (d) $2 + \sqrt{3}$

Answer. $\tan 75^\circ = 2 + \sqrt{3}$, and $\cot 75^\circ = \tan 15^\circ = 2 - \sqrt{3}$. When I add them the roots cancel: (c) 4. Another way is $\tan\theta + \cot\theta = 2\operatorname{cosec} 2\theta = 2\operatorname{cosec} 150^\circ = 2 \times 2$.

Question 13

What is the value of $(1 + \tan 13^\circ)(1 + \tan 32^\circ)$? (a) 2 (b) 1 (c) 4 (d) $\sqrt{2}$

Answer. 13° + 32° = 45°, so $\tan(13^\circ + 32^\circ) = 1$. Cross-multiplying the tangent formula gives $\tan 13^\circ + \tan 32^\circ + \tan 13^\circ\tan 32^\circ = 1$. Opening the brackets gives 1 plus exactly these three terms, so the product is 1 + 1 = (a) 2. I never needed the value of $\tan 13^\circ$ itself.

Question 14

$\tan 65^\circ - \tan 25^\circ$ is equal to (a) $\tan 40^\circ$ (b) $2\tan 40^\circ$ (c) $2\tan 50^\circ$ (d) $\frac{1}{2}\tan 40^\circ$

Answer. 65° and 25° add up to 90°, so $\tan 65^\circ\tan 25^\circ = \tan 65^\circ\cot 65^\circ = 1$. Now $\tan 40^\circ = \tan(65^\circ - 25^\circ) = \frac{\tan 65^\circ - \tan 25^\circ}{1 + 1}$. Multiplying both sides by 2: $\tan 65^\circ - \tan 25^\circ$ = (b) $2\tan 40^\circ$.

Question 15

What is the value of $\frac{\cos 12^\circ + \sin 12^\circ}{\cos 12^\circ - \sin 12^\circ} - \tan 57^\circ$? (a) 1 (b) $2\tan 57^\circ$ (c) $-1$ (d) 0

Answer. I divide the top and the bottom of the fraction by $\cos 12^\circ$. It becomes $\frac{1 + \tan 12^\circ}{1 - \tan 12^\circ}$, which is $\tan(45^\circ + 12^\circ) = \tan 57^\circ$. Taking away $\tan 57^\circ$ leaves (d) 0.