Trigonometry

An angle can be measured in two common units. You already know degrees: a full turn is 360°, a straight line is 180°, and 1° = 60′ (minutes), 1′ = 60″ (seconds).

The second unit is the radian. Draw a circle of radius r and mark an arc whose length is exactly r. The angle this arc makes at the centre is 1 radian. The full circumference is 2πr, so a full turn holds 2π such arcs. Therefore:

[[fig:0]]

2π radians = 360°, that is, π radians = 180°.

Both conversions come from this single line:

Formula (Degree and radian):
>> - Degree → radian: multiply by π180
>> - Radian → degree: multiply by 180π

[[fig:1]]

So 1 radian = 180π ≈ 57.3° (more precisely 57°17′45″), and 1° = π180 ≈ 0.01745 radian.

Radhika converts 210° into radians: 210 × π180 = 7π6. Then she turns 5π12 into degrees: 5 × 18012 = 75°. Whenever π is written in the radian measure, put π = 180° and simplify.

[[fig:2]]

When there is no π (say 1.2 radians), multiply by 180π and use whichever value of π the question gives, 3.1416 or 227: 1.2 × 180π ≈ 68.75°.

The type comes in two more shapes:

• Negative angles: −3π4 = −135°. The sign stays as it is.
• Part of a full circle: an angle of θ radians is θ2π of a full turn. Multiply by 100 to get the percentage.

This is the most familiar question of the chapter: one ratio is given, and you have to find another ratio or an expression. The method does not change.

Step 1. Turn the ratio into two sides. If sin θ = 817, then P = 8k and H = 17k. We can take k = 1, because ratios do not depend on the size of the triangle.

Step 2. Find the third side by Pythagoras. B = √(17² − 8²) = √(289 − 64) = √225 = 15. A triplet gives it at once: 8, 15, 17.

Step 3. Read any ratio you want. cos θ = 1517, tan θ = 815, cot θ = 158, sec θ = 1715, cosec θ = 178.

[[fig:0]]

The same question also comes in other forms.

The triangle itself is given. "In △ABC, ∠B = 90°, AB = 20, BC = 21." First find AC = 29 (triplet 20, 21, 29). For angle A, P = BC = 21 and B = AB = 20, so tan A = 2120. For angle C, P = AB = 20, so tan C = 2021. Always pick P and B for the angle that is asked. If three sides are given and no right angle is mentioned, check which side satisfies H² = P² + B²; the right angle is opposite the longest side.

A condition like 3 sin θ = 5 cos θ. Divide by cos θ: 3 tan θ = 5, so tan θ = 53. It is a ratio in disguise.

An expression like a sin θ + b cos θc sin θ + d cos θ. Here you do not even need the triangle. Divide the numerator and the denominator by cos θ, and everything becomes tan θ, which you already know. If cot θ is given, divide by sin θ instead.

[[fig:1]]

Suppose 4 tan θ = 3 and the question asks for 2 sin θ + cos θ2 sin θ − cos θ. Dividing by cos θ gives 2 tan θ + 12 tan θ − 1 = 32 + 132 − 1 = 5212 = 5, with no square root anywhere.

Unless the question says otherwise, θ is acute in this type, so all ratios are positive. Angles in other quadrants come in a later type.

Once the table of 0°, 30°, 45°, 60°, 90° is in your head, three kinds of questions become direct.

1. Evaluate an expression. Put the values and simplify. Write each value clearly before multiplying, and simplify the roots at the end.

For example, tan²60° + 2 sin²45° − cos 0° = 3 + 2 × 12 − 1 = 3. Here tan²60° = (√3)² = 3 and sin²45° = 12; the squares of standard values are easy numbers.

2. Find θ from an equation. Bring the equation to the form "one ratio = a standard value", then read θ from the table.

• tan θ = √3 ⇒ θ = 60° (for acute θ).
• A quadratic in sin θ or cos θ: solve it like any quadratic, then reject the root that is impossible. For example, 2 sin²θ − 3 sin θ + 1 = 0 gives (2 sin θ − 1)(sin θ − 1) = 0, so sin θ = 12 or 1, that is, θ = 30° or 90°. If a root comes out as sin θ = 3 or cos θ = −2, throw it away, because sin θ and cos θ always lie between −1 and 1.
• Two equations such as sin(A + B) = 1 and tan(A − B) = 1√3 give A + B = 90° and A − B = 30°; then add and subtract.

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3. Compare ratios. From 0° to 90°, sin θ, tan θ and sec θ increase while cos θ, cot θ and cosec θ decrease. So a question like which is greater or which increases continuously needs no calculation.

[[fig:1]]

The options help a lot here. When the question asks for θ and offers four standard angles, put each option into the equation. The one that satisfies it is the answer, and this is often quicker than solving.

Now we come to the heart of the chapter. Take the right triangle again: P² + B² = H². Divide every term by H²:

P²H² + B²H² = 1, that is, sin²θ + cos²θ = 1.

[[fig:0]]

Divide P² + B² = H² by B² instead: P²B² + 1 = H²B², that is, 1 + tan²θ = sec²θ.

Divide by P²: 1 + B²P² = H²P², that is, 1 + cot²θ = cosec²θ.

[[fig:1]]

These three are called identities because they are true for every angle θ (wherever the ratios are defined), not just for one particular angle. Learn each of them in all its forms, because a question uses whichever form suits it:

Formula (The three identities):
>> sin²θ + cos²θ = 1
>> 1 + tan²θ = sec²θ
>> 1 + cot²θ = cosec²θ

How to simplify any expression of this type

• Step 1: Look for a ready identity: (1 − cos²θ) is sin²θ, (sec²θ − 1) is tan²θ, (1 + cot²θ) is cosec²θ.
• Step 2: If nothing is ready, convert everything into sin θ and cos θ.
• Step 3: Take the LCM, use sin² + cos² = 1 in the numerator, and cancel.

Radhika simplifies (1 − sin²θ) tan²θ. Since 1 − sin²θ = cos²θ, the expression is cos²θ × sin²θcos²θ = sin²θ.

One more: 11 + sin θ + 11 − sin θ. With the LCM it becomes (1 − sin θ) + (1 + sin θ)1 − sin²θ = 2cos²θ = 2 sec²θ. Keep an eye on this shape: (1 + x)(1 − x) = 1 − x² nearly always turns into an identity.

Questions of this type look frightening, but they use only the three identities and a little algebra. These tools open almost all of them.

1. Algebraic identities in trigonometric form

• a² − b² = (a + b)(a − b): sin⁴θ − cos⁴θ = (sin²θ + cos²θ)(sin²θ − cos²θ) = sin²θ − cos²θ.
• a³ ± b³ = (a ± b)(a² ∓ ab + b²): sin³θ + cos³θsin θ + cos θ = 1 − sin θ cos θ.
• (a + b)² + (a − b)² = 2(a² + b²).

2. Higher powers of sin and cos. Put x = sin²θ cos²θ:

• sin⁴θ + cos⁴θ = (sin²θ + cos²θ)² − 2 sin²θ cos²θ = 1 − 2x
• sin⁶θ + cos⁶θ = (sin²θ + cos²θ)³ − 3 sin²θ cos²θ(sin²θ + cos²θ) = 1 − 3x

So a combination like a(sin⁶θ + cos⁶θ) + b(sin⁴θ + cos⁴θ) + c sin²θ cos²θ becomes a + b + (c − 3a − 2b)x. When the question asks for a fixed value, the x-part has to vanish, and the answer is just a + b.

[[fig:0]]

3. Fractions with 1 ± sin θ or 1 ± cos θ. Take the LCM and the denominator becomes cos²θ or sin²θ. Or multiply by the conjugate: cos θ1 + sin θ = cos θ(1 − sin θ)1 − sin²θ = 1 − sin θcos θ.

4. Two well-known products

• (1 + tan θ + sec θ)(1 + cot θ − cosec θ) = 2
• (1 + tan θ − sec θ)(1 + cot θ + cosec θ) = 2

Write each bracket over sin θ or cos θ and multiply: (sin θ + cos θ + 1)(sin θ + cos θ − 1)sin θ cos θ = (sin θ + cos θ)² − 1sin θ cos θ = 2 sin θ cos θsin θ cos θ = 2.

A small one to warm up: (sin θ + cos θ)² − 2 sin θ cos θ. Expanding gives sin²θ + cos²θ + 2 sin θ cos θ − 2 sin θ cos θ = 1.

Value-putting works best here. If the answer is a number, the expression has the same value for every θ, so put the most convenient angle. For expressions with only sin and cos, θ = 0° is quickest (sin 0° = 0, cos 0° = 1). For example, 3(sin⁴θ + cos⁴θ) − 2(sin⁶θ + cos⁶θ) at θ = 0° is 3 − 2 = 1, and that is the answer for every θ.

[[fig:1]]

Take the identity sec²θ − tan²θ = 1 and factorise it:

Formula (Reciprocal pair):
>> (sec θ + tan θ)(sec θ − tan θ) = 1

So sec θ − tan θ is the reciprocal of sec θ + tan θ. If sec θ + tan θ = k, then sec θ − tan θ = 1k. The other pair behaves the same way: cosec²θ − cot²θ = 1 gives (cosec θ + cot θ)(cosec θ − cot θ) = 1.

Once you have both the sum and the difference, add and subtract:

• sec θ = 12(k + 1k) = k² + 12k, tan θ = 12(k − 1k) = k² − 12k
• and then sin θ = tan θsec θ = k² − 1k² + 1.

[[fig:0]]

In the same way, cosec θ + cot θ = k gives cosec θ = k² + 12k, cot θ = k² − 12k and cos θ = cot θcosec θ = k² − 1k² + 1.

Say Radhika is told sec θ + tan θ = 4. Then sec θ − tan θ = 14. Adding, 2 sec θ = 174, so sec θ = 178. Subtracting, 2 tan θ = 154, so tan θ = 158, and sin θ = 1517. The triplet 8, 15, 17 turns up by itself.

The root forms. The same pairs hide inside square roots:

• √1 + sin θ1 − sin θ = 1 + sin θcos θ = sec θ + tan θ (multiply inside by (1 + sin θ) and use 1 − sin²θ = cos²θ)
• √1 − sin θ1 + sin θ = sec θ − tan θ
• √1 + cos θ1 − cos θ = cosec θ + cot θ, √1 − cos θ1 + cos θ = cosec θ − cot θ

So a sum like √1 + sin θ1 − sin θ + √1 − sin θ1 + sin θ is (sec θ + tan θ) + (sec θ − tan θ) = 2 sec θ. For acute θ all these quantities are positive, so the square roots open cleanly.

[[fig:1]]

In this type the question gives a value like sin θ + cos θ = k or tan θ + cot θ = k and asks for something bigger: a square, a cube or a fourth power. You do not need θ itself. Algebra is enough.

1. sin θ + cos θ and sin θ − cos θ. Square them:

• (sin θ + cos θ)² = 1 + 2 sin θ cos θ
• (sin θ − cos θ)² = 1 − 2 sin θ cos θ
• Adding: (sin θ + cos θ)² + (sin θ − cos θ)² = 2

So if sin θ + cos θ = k, then sin θ cos θ = k² − 12 and (sin θ − cos θ)² = 2 − k². If instead sin θ − cos θ = k, then sin θ cos θ = 1 − k²2 and (sin θ + cos θ)² = 2 − k²; do not use the (k² − 1) form for the minus sign. For real θ, |sin θ ± cos θ| is at most √2 (exactly), so a larger k is impossible. From sin θ cos θ you can go further: sin⁴θ + cos⁴θ = 1 − 2(sin θ cos θ)², and tan θ + cot θ = 1sin θ cos θ.

[[fig:0]]

2. tan θ + cot θ. Since tan θ · cot θ = 1, it behaves exactly like x + 1x:

• tan²θ + cot²θ = k² − 2
• tan³θ + cot³θ = k³ − 3k
• Also tan θ + cot θ = sin²θ + cos²θsin θ cos θ = sec θ cosec θ, so sec²θ cosec²θ = k².

The same algebra works for sin θ + cosec θ and cos θ + sec θ. sec θ + cosec θ has no product equal to 1, so it needs its own handling.

[[fig:1]]

3. Elimination. Sometimes x and y are given in terms of θ, and you must find a relation that does not contain θ. Square and add, so that sin² + cos² = 1 (or sec² − tan² = 1) removes θ.

• x = a cos θ + b sin θ, y = a sin θ − b cos θ ⇒ x² + y² = a² + b² (the cross terms 2ab sin θ cos θ cancel).
• x = a sec θ, y = b tan θ ⇒ x²a² − y²b² = 1.

[[fig:2]]

Let us see it with numbers. Radhika is told sin θ + cos θ = 75. Squaring, 1 + 2 sin θ cos θ = 4925, so sin θ cos θ = 1225. Then tan θ + cot θ = 1sin θ cos θ = 2512. She never had to find θ.

One more: if x = 2 cos θ + 3 sin θ and y = 2 sin θ − 3 cos θ, then x² + y² = 4 + 9 = 13, whatever θ is.

Here the question puts a condition on θ, and that condition quietly fixes θ, or at least fixes one ratio. Learn to recognise these conditions; each one leads to a fixed result.

1. x + 1x = 2 forces x = 1. This is because x + 1x − 2 = (x − 1)²x, which is zero only at x = 1.

• tan θ + cot θ = 2 ⇒ tan θ = 1 ⇒ θ = 45°. Then every power of tan θ and cot θ is 1, and sec²θ = 2.
• sin θ + cosec θ = 2 ⇒ sin θ = 1 ⇒ θ = 90°.
• cos θ + sec θ = 2 ⇒ cos θ = 1 ⇒ θ = 0°.

[[fig:0]]

2. sin θ + sin²θ = 1. Then sin θ = 1 − sin²θ = cos²θ. So cos²θ + cos⁴θ = cos²θ + sin²θ = 1. In the same way, cos θ + cos²θ = 1 gives sin²θ + sin⁴θ = 1. Replace the single power by the square of the other ratio.

[[fig:1]]

3. a sin²θ + b cos²θ = c. Write cos²θ = 1 − sin²θ to get (a − b) sin²θ = c − b. So sin²θ = c − ba − b, and dividing the original equation by cos²θ gives tan²θ = c − ba − c.

Suppose Radhika gets 6 sin²θ + 2 cos²θ = 3. Then 4 sin²θ = 1, so sin θ = 12 and θ = 30°. The formula agrees: tan²θ = 3 − 26 − 3 = 13.

[[fig:2]]

4. A linear condition hides a ratio. sin θ + cos θ = k sin θ ⇒ cos θ = (k − 1) sin θ ⇒ cot θ = k − 1. Divide by the ratio that appears on both sides.

5. An expression equal to a number. First simplify the expression with the identities of the earlier types. It usually becomes something like 2 sec θ = 4, which gives θ at once.

Be careful with value-putting in this type. θ is no longer just any angle; the condition decides it.

Two angles are complementary if they add up to 90°. In a right triangle the two acute angles are always complementary: if one is θ, the other is 90° − θ.

Now look at the triangle from the other acute angle. The side that was the perpendicular for θ becomes the base for 90° − θ, and the base becomes the perpendicular. So the ratios swap places:

[[fig:0]]

That is where the names sin–cosin, tan–cot and sec–cosec come from: "co" stands for complementary. A ratio of an angle equals the co-ratio of its complement.

How questions use this

• Ratio of two complementary angles: sin 37°cos 53° = sin 37°sin 37° = 1, because cos 53° = sin(90° − 53°) = sin 37°.
• Products: tan 17° · tan 73° = tan 17° · cot 17° = 1. So a product like tan 1° · tan 2° · … · tan 89° pairs up into 1s, with tan 45° = 1 left in the middle, and the whole product is 1.
• Sums of squares: sin²12° + sin²78° = sin²12° + cos²12° = 1. So sin²1° + sin²2° + … + sin²89° has 44 such pairs plus sin²45° = 12, a total of 4412.
• Angles of a triangle: A + B + C = 180°, so B + C2 = 90° − A2. Hence sin B + C2 = cos A2, and so on.
• Writing in angles between 0° and 45°: sin 70° = cos 20°, tan 58° = cot 32°.

[[fig:1]]

Take sec 50° · sin 40° + cos 25° · cosec 65°. Since sec 50° = cosec 40°, the first term is 1; since cosec 65° = sec 25°, the second term is also 1. The answer is 2.

From the last type we know sin θ = cos(90° − θ). Now read it the other way. If two acute angles satisfy sin A = cos B, then, since cos B = sin(90° − B), we get sin A = sin(90° − B). For acute angles the sine takes each value only once, so A = 90° − B, that is:

sin A = cos B ⇒ A + B = 90°

[[fig:0]]

The other pairs work in exactly the same way: tan A = cot B and sec A = cosec B each give A + B = 90°.

So these questions reduce to a one-line linear equation. Add the two angles and put the sum equal to 90°.

For example, tan 2x = cot(x + 15°) gives 2x + (x + 15°) = 90°, so 3x = 75° and x = 25°. Similarly, sin 4θ = cos 2θ gives 6θ = 90°, so θ = 15°.

Watch two things:

• The rule needs co-ratios on the two sides (sin with cos, tan with cot, sec with cosec). If both sides have the same ratio, like cos x = cos(2x − 30°), the angles are equal, not complementary: x = 2x − 30°, so x = 30°. A question may hide this by writing sin(90° − x) in place of cos x, so simplify first and then decide.
• The angles must be acute (the question usually says so, and it is stated in the stem when it matters). That is what makes the answer unique. For other angles, sin A = cos B has more solutions: sin 30° = sin 150°, so A = 90° − B + 360°n or A = 90° + B + 360°n.

[[fig:1]]

After finding the angle, the question often asks one step more, such as tan(x + y). Put the standard value and you are done.

Every maximum–minimum question in trigonometry stands on the natural limits of the ratios:

[[fig:0]]

1. a sin θ + b cos θ. Its maximum is √a² + b² and its minimum is −√a² + b². The reason: write a = r cos α and b = r sin α with r = √a² + b². Then a sin θ + b cos θ = r(sin θ cos α + cos θ sin α) = r sin(θ + α), and sin can move only from −1 to 1. A constant added outside shifts both values by the same amount. These limits hold for θ unrestricted over real angles. If θ is restricted (say acute), check the allowed interval and its end points: 3 sin θ + 4 cos θ + 7 has maximum 12 for acute θ, but it only comes close to 10 (as θ → 90°) and never reaches it, so its infimum is 10, not 2.

Radhika wants the range of 5 sin θ + 12 cos θ. Since √25 + 144 = 13, it runs from −13 to 13, and 5 sin θ + 12 cos θ + 4 runs from −9 to 17.

[[fig:1]]

2. a sin²θ + b cos²θ always lies between a and b, because it equals b + (a − b) sin²θ. It is b at θ = 0° and a at θ = 90°.

3. AM ≥ GM for a term plus a reciprocal-type term. For positive x and k, x + kx ≥ 2√k. So for positive a and b, a tan²θ + b cot²θ ≥ 2√ab. With sin²θ, which cannot exceed 1, take care: a sin²θ + b cosec²θ has minimum 2√ab only if the equality point sin²θ = √ba can actually be reached, that is, when a ≥ b. Otherwise the minimum comes at sin²θ = 1 and equals a + b.

[[fig:2]]

4. Powers of sin and cos. sin⁴θ + cos⁴θ = 1 − 2 sin²θ cos²θ lies between 12 and 1. More generally, for powers of 2 or more, sinⁿθ + cosᵐθ never goes above 1, because each part is at most the matching square and sin²θ + cos²θ = 1.

5. A sum of sines equal to the number of terms. If sin α₁ + sin α₂ + … + sin αₙ = n, every sine must be 1 (none can exceed 1), so every angle is 90°.

So far θ was acute. But an angle can be 120°, 225°, 570° or even negative. To handle these, imagine the angle measured from the positive x-axis, turning anticlockwise. The two axes cut the plane into four quadrants:

• Quadrant I: 0° to 90°
• Quadrant II: 90° to 180°
• Quadrant III: 180° to 270°
• Quadrant IV: 270° to 360°

Which ratios are positive where? Remember "Add Sugar To Coffee":

• I: All ratios positive
• II: only Sin (and cosec) positive
• III: only Tan (and cot) positive
• IV: only Cos (and sec) positive

[[fig:0]]

Reducing a big angle to an acute one. Write the angle as (n × 90°) ± θ with θ acute.

• If n is even (180°, 360°), the ratio stays the same: sin(180° − θ) = sin θ.
• If n is odd (90°, 270°), the ratio changes to its co-ratio: cos(90° + θ) = −sin θ.
• The sign is the sign of the original ratio in the quadrant where the original angle lies.

[[fig:1]]

A few worked lines: sin 120° = sin(180° − 60°) = +sin 60° = √32 (quadrant II, sin positive). cos 150° = −cos 30° = −√32. tan 225° = tan(180° + 45°) = +1 (quadrant III, tan positive). sin 300° = −sin 60°.

[[fig:2]]

Negative angles: sin(−θ) = −sin θ, cos(−θ) = cos θ, tan(−θ) = −tan θ. Only cos (and sec) ignore the minus sign.

Very large angles: first remove full turns of 360° (for tan and cot you may remove multiples of 180°). sin 750° = sin(720° + 30°) = sin 30° = 12.

One ratio given with its quadrant. Find the size of the other ratio from the triangle as before, then attach the sign of that quadrant. If sin θ = 817 with θ in quadrant II, then cos θ = −1517 and tan θ = −815.

Radhika finds cos 120° this way: 120° = 180° − 60° is in quadrant II, where cos is negative, so cos 120° = −12. Likewise tan 315° = tan(360° − 45°) = −1, because 315° is in quadrant IV, where tan is negative.

Period and range. Because an angle can go round the circle again and again, every ratio repeats. The period is the smallest turn after which the ratio takes the same values again:

If the angle is multiplied by k, the wave is squeezed by k: the period of sin kθ or cos kθ is 360°k, and the period of tan kθ is 180°k. So sin 2θ repeats every 180°, and cos θ2 every 720°. The multiplier changes the period, never the range: sin 3θ still stays between −1 and 1.

Because of the repeat, an equation such as sin θ = 12 has two answers in every full turn of 0° to 360°: θ = 30° (quadrant I) and θ = 150° (quadrant II).

Radhika wants the period of tan 4θ: it is 180°4 = 45°.

[[fig:3]]

The last type deals with ratios of sums and differences of angles. These formulas do not come out of a single triangle, so learn them carefully. Once they are learnt, many questions open in one line.

Formula — Compound angles:
>> - sin(A ± B) = sin A cos B ± cos A sin B
>> - cos(A ± B) = cos A cos B ∓ sin A sin B (the sign flips here)
>> - tan(A ± B) = tan A ± tan B1 ∓ tan A tan B

[[fig:0]]

They give the values sin 15° = √3 − 12√2, cos 15° = sin 75° = √3 + 12√2, tan 15° = 2 − √3 and tan 75° = 2 + √3.

[[fig:1]]

Formula — Double angles (put B = A):
>> - sin 2A = 2 sin A cos A = 2 tan A1 + tan²A
>> - cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A = 1 − tan²A1 + tan²A
>> - tan 2A = 2 tan A1 − tan²A

From cos 2A we also get 1 + cos 2A = 2 cos²A and 1 − cos 2A = 2 sin²A. So √2 + 2 cos 2θ = 2 cos θ (when cos θ ≥ 0).

[[fig:2]]

Formula — Triple angles:
>> - sin 3A = 3 sin A − 4 sin³A
>> - cos 3A = 4 cos³A − 3 cos A
>> - tan 3A = 3 tan A − tan³A1 − 3 tan²A

Useful special results

• tan(45° + A) = 1 + tan A1 − tan A = cos A + sin Acos A − sin A
• If A + B = 45°: (1 + tan A)(1 + tan B) = 2, and tan A + tan B + tan A tan B = 1.
• If A − B = 45°: tan A − tan B − tan A tan B = 1.
• cos(A + B) cos(A − B) = cos²A − sin²B; sin(A + B) sin(A − B) = sin²A − sin²B.
• If A + B = C: tan A tan B tan C = tan C − tan A − tan B.

Formula — Product and sum forms (for longer expressions):
>> - 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)
>> - sin C − sin D = 2 cos C + D2 sin C − D2; cos C + cos D = 2 cos C + D2 cos C − D2

Radhika finds cos 15° as cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30° = 1√2 · √32 + 1√2 · 12 = √3 + 12√2. Then, with tan A = 12, she gets sin 2A = 2 × 121 + 14 = 45 without drawing any triangle.

A few angles outside the standard table come back again and again: 15°, 75°, 22.5°, 18°, 36°, 54° and 72°. Nobody expects you to derive them from scratch in the hall. You are expected to know the values (or to rebuild them in a minute) and to use the neat relations they give.

15° and 75°. Write 15° = 45° − 30° and use the compound-angle formula:

Formula (Values of 15° and 75°):
>> - sin 15° = cos 75° = √6 − √24
>> - cos 15° = sin 75° = √6 + √24
>> - tan 15° = cot 75° = 2 − √3, tan 75° = cot 15° = 2 + √3

[[fig:0]]

Because 15° and 75° are complementary, all the neat results follow without any root: sin 15° cos 15° = 12 sin 30° = 14, tan 15° · tan 75° = 1, and tan 15° + cot 15° = (2 − √3) + (2 + √3) = 4. Also sin 15° + cos 15° = √62 and cos 15° − sin 15° = √22.

22.5°. It is half of 45°, so use tan θ2 = 1 − cos θsin θ:

tan 22.5° = 1 − 1√21√2 = √2 − 1, cot 22.5° = √2 + 1

The figure shows the same result without any formula: the right triangle has legs 1 and 1 + √2. Also sin²22.5° = 2 − √24 and cos²22.5° = 2 + √24.

[[fig:1]]

18°, 36°, 54° and 72°. Put θ = 18°. Then 5θ = 90°, so 2θ = 90° − 3θ and sin 2θ = cos 3θ. That is 2 sin θ cos θ = 4 cos³θ − 3 cos θ. Divide by cos θ and change cos²θ into 1 − sin²θ:

4 sin²θ + 2 sin θ − 1 = 0 ⇒ sin 18° = √5 − 14

Then cos 36° = 1 − 2 sin²18° = √5 + 14. The complements come free: sin 18° = cos 72° and cos 36° = sin 54°. The figure shows where the numbers sit: in the triangle with angles 36°–72°–72° and base 1, the equal sides are 1 + √52.

[[fig:2]]

The results that exams like: sin 18° × cos 36° = 14, cos 36° − cos 72° = 12, sin 18° + cos 36° = √52.

Radhika checks the value of sin 18° by calculator sense: √5 ≈ 2.236, so √5 − 14 ≈ 0.309, which is indeed a little less than sin 30° = 0.5.

When A, B and C are the three angles of a triangle, they are not three free angles: A + B + C = 180°. One fact, used in the right place, turns many long expressions into a single line. The trick is always the same. Write A + B as 180° − C and let the quadrant rules of Type 12 do the work.

[[fig:0]]

Two angles together. Since A + B = 180° − C:

Formula (Angles of a triangle, A + B + C = 180°):
>> - sin(A + B) = sin(180° − C) = sin C
>> - cos(A + B) = cos(180° − C) = −cos C
>> - tan(A + B) = −tan C

Half angles. A + B2 = 90° − C2, a complementary pair, so sin A + B2 = cos C2, cos A + B2 = sin C2 and tan A + B2 = cot C2.

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Sum and product of tangents. From tan(A + B) = −tan C we get tan A + tan B1 − tan A tan B = −tan C. Multiply across and rearrange:

Theorem (Sum of tangents in a triangle):
>> If A + B + C = 180°, then tan A + tan B + tan C = tan A tan B tan C

In the same way, cot A cot B + cot B cot C + cot C cot A = 1. These two results solve questions that give the sum of the tangents and ask for their product, or the other way round.

If A + B + C = 90° (three angles that together make a right angle), the roles swap: tan A tan B + tan B tan C + tan C tan A = 1, and cot A + cot B + cot C = cot A cot B cot C.

Sum of double angles. sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C for the angles of a triangle. In an equilateral triangle this is 3 sin 120° = 3√32.

When two tangents are given. If tan A and tan B are known, find tan(A + B) = tan A + tan B1 − tan A tan B. Then tan C = −tan(A + B). For tan A = 12 and tan B = 13 we get tan(A + B) = 1, so A + B = 45° and C = 135°.

Angles in a ratio. If A : B : C = 1 : 2 : 3, then A + B + C = 6 parts = 180°, so one part is 30° and the angles are 30°, 60°, 90°. Now the standard table does the rest.

Radhika has a triangle with A = 40° and B = 50°. Then C = 90°, and sin(A + B) = sin 90° = 1 = sin C, exactly as the rule says.