Chapter 20 — Trigonometry

Suppose a student stands some distance from a tall tree and looks up at its top. The line of sight, the vertical tree and the horizontal ground form a triangle with a right angle at the foot of the tree. If the distance to the tree and the angle at which the student looks up are known, can the height of the tree be found without climbing it? Trigonometry answers yes, and this chapter shows how.

Everything starts with a right-angled triangle. Let triangle ABC have its right angle at B. The side AC opposite the right angle is the hypotenuse, always the longest side. The other two sides are named with reference to one of the acute angles. Take the angle at A, and call it θ (theta). The side BC, which lies across from A, is the side opposite to θ. The side AB, which together with the hypotenuse forms the angle θ, is the side adjacent to θ.

The names change when the angle changes. For the angle at C, the side AB is opposite and BC is adjacent. The hypotenuse is the hypotenuse for either angle. A student should practise this naming until it is automatic, because every ratio in the chapter is defined by it: choose the angle first, then find the opposite side, then the adjacent side.

Since the angles of a triangle add to 180° and one angle is 90°, the two acute angles add to 90°: they are complementary. If ∠A = θ then ∠C = 90° − θ. This fact returns later in the chapter.

Worked example. In triangle PQR right-angled at Q, PQ = 3 cm, QR = 4 cm and PR = 5 cm. For the angle at P: hypotenuse PR = 5, opposite QR = 4, adjacent PQ = 3. For the angle at R: hypotenuse PR = 5, opposite PQ = 3, adjacent QR = 4. Notice that 3² + 4² = 5², as Pythagoras requires.

A note on Greek letters. Angles in trigonometry are usually denoted by Greek letters: θ (theta), α (alpha), β (beta), φ (phi). They are just names, like x and y for numbers.

History. The Indian astronomer Aryabhata (5th century) tabulated half-chords of a circle, which are what we now call sines; the word sine itself descends, through Arabic and Latin, from his Sanskrit term jya. Trigonometry was used in India to predict eclipses and in Europe for navigation, and today it is the basis of surveying, engineering drawing, sound and light waves, and computer graphics.

In a right triangle ABC with the right angle at B and an acute angle θ at A, six ratios of pairs of sides can be formed. Each has a name and an abbreviation.

The first three are the basic ratios; the last three are their reciprocals: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ. A memory aid used by many students is the phrase Some People Have Curly Brown Hair Turned Permanently Black: Sine = Perpendicular/Hypotenuse, Cosine = Base/Hypotenuse, Tangent = Perpendicular/Base, where perpendicular means the opposite side and base the adjacent side.

Important points. (1) sin θ is one symbol, not 'sin' multiplied by θ; sin θ without the angle has no meaning. (2) Because the hypotenuse is the longest side, sin θ and cos θ are always less than 1 for an acute angle, while cosec θ and sec θ are always greater than 1. tan θ and cot θ can take any positive value. (3) (sin θ)² is written sin²θ, and it is not the same as sin θ², which would mean the sine of the angle θ².

Worked example 1. In triangle ABC right-angled at B, AB = 24 cm and BC = 7 cm. Then AC = √(576 + 49) = √625 = 25 cm. For the angle at A: sin A = 7/25, cos A = 24/25, tan A = 7/24, cosec A = 25/7, sec A = 25/24, cot A = 24/7. For the angle at C: sin C = 24/25, cos C = 7/25, tan C = 24/7.

Worked example 2. In triangle PQR right-angled at Q, PR = 13 cm and PQ = 12 cm. QR = √(169 − 144) = 5 cm. sin P = 5/13, cos P = 12/13, tan P = 5/12; sin R = 12/13, cos R = 5/13.

The ratios are pure numbers; the units of the sides cancel. This is why a ratio measured in a small drawing applies to a real tree or tower of the same angle, which the next topic explains.

A trigonometric ratio is defined from a particular right triangle, so it is fair to ask whether a different triangle with the same angle θ would give a different value. It does not, and the reason is similarity.

Draw an acute angle θ with vertex A. On one arm mark points B₁, B₂, B₃ at different distances from A, and from each drop a perpendicular to the other arm, meeting it at C₁, C₂, C₃. This gives three right triangles AB₁C₁, AB₂C₂, AB₃C₃, all sharing the angle θ at A and each having a right angle. By the AA criterion they are all similar. In similar triangles corresponding sides are in the same ratio, so B₁C₁/AB₁ = B₂C₂/AB₂ = B₃C₃/AB₃. Each of these is sin θ (opposite over hypotenuse) computed in a different triangle, and they are all equal. The same argument works for every ratio. Hence the trigonometric ratios of an angle depend only on the angle and not on the size of the triangle used to compute them.

This is exactly what makes trigonometry useful. The ratio can be found once, from any convenient triangle or from a table, and then applied to a triangle of any size, a survey of a mountain or the design of a roof truss. Conversely, if two right triangles have the same value of, say, tan θ, their acute angles are equal.

Worked example 1. A right triangle has legs 3 cm and 4 cm; another has legs 6 cm and 8 cm. For the angle opposite the shorter leg, tan θ = 3/4 in the first triangle and 6/8 = 3/4 in the second; the angles are the same, and sin θ = 3/5 = 6/10 in both.

Worked example 2. A ladder leaning against a wall makes an angle θ with the ground, with its foot 2 m from the wall and its top 4 m up the wall. A second ladder at the same angle has its foot 3 m from the wall; its top is at height 3 × tan θ = 3 × 2 = 6 m. The ratio tan θ = 2 transfers from one triangle to the other.

Worked example 3. A ramp rises 1 m for every 12 m along the ground, so tan θ = 1/12 for its angle. A ramp of the same angle that rises 2.5 m must run 2.5 × 12 = 30 m along the ground.

Because of this independence, sin θ, cos θ and tan θ can be regarded as functions of the angle alone, and their values for particular angles can be tabulated once and for all. The next topics find the values for the angles that arise most often.

If one trigonometric ratio of an acute angle is known, the triangle can be drawn up to similarity, and every other ratio follows. The method is: write the given ratio as a fraction of two sides, choose those sides (or any multiple), find the third side by Pythagoras, then write the remaining ratios.

Worked example 1. Given tan A = 4/3, find sin A and cos A. Take the opposite side 4k and the adjacent side 3k for some positive k. The hypotenuse is √(16k² + 9k²) = 5k. So sin A = 4k/5k = 4/5 and cos A = 3/5. The factor k cancels, which is the similarity argument again; in practice one takes k = 1.

Worked example 2. Given sin θ = 3/4, find cos θ and tan θ. Opposite 3, hypotenuse 4, adjacent √(16 − 9) = √7. cos θ = √7/4, tan θ = 3/√7 = 3√7/7.

Worked example 3. Given cot θ = 7/8, evaluate (1 + sin θ)(1 − sin θ) / ((1 + cos θ)(1 − cos θ)). The expression is (1 − sin²θ)/(1 − cos²θ) = cos²θ/sin²θ = cot²θ = 49/64. This can also be done by drawing the triangle with adjacent 7, opposite 8, hypotenuse √113, but the identity route is shorter.

Worked example 4. Given sec θ = 13/5, find the other five ratios. Hypotenuse 13, adjacent 5, opposite √(169 − 25) = 12. cos θ = 5/13, sin θ = 12/13, tan θ = 12/5, cot θ = 5/12, cosec θ = 13/12.

Worked example 5. If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A. cot A = 4/3, so tan A = 3/4; adjacent 4, opposite 3, hypotenuse 5. Left side: (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25. Right side: 16/25 − 9/25 = 7/25. The statement is true.

Worked example 6. In triangle ABC right-angled at C, if tan A = 1/√3, find sin A cos B + cos A sin B. Opposite 1, adjacent √3, hypotenuse 2: sin A = 1/2, cos A = √3/2. Since B = 90° − A, sin B = cos A = √3/2 and cos B = sin A = 1/2. The expression = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.

Rationalising. When a ratio comes out with a surd in the denominator, such as 3/√7, multiply numerator and denominator by the surd: 3√7/7. Examiners accept either form, but the rationalised form is standard.

The six ratios are not independent. Three relations follow at once from the definitions and are used in almost every problem.

Reciprocal relations. Since cosec θ = hyp/opp and sin θ = opp/hyp, their product is 1: sin θ × cosec θ = 1. Likewise cos θ × sec θ = 1 and tan θ × cot θ = 1. So cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ, and in any expression a reciprocal ratio may be replaced by 1 over the basic one.

Quotient relations. tan θ = opp/adj = (opp/hyp)/(adj/hyp) = sin θ/cos θ. Hence tan θ = sin θ/cos θ and cot θ = cos θ/sin θ. These allow every expression to be written in terms of sine and cosine alone, which is the usual first step in proving an identity.

Worked example 1. If sin θ = 5/13 and cos θ = 12/13, then tan θ = (5/13)/(12/13) = 5/12, cot θ = 12/5, sec θ = 13/12 and cosec θ = 13/5, all without drawing a triangle.

Worked example 2. Simplify sec θ × cot θ. sec θ cot θ = (1/cos θ)(cos θ/sin θ) = 1/sin θ = cosec θ.

Worked example 3. Simplify (sin θ cot θ + cos θ tan θ)/(cosec θ sec θ). Numerator: sin θ (cos θ/sin θ) + cos θ (sin θ/cos θ) = cos θ + sin θ. Denominator: 1/(sin θ cos θ). So the expression equals (cos θ + sin θ) sin θ cos θ.

Worked example 4. Show that tan θ + cot θ = sec θ cosec θ. Left side = sin θ/cos θ + cos θ/sin θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/(sin θ cos θ) = sec θ cosec θ, using sin²θ + cos²θ = 1, proved later in the chapter.

Worked example 5. If tan θ = 1, find sin θ and cos θ. tan θ = 1 means opposite = adjacent, so the triangle is isosceles right-angled; hypotenuse = √2 × leg, sin θ = cos θ = 1/√2. (This is the case θ = 45°.)

Sign and range for acute angles. All six ratios of an acute angle are positive. As θ increases from 0° to 90°, sin θ increases from 0 to 1, cos θ decreases from 1 to 0, and tan θ increases from 0 without bound. These trends are visible in the table of the next topics and are useful for checking answers: for an angle less than 45°, sin θ is less than cos θ and tan θ is less than 1.

The values of the trigonometric ratios for most angles can only be found approximately, but for a few special angles they are exact and easy to derive. The angles 30° and 60° come from the equilateral triangle.

Take an equilateral triangle ABC of side 2a. Each angle is 60°. Draw the altitude AD from A to BC. In an equilateral triangle the altitude bisects the base and the vertical angle, so BD = a and ∠BAD = 30°. Triangle ABD is right-angled at D with ∠ABD = 60° and ∠BAD = 30°. By Pythagoras, AD = √((2a)² − a²) = √(3a²) = a√3.

Ratios of 30° (angle at A in triangle ABD: opposite BD = a, adjacent AD = a√3, hypotenuse AB = 2a): sin 30° = a/2a = 1/2; cos 30° = a√3/2a = √3/2; tan 30° = a/a√3 = 1/√3; cosec 30° = 2; sec 30° = 2/√3; cot 30° = √3.

Ratios of 60° (angle at B: opposite AD = a√3, adjacent BD = a, hypotenuse 2a): sin 60° = √3/2; cos 60° = 1/2; tan 60° = √3; cosec 60° = 2/√3; sec 60° = 2; cot 60° = 1/√3.

Notice that sin 30° = cos 60° and cos 30° = sin 60°, and tan 30° = cot 60°: the ratios of 30° and 60° are exchanged in pairs. This is the complementary-angle relation, since 30° + 60° = 90°.

Worked example 1. Evaluate sin 30° cos 60° + cos 30° sin 60° = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.

Worked example 2. Evaluate 2 tan²30° + sin²60° − cos²60° = 2(1/3) + 3/4 − 1/4 = 2/3 + 1/2 = 7/6.

Worked example 3. A ladder 10 m long leans against a wall making 60° with the ground. Height reached = 10 sin 60° = 10 × √3/2 = 5√3 ≈ 8.66 m; distance of the foot from the wall = 10 cos 60° = 5 m.

Worked example 4. In a right triangle the hypotenuse is 12 cm and one angle is 30°. The side opposite 30° is 12 sin 30° = 6 cm, half the hypotenuse, and the other leg is 12 cos 30° = 6√3 cm. The fact that the side opposite 30° is half the hypotenuse is worth remembering on its own.

45°. In a right triangle with one acute angle 45°, the other is also 45°, so the triangle is isosceles and the two legs are equal, say a each. The hypotenuse is √(a² + a²) = a√2. Hence sin 45° = a/a√2 = 1/√2, cos 45° = 1/√2, tan 45° = a/a = 1, cosec 45° = sec 45° = √2, cot 45° = 1.

0° and 90°. These are not angles of an actual right triangle, but the ratios can be defined by seeing what happens as the angle approaches them. In triangle ABC right-angled at B, let the angle θ at A shrink towards 0°. The opposite side BC becomes smaller and smaller while the hypotenuse AC becomes almost equal to AB. So sin θ = BC/AC tends to 0 and cos θ = AB/AC tends to 1. We define sin 0° = 0, cos 0° = 1, tan 0° = 0; sec 0° = 1, cot 0° and cosec 0° are not defined (they would need division by zero). Now let θ grow towards 90°: AB shrinks to nothing and BC becomes almost the hypotenuse, so sin 90° = 1, cos 90° = 0, cosec 90° = 1, cot 90° = 0, and tan 90° and sec 90° are not defined.

The table.

A way to remember the sine row. Write 0, 1, 2, 3, 4; divide each by 4; take the square root: √0/2 = 0, √1/2 = 1/2, √2/2 = 1/√2, √3/2, √4/2 = 1. The cosine row is the sine row reversed. The tangent row is sine divided by cosine.

Worked example 1. sin 45° cos 45° = (1/√2)(1/√2) = 1/2. tan 45° + cot 45° = 2. sin²45° + cos²45° = 1/2 + 1/2 = 1.

Worked example 2. Evaluate (sin 30° + tan 45° − cosec 60°)/(sec 30° + cos 60° + cot 45°). Numerator = 1/2 + 1 − 2/√3; denominator = 2/√3 + 1/2 + 1. With 2/√3 = 2√3/3: numerator = 3/2 − 2√3/3 = (9 − 4√3)/6, denominator = (9 + 4√3)/6, so the value is (9 − 4√3)/(9 + 4√3). Rationalising gives (9 − 4√3)²/(81 − 48) = (81 − 72√3 + 48)/33 = (129 − 72√3)/33 = (43 − 24√3)/11.

The table must be memorised; almost every question in this chapter and the next uses it.

With the table in hand, numerical expressions in the standard angles can be evaluated exactly, and simple equations involving a ratio of an unknown angle can be solved by reading the table backwards.

Worked example 1. Evaluate 2 tan²45° + cos²30° − sin²60°. tan 45° = 1, cos 30° = sin 60° = √3/2, so the value is 2(1) + 3/4 − 3/4 = 2.

Worked example 2. Evaluate cos 45°/(sec 30° + cosec 30°). cos 45° = 1/√2, sec 30° = 2/√3, cosec 30° = 2. Denominator = 2/√3 + 2 = (2 + 2√3)/√3. Value = (1/√2) × √3/(2 + 2√3) = √3/(2√2(1 + √3)) = √3/(2√2 + 2√6). Rationalising: multiply top and bottom by (√3 − 1): √3(√3 − 1)/(2√2 × 2) = (3 − √3)/(4√2) = (3√2 − √6)/8.

Worked example 3. Evaluate (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + cos²30°). Numerator = 5(1/4) + 4(4/3) − 1 = 5/4 + 16/3 − 1 = (15 + 64 − 12)/12 = 67/12. Denominator = 1. Value = 67/12.

Worked example 4: solving for an angle. If tan(A + B) = √3 and tan(A − B) = 1/√3, with 0° < A + B ≤ 90° and A > B, find A and B. tan 60° = √3 gives A + B = 60°; tan 30° = 1/√3 gives A − B = 30°. Adding, 2A = 90°, A = 45°, and B = 15°.

Worked example 5. If sin(A − B) = 1/2 and cos(A + B) = 1/2, find A and B. A − B = 30° and A + B = 60°, so A = 45°, B = 15°.

Worked example 6. If 2 sin 2θ = √3 with 2θ acute, then sin 2θ = √3/2, 2θ = 60°, θ = 30°.

Worked example 7. If √3 tan θ = 1, then tan θ = 1/√3 and θ = 30°. Then sin 3θ = sin 90° = 1 and cos 2θ = cos 60° = 1/2.

Worked example 8: true or false. Is sin(A + B) = sin A + sin B? Take A = B = 30°: sin 60° = √3/2 ≈ 0.866, but sin 30° + sin 30° = 1. So it is false. Is cos 2A = 2 cos A? At A = 30°: cos 60° = 1/2 but 2 cos 30° = √3. False. These counterexamples show that sin is not a multiplier and that the angle cannot be pulled out of the function.

In all such work write the standard values first, then do the algebra; keep surds exact and rationalise only at the end.

Two angles are complementary if their sum is 90°. In a right triangle ABC with the right angle at B, the acute angles A and C are complementary: C = 90° − A.

Look at the two angles in turn. For angle A, the opposite side is BC and the adjacent side is AB. For angle C, the opposite side is AB and the adjacent side is BC: the roles of the two legs are swapped, while the hypotenuse AC is the same. Therefore sin C = AB/AC = cos A, cos C = BC/AC = sin A, tan C = AB/BC = cot A, and so on. Writing C as 90° − A:

sin(90° − A) = cos A, cos(90° − A) = sin A, tan(90° − A) = cot A, cot(90° − A) = tan A, sec(90° − A) = cosec A, cosec(90° − A) = sec A.

The 'co' in cosine, cotangent and cosecant means 'of the complement': the cosine of an angle is the sine of its complement. The table of standard values shows the rule at work: sin 30° = cos 60°, tan 30° = cot 60°, sec 0° = cosec 90°.

Worked example 1. Evaluate sin 18°/cos 72°. Since 72° = 90° − 18°, cos 72° = sin 18°, and the value is 1.

Worked example 2. Evaluate tan 26°/cot 64°. cot 64° = tan(90° − 64°) = tan 26°, so the value is 1.

Worked example 3. Evaluate cos 48° − sin 42°. sin 42° = cos 48°, so the difference is 0.

Worked example 4. Show that tan 48° tan 23° tan 42° tan 67° = 1. tan 48° = cot 42° and tan 23° = cot 67°, so the product is cot 42° tan 42° × cot 67° tan 67° = 1 × 1 = 1.

Worked example 5. Evaluate sin 25° cos 65° + cos 25° sin 65°. cos 65° = sin 25° and sin 65° = cos 25°, so the expression is sin²25° + cos²25° = 1.

Worked example 6. If tan 2A = cot(A − 18°), where 2A is acute, find A. cot(A − 18°) = tan(90° − (A − 18°)) = tan(108° − A). So 2A = 108° − A, 3A = 108°, A = 36°.

Worked example 7. If sec 4A = cosec(A − 20°), find A. cosec(A − 20°) = sec(90° − A + 20°) = sec(110° − A). So 4A = 110° − A, A = 22°.

Worked example 8. Express sin 67° + cos 75° in terms of ratios of angles between 0° and 45°: sin 67° = cos 23° and cos 75° = sin 15°, so the sum is cos 23° + sin 15°.

The general strategy: wherever two angles in an expression add to 90°, convert one ratio into the co-ratio of the other, and the expression collapses to a standard value or a known identity.

An identity is an equation that is true for every value of the variable for which both sides are defined. The three trigonometric identities of this chapter all come from the Pythagoras theorem.

Identity 1: sin²θ + cos²θ = 1. In triangle ABC right-angled at B with θ at A, Pythagoras gives AB² + BC² = AC². Divide every term by AC²: (AB/AC)² + (BC/AC)² = 1, that is cos²θ + sin²θ = 1. This holds for every acute θ, and also for θ = 0° and 90° (1 + 0 = 1). It gives cos θ = √(1 − sin²θ) and sin θ = √(1 − cos²θ) for acute angles.

Identity 2: 1 + tan²θ = sec²θ. Divide AB² + BC² = AC² by AB²: 1 + (BC/AB)² = (AC/AB)², that is 1 + tan²θ = sec²θ. It fails to be defined at θ = 90°, where tan and sec are undefined. Equivalently sec²θ − tan²θ = 1, which factorises as (sec θ − tan θ)(sec θ + tan θ) = 1.

Identity 3: 1 + cot²θ = cosec²θ. Divide by BC²: (AB/BC)² + 1 = (AC/BC)², that is cot²θ + 1 = cosec²θ, defined for 0° < θ ≤ 90°. Equivalently (cosec θ − cot θ)(cosec θ + cot θ) = 1.

Worked example 1. Express sin θ, tan θ and sec θ in terms of cos θ. sin θ = √(1 − cos²θ); tan θ = √(1 − cos²θ)/cos θ; sec θ = 1/cos θ.

Worked example 2. Express all ratios in terms of tan θ. sec θ = √(1 + tan²θ); cos θ = 1/√(1 + tan²θ); sin θ = tan θ cos θ = tan θ/√(1 + tan²θ); cot θ = 1/tan θ; cosec θ = √(1 + tan²θ)/tan θ.

Worked example 3. If sec θ + tan θ = p, find sec θ − tan θ. Since (sec θ + tan θ)(sec θ − tan θ) = 1, sec θ − tan θ = 1/p. Adding and subtracting: sec θ = (p + 1/p)/2 = (p² + 1)/2p and tan θ = (p² − 1)/2p, so sin θ = tan θ/sec θ = (p² − 1)/(p² + 1).

Worked example 4. If cosec θ − cot θ = 1/3, then cosec θ + cot θ = 3, so cosec θ = 5/3 and cot θ = 4/3; sin θ = 3/5, cos θ = 4/5.

Worked example 5. If sin θ + cos θ = √2, find sin θ cos θ. Squaring: sin²θ + cos²θ + 2 sin θ cos θ = 2, so 1 + 2 sin θ cos θ = 2 and sin θ cos θ = 1/2. (This is the case θ = 45°.)

Worked example 6. Simplify (1 − cos²θ) sec²θ = sin²θ/cos²θ = tan²θ. And (1 + tan²θ)(1 − sin θ)(1 + sin θ) = sec²θ (1 − sin²θ) = sec²θ cos²θ = 1.

These three identities, with the reciprocal and quotient relations, are the complete toolkit for the proofs in the next topic.

To prove an identity means to show that the left-hand side (LHS) and the right-hand side (RHS) are equal for every admissible angle. The accepted methods are: start from one side (usually the more complicated) and transform it step by step into the other; or reduce both sides separately to the same expression. It is not acceptable to assume the identity, cross-multiply and arrive at a true statement, because that proves nothing.

Standard moves. Convert everything to sin and cos. Replace 1 by sin²θ + cos²θ where it helps. Use a² − b² = (a − b)(a + b) on 1 − sin²θ, sec²θ − 1, cosec²θ − 1. Multiply numerator and denominator by a conjugate such as 1 + sin θ or sec θ + tan θ. Take a common denominator.

Proof 1. (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ). LHS = (1/sin θ − cos θ/sin θ)² = (1 − cos θ)²/sin²θ = (1 − cos θ)²/(1 − cos²θ) = (1 − cos θ)²/((1 − cos θ)(1 + cos θ)) = (1 − cos θ)/(1 + cos θ) = RHS.

Proof 2. cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A. LHS = (cos²A + (1 + sin A)²)/((1 + sin A) cos A) = (cos²A + 1 + 2 sin A + sin²A)/((1 + sin A) cos A) = (2 + 2 sin A)/((1 + sin A) cos A) = 2/cos A = 2 sec A.

Proof 3. tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ. Write t = tan θ, so cot θ = 1/t. LHS = t/(1 − 1/t) + (1/t)/(1 − t) = t²/(t − 1) + 1/(t(1 − t)) = t²/(t − 1) − 1/(t(t − 1)) = (t³ − 1)/(t(t − 1)) = (t − 1)(t² + t + 1)/(t(t − 1)) = (t² + t + 1)/t = t + 1 + 1/t = tan θ + cot θ + 1 = sec θ cosec θ + 1, using tan θ + cot θ = sec θ cosec θ.

Proof 4. (sin θ − 2 sin³θ)/(2 cos³θ − cos θ) = tan θ. LHS = sin θ(1 − 2 sin²θ)/(cos θ(2 cos²θ − 1)). Now 1 − 2 sin²θ = 1 − 2(1 − cos²θ) = 2 cos²θ − 1, so LHS = sin θ/cos θ = tan θ.

Proof 5. √((1 + sin A)/(1 − sin A)) = sec A + tan A. Multiply inside the root by (1 + sin A)/(1 + sin A): √((1 + sin A)²/(1 − sin²A)) = √((1 + sin A)²/cos²A) = (1 + sin A)/cos A = sec A + tan A.

Proof 6. (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A. Expand: sin²A + 2 + cosec²A + cos²A + 2 + sec²A = (sin²A + cos²A) + 4 + (1 + cot²A) + (1 + tan²A) = 1 + 4 + 2 + tan²A + cot²A = 7 + tan²A + cot²A.

Proof 7. (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A). LHS = (1/sin A − sin A)(1/cos A − cos A) = ((1 − sin²A)/sin A)((1 − cos²A)/cos A) = (cos²A/sin A)(sin²A/cos A) = sin A cos A. RHS = 1/((sin²A + cos²A)/(sin A cos A)) = sin A cos A. Both sides equal sin A cos A.

Write 'LHS =' at the start, one transformation per line with the identity used named in brackets, and end with '= RHS'. That layout is what the examiner marks.

Definitions. In a right triangle with acute angle θ: sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj, and cosec, sec, cot are their reciprocals. The ratios depend only on θ, by the similarity of right triangles with a common acute angle.

Relations. tan θ = sin θ/cos θ, cot θ = cos θ/sin θ; sin θ cosec θ = cos θ sec θ = tan θ cot θ = 1.

Standard values. sin: 0, 1/2, 1/√2, √3/2, 1 and cos: 1, √3/2, 1/√2, 1/2, 0 for 0°, 30°, 45°, 60°, 90°; tan: 0, 1/√3, 1, √3, undefined.

Complementary angles. sin(90° − θ) = cos θ, tan(90° − θ) = cot θ, sec(90° − θ) = cosec θ, and the three converse forms.

Identities. sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.

How the board asks. One-mark questions: the value of a standard ratio, whether a statement like sin θ = 5/3 is possible (it is not, since sin θ ≤ 1), the value of an expression like tan 36°/cot 54°, or which identity is being used. Two-mark questions: given one ratio, find the others; evaluate an expression in standard values; find A and B from two conditions. Four-mark questions: prove an identity, typically one of the seven proved above or a close relative; or a combined problem, such as: if sec θ + tan θ = p, find sin θ. Occasionally the board asks for the derivation of the standard values of 30° and 60° from the equilateral triangle, or for the proof of sin²θ + cos²θ = 1 from Pythagoras.

Common errors. Confusing opposite and adjacent by not first fixing the angle. Writing sin θ² for sin²θ. Treating sin(A + B) as sin A + sin B. Forgetting that tan 90° is undefined. Cross-multiplying to 'prove' an identity. Leaving a surd in the denominator when the question asks for a simplified answer. Dropping the square root when using cos θ = √(1 − sin²θ). Using degrees where a value is a ratio: cos 60° is 1/2, not 60.

Checking. A value of sin or cos greater than 1 is impossible. For an angle less than 45°, sin is less than cos. An identity can be tested numerically at θ = 30° before attempting the proof; if the two sides differ at 30°, the identity is misread or misprinted.

What lies ahead. The next chapter applies these ratios to heights and distances with angles of elevation and depression. The Intermediate course extends the ratios to all angles using the unit circle, introduces radian measure, the compound-angle formulas sin(A + B) = sin A cos B + cos A sin B, and the graphs of the functions, and uses trigonometry in calculus and in physics for waves, oscillations and vectors.