Shortcuts & Speed Methods

Rebuild the Standard Table in Ten Seconds

Most trigonometry questions in the exam need only three or four lines of working, and the options sit close together, so the time goes in choosing the route. The methods below are the quick routes for this chapter, from the table of standard values to maximum values and allied angles, and each one says where it stops working. Two habits help with all of them: know the standard values without looking anything up, and read the four options before you start, because their shape often tells you which method the question wants.

Fast way: write 0, 1, 2, 3, 4 under $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$. Divide each by 4 and take the square root: that is the sin row. The cos row is the same five values read backwards, and tan is sin divided by cos.

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin\theta$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
$\cos\theta$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0
$\tan\theta$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	not defined

Cosec, sec and cot are the reciprocals of sin, cos and tan, so they need no rows of their own. When an expression has only squares, skip the roots: $\sin^2\theta$ runs 0, $\frac{1}{4}$, $\frac{1}{2}$, $\frac{3}{4}$, 1, and $\tan^2\theta$ runs 0, $\frac{1}{3}$, 1, 3.

If a value slips your mind in the exam, draw the half square (sides 1, 1 and $\sqrt{2}$) or the half equilateral triangle (sides 1, $\sqrt{3}$ and 2) and read it off.

For example: $2\sin^2 45^\circ + 3\tan^2 60^\circ - 4\cos^2 30^\circ = 2 \times \frac{1}{2} + 3 \times 3 - 4 \times \frac{3}{4} = 1 + 9 - 3 = 7$.

Where it fails:

• $\tan 90^\circ$, $\sec 90^\circ$, $\cot 0^\circ$ and $\operatorname{cosec} 0^\circ$ are not defined. If one of them turns up in your working, read the question again.
• An angle given in radians must be changed first: $\frac{\pi}{6} = 30^\circ$, $\frac{\pi}{4} = 45^\circ$, $\frac{\pi}{3} = 60^\circ$, $\frac{\pi}{2} = 90^\circ$.

Degrees, Radians and Clock Hands

Fast way:

• $\pi$ radians $= 180^\circ$. To go from degrees to radians multiply by $\frac{\pi}{180}$, and back by $\frac{180}{\pi}$. Cancel the 180 against the angle first: $135^\circ = \frac{135}{180}\pi = \frac{3\pi}{4}$.
• One radian is about $57.3^\circ$ (it is $57^\circ 16' 22''$ if $\pi$ is taken as $\frac{22}{7}$).
• The minute hand of a clock turns $6^\circ$ a minute and the hour hand $\frac{1}{2}^\circ$ a minute. At h hours and m minutes the hands are $|30h - 5.5m|$ degrees apart; if that is more than $180^\circ$, take it away from $360^\circ$.

For example: at 5:12 the hands are $|150 - 66| = 84^\circ$ apart, and $84^\circ = \frac{84}{180}\pi = \frac{7\pi}{15}$ radians.

Where it fails:

• Minutes are sixtieths of a degree: $40^\circ 30'$ is $40.5^\circ$, not $40.3^\circ$.
• The arc $l = r\theta$ and the sector area $\frac{1}{2}r^2\theta$ need $\theta$ in radians. With degrees, use $\frac{\theta}{360^\circ} \times 2\pi r$ (see Chapter 13).

One Ratio Given: Draw the Triangle

Fast way: put the two numbers of the ratio on the two sides it names (sin: opposite over hypotenuse, cos: adjacent over hypotenuse, tan: opposite over adjacent), find the third side, and read off every other ratio. These Pythagorean triplets save the square root:

3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29, 9-40-41, 12-35-37, 11-60-61, 28-45-53, 33-56-65, 48-55-73, 13-84-85, 39-80-89, 65-72-97, and their multiples.

For example: $\tan\theta = \frac{48}{55}$. The hypotenuse is 73, so $\sin\theta = \frac{48}{73}$, $\cos\theta = \frac{55}{73}$, and $\sec\theta + \tan\theta = \frac{73 + 48}{55} = \frac{121}{55} = \frac{11}{5}$.

Where it fails:

• Read which sides the ratio uses. $\operatorname{cosec}\theta = \frac{13}{5}$ puts 13 on the hypotenuse and 5 on the opposite side, so the adjacent side is 12. Adding the squares would treat 13 as a short side.
• If the third side is not a whole number, keep the root: $\sin\theta = \frac{3}{7}$ gives an adjacent side of $\sqrt{49 - 9} = 2\sqrt{10}$.
• The triangle gives sizes only. For an angle beyond $90^\circ$, the sign comes from the quadrant (see the method for allied angles below).

When tan Is Given, Divide by cos

When to use it: $\tan\theta$ (or $\cot\theta$) is given, and every term of the expression has sin and cos to the same total power.

Fast way: divide the top and the bottom by $\cos\theta$. Every $\frac{\sin\theta}{\cos\theta}$ becomes $\tan\theta$:

$$\frac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta} = \frac{a\tan\theta + b}{c\tan\theta + d}$$

With squares, divide by $\cos^2\theta$; a plain number k then becomes $k\sec^2\theta = k(1 + \tan^2\theta)$. When $\cot\theta$ is given, divide by $\sin\theta$ instead.

For example: $3\tan\theta = 2$. Then $\frac{6\sin\theta - \cos\theta}{3\sin\theta + 2\cos\theta} = \frac{6 \times \frac{2}{3} - 1}{3 \times \frac{2}{3} + 2} = \frac{3}{4}$.

With squares, only $\tan^2\theta$ is needed. For $\tan\theta = \frac{1}{\sqrt{2}}$, $\tan^2\theta = \frac{1}{2}$ and $\frac{\sin^2\theta + 2\cos^2\theta}{3 - \sin^2\theta} = \frac{\tan^2\theta + 2}{3(1 + \tan^2\theta) - \tan^2\theta} = \frac{\frac{5}{2}}{4} = \frac{5}{8}$.

Where it fails: if the powers do not match, as in $\frac{\sin\theta + 1}{\cos\theta}$, dividing does not clear it. Draw the triangle instead.

Complementary Pairs in Long Products and Sums

Fast way: pair each angle with the one that makes $90^\circ$ with it.

• $\tan\theta\tan(90^\circ - \theta) = \tan\theta\cot\theta = 1$, so a product such as $\tan 1^\circ\tan 2^\circ \cdots \tan 89^\circ$ is 1. The middle term is $\tan 45^\circ = 1$.
• $\sin^2\theta + \sin^2(90^\circ - \theta) = 1$, and the same for $\cos^2$. So a sum of squares is the number of pairs, plus $\frac{1}{2}$ if $45^\circ$ is one of the angles.
• Count the terms first: from a to b in steps of d there are $\frac{b - a}{d} + 1$ terms.
• A product with $\cos 90^\circ$ (or $\sin 0^\circ$) anywhere in it is 0.

For example: $\sin^2 10^\circ + \sin^2 15^\circ + \cdots + \sin^2 80^\circ$ has $\frac{80 - 10}{5} + 1 = 15$ terms: 7 pairs and the middle term $\sin^2 45^\circ$. The sum is $7 + \frac{1}{2} = 7\frac{1}{2}$.

Where it fails:

• Every angle needs its partner. In $\cos^2 15^\circ + \cos^2 25^\circ + \cdots + \cos^2 85^\circ$ the $85^\circ$ has no $5^\circ$, so $\cos^2 85^\circ$ is left over with no standard value. If that happens, check the first and last angles again.
• A sum running up to $90^\circ$ or starting at $0^\circ$ has one more term to add: $\sin^2 90^\circ = 1$, $\sin^2 0^\circ = 0$.

[SSC Tip] Long products of tan or cot pair up to 1. The same product of sin or cos does not: $\sin 1^\circ\sin 89^\circ$ is $\sin 1^\circ\cos 1^\circ$, which is not 1. If a long product of sines or cosines is asked, look for a $\cos 90^\circ$ or $\sin 0^\circ$ in it that makes the whole product 0.

Equal Co-ratios Mean the Angles Add to 90°

Fast way: if $\sin A = \cos B$, $\tan A = \cot B$ or $\sec A = \operatorname{cosec} B$, with both angles acute, then $A + B = 90^\circ$. Add the two angles and set the sum equal to $90^\circ$; there is no need to change either ratio.

In the figure, the side opposite $\theta$ is the side next to the angle $90^\circ - \theta$. That is why $\sin\theta = \cos(90^\circ - \theta)$ and $\tan\theta = \cot(90^\circ - \theta)$.

For example: $\sin(2\theta + 10^\circ) = \cos(3\theta - 25^\circ)$ gives $5\theta - 15^\circ = 90^\circ$, so $\theta = 21^\circ$. The two angles are $52^\circ$ and $38^\circ$, both acute.

Inside a triangle, $A + B + C = 180^\circ$, so $\frac{B + C}{2} = 90^\circ - \frac{A}{2}$. This turns $\sin\frac{B + C}{2}$ into $\cos\frac{A}{2}$ and $\tan\frac{A + B}{2}$ into $\cot\frac{C}{2}$.

Where it fails:

• $\sin A = \sin B$ means $A = B$ for acute angles. Add the angles only when the two ratios are a co-pair: sin with cos, tan with cot, sec with cosec.
• Check that both angles come out acute. A negative angle or one above $90^\circ$ means a slip in the working.

Test the Options at 45°, Then at 30° or 60°

When to use it: the question asks which option equals an expression in $\theta$, or asks for a value that does not depend on $\theta$.

Fast way: put $\theta = 45^\circ$, where $\sin\theta = \cos\theta = \frac{1}{\sqrt{2}}$ and $\tan\theta = 1$. Work out the question and the four options at that angle, and keep only the options that match.

When two options tie: at $45^\circ$, sin equals cos, tan equals cot and sec equals cosec, so two options that differ only by such a swap always agree there. Test again at $30^\circ$ or $60^\circ$.

For example: $\frac{1 + \sec\theta}{\tan\theta + \sin\theta}$, with options $\operatorname{cosec}\theta$, $\sec\theta$, $\sin\theta$ and $\cot\theta$.

• At $45^\circ$: $\frac{1 + \sqrt{2}}{1 + \frac{1}{\sqrt{2}}} = \frac{(1 + \sqrt{2})\sqrt{2}}{\sqrt{2} + 1} = \sqrt{2}$. Both $\operatorname{cosec}\theta$ and $\sec\theta$ give $\sqrt{2}$.
• At $30^\circ$: multiplying the top and the bottom by $2\sqrt{3}$, $\frac{1 + \frac{2}{\sqrt{3}}}{\frac{1}{\sqrt{3}} + \frac{1}{2}} = \frac{2\sqrt{3} + 4}{2 + \sqrt{3}} = 2$. Now $\operatorname{cosec} 30^\circ = 2$ but $\sec 30^\circ = \frac{2}{\sqrt{3}}$, so the answer is $\operatorname{cosec}\theta$.

Where it fails:

• With a condition in the question, only an angle that satisfies it may be used. If $\sin\theta = 2\cos\theta$ is given, $45^\circ$ is not allowed, since there sin and cos are equal.
• Some expressions break at $45^\circ$: anything with $1 - \tan\theta$ or $\sin\theta - \cos\theta$ in a denominator. Start at $30^\circ$ for those.
• $0^\circ$ and $90^\circ$ are quick, but they leave tan, sec or cot undefined. Use them only when every term stays defined.

Spot the Identity from the Shape of the Options

Many simplifications rest on four products:

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

$$(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 \qquad (\operatorname{cosec}\theta - \cot\theta)(\operatorname{cosec}\theta + \cot\theta) = 1$$

So the shape of the question tells you the shape of the answer: $1 \pm \sin\theta$ leads to sec and tan, and $1 \pm \cos\theta$ leads to cosec and cot.

Expression ($\theta$ acute)	Value
$\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta}$	$2\sec^2\theta$
$\frac{1}{1 - \cos\theta} + \frac{1}{1 + \cos\theta}$	$2\operatorname{cosec}^2\theta$
$\frac{1}{1 - \sin\theta} - \frac{1}{1 + \sin\theta}$	$2\tan\theta\sec\theta$
$\sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}}$	$\sec\theta + \tan\theta$
$\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$	$\operatorname{cosec}\theta - \cot\theta$
$\frac{1}{\sec\theta - \tan\theta}$	$\sec\theta + \tan\theta$

For example: $\sqrt{\frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta}}$. Dividing by $\sec\theta - \tan\theta$ is the same as multiplying by $\sec\theta + \tan\theta$, so the fraction inside is $(\sec\theta + \tan\theta)^2$ and the root is $\sec\theta + \tan\theta$. Multiplying the top and the bottom by $\cos\theta$ shows the same thing: the fraction is $\frac{1 + \sin\theta}{1 - \sin\theta}$, the fourth row of the table. Any option in cosec and cot can be crossed out at a glance.

Where it fails: the lines with a square root take $\theta$ as acute, so that $\sec\theta + \tan\theta$ and the others are positive. For an angle in another quadrant the root may need a minus sign.

Fourth and Sixth Powers in One Line

Write $x = \sin^2\theta\cos^2\theta$. Then:

• $\sin^4\theta + \cos^4\theta = 1 - 2x$
• $\sin^6\theta + \cos^6\theta = 1 - 3x$
• $\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta$, since the other factor $\sin^2\theta + \cos^2\theta$ is 1
• $\sec^4\theta - \tan^4\theta = \sec^2\theta + \tan^2\theta$, for the same reason

The first two give a result that holds for every $\theta$: $2(\sin^6\theta + \cos^6\theta) - 3(\sin^4\theta + \cos^4\theta) + 1 = 2 - 6x - 3 + 6x + 1 = 0$.

For example: if $\sin\theta\cos\theta = \frac{2}{7}$, then $x = \frac{4}{49}$ and $\sin^6\theta + \cos^6\theta = 1 - \frac{12}{49} = \frac{37}{49}$.

Where it fails: $\sin^4\theta + \cos^4\theta$ is not 1. It lies between $\frac{1}{2}$ and 1, and $\sin^6\theta + \cos^6\theta$ lies between $\frac{1}{4}$ and 1; the low ends come at $45^\circ$.

The Partner of sec + tan Is Its Reciprocal

Fast way: $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = \sec^2\theta - \tan^2\theta = 1$. So if $\sec\theta + \tan\theta = p$:

• $\sec\theta - \tan\theta = \frac{1}{p}$
• adding the two and halving, $\sec\theta = \frac{1}{2}\left(p + \frac{1}{p}\right) = \frac{p^2 + 1}{2p}$
• subtracting and halving, $\tan\theta = \frac{p^2 - 1}{2p}$, and then $\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{p^2 - 1}{p^2 + 1}$

The same works for cosec and cot: if $\operatorname{cosec}\theta + \cot\theta = q$, then $\operatorname{cosec}\theta - \cot\theta = \frac{1}{q}$ and $\cos\theta = \frac{q^2 - 1}{q^2 + 1}$.

For example: $\sec\theta + \tan\theta = 9$. Then $\sec\theta - \tan\theta = \frac{1}{9}$, $\sec\theta = \frac{1}{2}\left(9 + \frac{1}{9}\right) = \frac{41}{9}$, $\tan\theta = \frac{40}{9}$ and $\sin\theta = \frac{40}{41}$.

Where it fails:

• Only sec with tan, and cosec with cot, multiply to 1. A condition like $\sin\theta + \cos\theta = p$ needs the next method.
• For an acute angle, p is more than 1. With $p < 1$, $\tan\theta$ comes out negative.

sin + cos Given: Square It

Fast way: if $\sin\theta + \cos\theta = p$, squaring gives $1 + 2\sin\theta\cos\theta = p^2$, so

$$\sin\theta\cos\theta = \frac{p^2 - 1}{2}$$

Everything else follows from this one value:

• $\tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta} = \frac{2}{p^2 - 1}$
• $\sec\theta + \operatorname{cosec}\theta = \frac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = \frac{2p}{p^2 - 1}$
• $(\sin\theta - \cos\theta)^2 = 2 - p^2$
• $\sin^3\theta + \cos^3\theta = p(1 - \sin\theta\cos\theta)$

If $\sin\theta - \cos\theta = q$ is given instead, $\sin\theta\cos\theta = \frac{1 - q^2}{2}$.

For example: $\sin\theta + \cos\theta = \frac{11}{8}$. Then $\sin\theta\cos\theta = \frac{1}{2}\left(\frac{121}{64} - 1\right) = \frac{57}{128}$, so $\tan\theta + \cot\theta = \frac{128}{57}$, and $\sin\theta - \cos\theta = \pm\sqrt{2 - \frac{121}{64}} = \pm\frac{\sqrt{7}}{8}$.

Where it fails:

• p can never be more than $\sqrt{2}$, about 1.414. No angle has $\sin\theta + \cos\theta = \frac{10}{7}$, since $\frac{10}{7}$ is about 1.43.
• Squaring loses the sign: $\sin\theta - \cos\theta$ is negative below $45^\circ$ and positive above it.

Number Plus Its Reciprocal Equal to 2

A positive number plus its reciprocal is at least 2, and equals 2 only when the number is 1 (see Chapter 11). So:

• $\tan\theta + \cot\theta = 2$ gives $\tan\theta = 1$ and $\theta = 45^\circ$; then $\tan^n\theta + \cot^n\theta = 2$ for every power n.
• $\sin\theta + \operatorname{cosec}\theta = 2$ gives $\sin\theta = 1$ and $\theta = 90^\circ$; $\cos\theta + \sec\theta = 2$ gives $\theta = 0^\circ$.
• For any other value k, square it: $\tan^2\theta + \cot^2\theta = k^2 - 2$.

A related condition: $\sin\theta + \sin^2\theta = 1$ means $\sin\theta = 1 - \sin^2\theta = \cos^2\theta$. Then $\cos^2\theta + \cos^4\theta = \sin\theta + \sin^2\theta = 1$, and a long expression in powers of $\cos^2\theta$ can be built up from that.

For example: if $\tan\theta + \cot\theta = 2$, then $\theta = 45^\circ$, and $\tan^{13}\theta + \sin^2\theta = 1 + \frac{1}{2} = \frac{3}{2}$.

Where it fails: a sum such as $\tan\theta + \cot\theta = 1.5$ is impossible. With a positive tan the sum is at least 2, and with a negative tan it is at most $-2$.

Same a and b in Two Lines: Square and Add

Fast way: if $x = a\cos\theta + b\sin\theta$ and $y = a\sin\theta - b\cos\theta$, then

$$x^2 + y^2 = a^2 + b^2$$

because the cross terms $2ab\sin\theta\cos\theta$ cancel. So when one of the two is given, the other is $\pm\sqrt{a^2 + b^2 - (\text{the given value})^2}$.

The same idea removes $\theta$ from other pairs. $x = a\sec\theta + b\tan\theta$ and $y = a\tan\theta + b\sec\theta$ give $x^2 - y^2 = a^2 - b^2$, using $\sec^2\theta - \tan^2\theta = 1$.

For example: $7\cos\theta + 24\sin\theta = 15$. Then $7\sin\theta - 24\cos\theta = \pm\sqrt{49 + 576 - 225} = \pm\sqrt{400} = \pm 20$.

When the given value is exactly $\sqrt{a^2 + b^2}$, the partner is 0, and that fixes $\theta$: $8\cos\theta + 15\sin\theta = 17$ gives $8\sin\theta - 15\cos\theta = 0$, so $\tan\theta = \frac{15}{8}$.

Where it fails:

• The second line must swap sin and cos and change one sign. $6\cos\theta + 8\sin\theta$ and $6\sin\theta + 8\cos\theta$ do not square-add to 100.
• The data never fix the sign, so the answer stays $\pm$ unless the range of $\theta$ is given.

Largest and Smallest Value of a sin + b cos

Fast way:

$$-\sqrt{a^2 + b^2} \le a\sin\theta + b\cos\theta \le \sqrt{a^2 + b^2}$$

A constant added outside just moves both ends. With a triplet the root costs nothing: $28\sin\theta + 45\cos\theta$ runs from $-53$ to 53.

The graph shows $8\sin\theta + 6\cos\theta$ through a full turn. It touches 10 and $-10$ and never comes near 14.

For example: $8\sin\theta - 15\cos\theta + 3$ lies between $-17 + 3 = -14$ and $17 + 3 = 20$.

Where it fails:

• The maximum is not $a + b$, because sin and cos never reach 1 at the same angle.
• If $\theta$ must stay between $0^\circ$ and $90^\circ$, the low end changes. There $8\sin\theta + 6\cos\theta$ runs only from 6 (at $0^\circ$) up to 10.
• Both ratios must be of the same angle. $3\sin 2\theta + 4\cos\theta$ is a different problem.

[SSC Tip] The largest and smallest values sit the same distance either side of the constant, so their sum is twice the constant. In the example, $20 + (-14) = 6$.

Squared sin and cos Together: Between a and b

Fast way: write $a\sin^2\theta + b\cos^2\theta = b + (a - b)\sin^2\theta$. As $\sin^2\theta$ goes from 0 to 1, the value goes from b to a. So $5\sin^2\theta + 9\cos^2\theta$ always lies between 5 and 9.

For example: the same rewriting finds an angle. $6\sin^2\theta + 2\cos^2\theta = 5$ becomes $2 + 4\sin^2\theta = 5$, so $\sin^2\theta = \frac{3}{4}$, $\theta = 60^\circ$ and $\tan\theta = \sqrt{3}$.

Two more ranges worth knowing: $\sin\theta\cos\theta$ never goes above $\frac{1}{2}$, and $\sin^4\theta + \cos^4\theta$ lies between $\frac{1}{2}$ and 1.

Where it fails: both terms must be squares of the same angle. $a\sin^2\theta + b\cos\theta$ is a quadratic in $\cos\theta$, and its range has to be found the long way.

The AM-GM Least Value for tan and cot Squares

For positive numbers, $px + \frac{q}{x} \ge 2\sqrt{pq}$, with equality when $px = \frac{q}{x}$ (the AM-GM rule, see Chapter 11). Since $\tan^2\theta\cot^2\theta = 1$:

• $a\tan^2\theta + b\cot^2\theta \ge 2\sqrt{ab}$
• $a\sec^2\theta + b\operatorname{cosec}^2\theta = (a + b) + a\tan^2\theta + b\cot^2\theta \ge (\sqrt{a} + \sqrt{b})^2$
• $\sin^2\theta + \operatorname{cosec}^2\theta \ge 2$ and $\cos^2\theta + \sec^2\theta \ge 2$

For example: $16\tan^2\theta + 9\cot^2\theta \ge 2\sqrt{144} = 24$, and $4\sec^2\theta + 25\operatorname{cosec}^2\theta \ge (2 + 5)^2 = 49$.

Where it fails: the equality has to be possible. For $\cos^2\theta + 4\sec^2\theta$ the rule says 4, but that needs $\cos^2\theta = 2$. Since $\cos^2\theta$ cannot pass 1, the least value comes at $\cos^2\theta = 1$ (that is, $\theta = 0^\circ$) and is $1 + 4 = 5$. With tan and cot there is no such problem, because $\tan^2\theta$ can take any positive value. None of these expressions has a largest value.

Allied Angles: Sign First, Then the Ratio

Fast way:

1. Remove whole turns: take away $360^\circ$ as many times as you can. For a negative angle, use $\sin(-\theta) = -\sin\theta$, $\cos(-\theta) = \cos\theta$ and $\tan(-\theta) = -\tan\theta$.
2. Find the quadrant and write down the sign from the figure.
3. Write the angle as $180^\circ \pm \theta$ or $360^\circ - \theta$ and keep the same ratio, or as $90^\circ \pm \theta$ or $270^\circ \pm \theta$ and switch to the co-ratio (sin to cos, tan to cot, sec to cosec).

A common way to remember the signs, going anticlockwise from the first quadrant, is All, Sin, Tan, Cos.

For example: $\sin 1290^\circ$. Take off $3 \times 360^\circ = 1080^\circ$ to leave $210^\circ$, which is in the third quadrant, where sin is negative. As $210^\circ = 180^\circ + 30^\circ$, $\sin 1290^\circ = -\sin 30^\circ = -\frac{1}{2}$.

Where it fails:

• The sign belongs to the quadrant of the original angle, not to $30^\circ$. Writing $\sin 210^\circ = \frac{1}{2}$ is the usual slip.
• With $90^\circ + \theta$ the ratio switches and the sign still follows the quadrant: $\cos(90^\circ + \theta) = -\sin\theta$.

Values of 15° and 75°, and Pairs That Add to 45°

From $45^\circ - 30^\circ$ and $45^\circ + 30^\circ$:

Ratio	$15^\circ$	$75^\circ$
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	$2 - \sqrt{3}$	$2 + \sqrt{3}$

Quick facts from the table: $\tan 15^\circ + \tan 75^\circ = 4$, $\tan 15^\circ\tan 75^\circ = 1$ and $\sin 15^\circ\cos 15^\circ = \frac{1}{4}$. Also $\tan 22.5^\circ = \sqrt{2} - 1$ and $\tan 67.5^\circ = \sqrt{2} + 1$.

Pairs adding to 45°: if $A + B = 45^\circ$, then $\tan A + \tan B = 1 - \tan A\tan B$, which rearranges to $(1 + \tan A)(1 + \tan B) = 2$. In cot form, $(\cot A - 1)(\cot B - 1) = 2$.

For example: $\tan^2 75^\circ + \tan^2 15^\circ = (\tan 75^\circ + \tan 15^\circ)^2 - 2\tan 75^\circ\tan 15^\circ = 16 - 2 = 14$.

Where it fails:

• $\tan 15^\circ$ is not $\tan 45^\circ - \tan 30^\circ$, and $\sin 75^\circ$ is not $\sin 45^\circ + \sin 30^\circ$. A ratio of a sum needs the compound-angle formula.
• The product 2 needs $\tan(A + B) = 1$: $A + B = 45^\circ$ or $225^\circ$. For $A + B = 135^\circ$ it is $(1 - \tan A)(1 - \tan B) = 2$ instead.

Double Angles Straight from tan

With $t = \tan\theta$:

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

So with $t = \frac{p}{q}$, $\sin 2\theta = \frac{2pq}{p^2 + q^2}$ and $\cos 2\theta = \frac{q^2 - p^2}{p^2 + q^2}$, with no triangle to draw. Three short simplifications come from the same formulas: $\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan\theta$, $\frac{1 + \cos 2\theta}{\sin 2\theta} = \cot\theta$ and $\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2\theta$.

For example: $\tan\theta = \frac{2}{3}$ gives $\sin 2\theta = \frac{12}{4 + 9} = \frac{12}{13}$ and $\cos 2\theta = \frac{9 - 4}{13} = \frac{5}{13}$.

Where it fails:

• $\sin 2\theta$ is not $2\sin\theta$. At $\theta = 30^\circ$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$ while $2\sin 30^\circ = 1$.
• $\tan 2\theta$ is not defined when $t = 1$, since $2\theta$ is then $90^\circ$.

Throw Out Impossible Options First

Before any working, cross out what cannot be true:

• $\sin\theta$ and $\cos\theta$ lie between $-1$ and 1, while $\sec\theta$ and $\operatorname{cosec}\theta$ are never between $-1$ and 1. An option $\cos\theta = \frac{\sqrt{5}}{2}$ or $\sec\theta = \frac{3}{4}$ is out.
• For acute angles, sin and tan grow as the angle grows, and cos shrinks. Below $45^\circ$, $\cos\theta > \sin\theta$; above it, $\sin\theta > \cos\theta$.
• $\sin\theta + \cos\theta$ is at most $\sqrt{2}$, and for an acute angle it is more than 1.

Wrong options are usually made from a short list of slips: sin and cos values swapped, $\frac{1}{\sqrt{3}}$ written for $\sqrt{3}$, the reciprocal ratio, a minus sign lost in the second quadrant, an answer in radians without its $\pi$, $\cos(A + B)$ taken as $\cos A + \cos B$, and $a + b$ given as the maximum in place of $\sqrt{a^2 + b^2}$. When two options differ by one of these slips, the answer is usually one of that pair.

For example: which is larger, $\sin 1$ or $\cos 1$, with the angle in radians? One radian is about $57^\circ$, past $45^\circ$, so $\sin 1$ is larger.

Where it fails: these checks narrow the field but seldom pick the answer alone. If two options both pass, work the question out.

Solved Examples

Every question that follows is solved with one of the methods in this section. A few carry a time check; the rest should take about as long once the method is familiar.

Question 1

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

Answer. Everything is squared, so I use the squared values straight away: $\tan^2 30^\circ = \frac{1}{3}$, $\cos^2 45^\circ = \frac{1}{2}$, $\operatorname{cosec}^2 45^\circ = 2$, $\sec^2 30^\circ = \frac{4}{3}$. The top is $1 + 2 = 3$ and the bottom is $2 + \frac{4}{3} = \frac{10}{3}$, so the value is $3 \times \frac{3}{10}$ = $\frac{9}{10}$.

Time check: about 15 seconds.

Question 2

Find, in radians, the smaller angle between the hands of a clock at 7:20.

Answer. $|30 \times 7 - 5.5 \times 20| = |210 - 110| = 100^\circ$, which is less than $180^\circ$, so it stays. In radians, $\frac{100}{180}\pi$ = $\frac{5\pi}{9}$.

Question 3

If $\operatorname{cosec}\theta = \frac{85}{13}$ and $\theta$ is acute, find $\sec\theta - \tan\theta$.

Answer. Cosec is hypotenuse over opposite, so the opposite side is 13 and the hypotenuse 85. The triplet 13-84-85 gives the adjacent side as 84. Then $\sec\theta - \tan\theta = \frac{85}{84} - \frac{13}{84} = \frac{72}{84}$ = $\frac{6}{7}$.

Question 4

If $5\cot\theta = 2$, find $\frac{10\cos\theta + 3\sin\theta}{5\cos\theta - \sin\theta}$.

Answer. Here cot is given, so I divide the top and the bottom by $\sin\theta$: $\frac{10\cot\theta + 3}{5\cot\theta - 1} = \frac{4 + 3}{2 - 1}$ = 7.

Question 5

Find $\cos^2 9^\circ + \cos^2 18^\circ + \cos^2 27^\circ + \cdots + \cos^2 81^\circ + \cos^2 90^\circ$.

Answer. From $9^\circ$ to $81^\circ$ in steps of 9 there are $\frac{81 - 9}{9} + 1 = 9$ terms: four pairs adding to $90^\circ$ ($9^\circ$ with $81^\circ$, $18^\circ$ with $72^\circ$, $27^\circ$ with $63^\circ$, $36^\circ$ with $54^\circ$) and the middle term $\cos^2 45^\circ = \frac{1}{2}$. The last term $\cos^2 90^\circ$ is 0, not 1. The sum is $4\frac{1}{2}$.

Time check: about 20 seconds.

Question 6

If $\sin(5\theta - 8^\circ) = \cos(3\theta + 18^\circ)$, where both angles are acute, find $\sec(3\theta + 30^\circ)$.

Answer. A sin equal to a cos means the angles add to $90^\circ$: $8\theta + 10^\circ = 90^\circ$, so $\theta = 10^\circ$. (The angles are $42^\circ$ and $48^\circ$.) Then $\sec(30^\circ + 30^\circ) = \sec 60^\circ$ = 2.

Question 7

Which of these equals $(\operatorname{cosec}\theta - \sin\theta)(\sec\theta - \cos\theta)(\tan\theta + \cot\theta)$: 1, $\tan\theta$, $\sin\theta\cos\theta$ or 2?

Answer. I put $\theta = 45^\circ$: $\left(\sqrt{2} - \frac{1}{\sqrt{2}}\right)^2 \times 2 = \frac{1}{2} \times 2 = 1$. That rules out $\sin\theta\cos\theta$ (which is $\frac{1}{2}$ there) and 2, but $\tan 45^\circ$ is also 1, so I try $60^\circ$: $\left(\frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2}\right)\left(2 - \frac{1}{2}\right)\left(\sqrt{3} + \frac{1}{\sqrt{3}}\right) = \frac{1}{2\sqrt{3}} \times \frac{3}{2} \times \frac{4}{\sqrt{3}} = 1$. It is 1 again while $\tan 60^\circ = \sqrt{3}$, so the expression equals 1.

Question 8

If $\frac{1}{1 + \sin\theta} + \frac{1}{1 - \sin\theta} = 8$ and $\theta$ is acute, find $\theta$.

Answer. The two denominators are a conjugate pair, so the sum is $\frac{2}{1 - \sin^2\theta} = 2\sec^2\theta$. Then $2\sec^2\theta = 8$, $\sec\theta = 2$ and $\theta$ = $60^\circ$.

Question 9

If $\sin\theta\cos\theta = \frac{2}{5}$, find $\sin^4\theta + \cos^4\theta$.

Answer. $\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - 2 \times \frac{4}{25}$ = $\frac{17}{25}$.