Chapter 21 — Trigonometry Optional

Every problem in the optional exercise is solved with the same small toolkit: the six definitions, the reciprocal relations (sin θ cosec θ = 1 and its two companions), the quotient relations (tan θ = sin θ/cos θ, cot θ = cos θ/sin θ), the three identities (sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ), the standard values and the complementary relations. The exercise adds no new fact. What it adds is judgement: which tool to reach for first, and how to lay out an argument of six or eight lines so that each step is justified.

The four families of problems. (1) Prove an identity: transform the more complicated side until it becomes the other side, or reduce both sides to a common expression. (2) Given one relation, find another: for instance, sec θ + tan θ = p gives sec θ − tan θ = 1/p by the identity sec²θ − tan²θ = 1, and then any ratio follows. (3) Eliminate θ: two equations such as x = a sec θ + b tan θ and y = a tan θ + b sec θ are squared and combined so that the identity removes θ, leaving a relation in x, y, a, b. (4) Express in terms of one ratio: using the identities to write, say, cos θ, tan θ and sec θ in terms of sin θ.

The first move. When in doubt, write every ratio in terms of sin θ and cos θ. Products and fractions then simplify by ordinary algebra, and 1 can be replaced by sin²θ + cos²θ wherever a 1 stands alone. This method always works, though it is not always the shortest.

The shortest moves. Recognise a difference of squares: 1 − sin²θ, sec²θ − 1, cosec²θ − 1, 1 − cos²θ. Recognise a conjugate pair: (1 + sin θ)(1 − sin θ) = cos²θ, (sec θ + tan θ)(sec θ − tan θ) = 1, (cosec θ + cot θ)(cosec θ − cot θ) = 1. Recognise tan θ + cot θ = sec θ cosec θ and sin θ cos θ = 1/(tan θ + cot θ). Recognise that a 1 in a numerator can be replaced by sec²θ − tan²θ or cosec²θ − cot²θ to make a factor appear.

What is not allowed. A proof that starts from the identity to be proved, cross-multiplies and reaches a true statement is not a proof; the examiner gives it no credit. Start from one side and end at the other.

A check that saves marks. Before proving an identity, test it at θ = 30° or 45°. If the two sides give different numbers, the question has been misread, and it is better to know that at once.

The topics that follow take the families in turn, state the method, and work the standard problems of the exercise in full.

An expression like (cosec θ − sin θ)(sec θ − cos θ) looks awkward but becomes simple once each bracket is written over a single denominator.

Proof 1. (cosec θ − sin θ)(sec θ − cos θ) = 1/(tan θ + cot θ).

LHS: cosec θ − sin θ = 1/sin θ − sin θ = (1 − sin²θ)/sin θ = cos²θ/sin θ. Similarly sec θ − cos θ = (1 − cos²θ)/cos θ = sin²θ/cos θ. The product is (cos²θ/sin θ)(sin²θ/cos θ) = sin θ cos θ.

RHS: tan θ + cot θ = sin θ/cos θ + cos θ/sin θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/(sin θ cos θ). So 1/(tan θ + cot θ) = sin θ cos θ. Both sides equal sin θ cos θ; the identity is proved.

Proof 2. (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ. Expand each square: sin²θ + 2 sin θ cosec θ + cosec²θ + cos²θ + 2 cos θ sec θ + sec²θ. Now sin θ cosec θ = 1 and cos θ sec θ = 1, so the middle terms give 2 + 2 = 4. Group: (sin²θ + cos²θ) + 4 + cosec²θ + sec²θ = 1 + 4 + (1 + cot²θ) + (1 + tan²θ) = 7 + tan²θ + cot²θ.

Proof 3. (1 + cot θ − cosec θ)(1 + tan θ + sec θ) = 2. Convert: (1 + cos θ/sin θ − 1/sin θ)(1 + sin θ/cos θ + 1/cos θ) = ((sin θ + cos θ − 1)/sin θ)((cos θ + sin θ + 1)/cos θ). The numerators multiply as ((sin θ + cos θ) − 1)((sin θ + cos θ) + 1) = (sin θ + cos θ)² − 1 = sin²θ + 2 sin θ cos θ + cos²θ − 1 = 2 sin θ cos θ. Dividing by sin θ cos θ gives 2.

Proof 4. (sec θ − cos θ)(cosec θ − sin θ)(tan θ + cot θ) = 1. From Proof 1 the first two factors multiply to sin θ cos θ, and tan θ + cot θ = 1/(sin θ cos θ), so the product is 1.

Proof 5. (1 + tan²A)/(1 + cot²A) = tan²A. LHS = sec²A/cosec²A = (1/cos²A)/(1/sin²A) = sin²A/cos²A = tan²A. The exercise also asks to show this equals ((1 − tan A)/(1 − cot A))²: since 1 − cot A = 1 − 1/tan A = (tan A − 1)/tan A, the fraction (1 − tan A)/(1 − cot A) = (1 − tan A) tan A/(tan A − 1) = −tan A, and its square is tan²A.

The pattern in all five: bring each bracket to a single fraction, cancel, and use sin²θ + cos²θ = 1 where a 1 appears with a square.

Identities in which the two sides are sums of fractions are handled by taking a common denominator, and identities in which 1 − tan θ or 1 − cot θ appears are often quickest with the substitution t = tan θ, cot θ = 1/t.

Proof 1. tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ. Put t = tan θ. Then 1 − cot θ = 1 − 1/t = (t − 1)/t, so the first fraction is t × t/(t − 1) = t²/(t − 1). The second is (1/t)/(1 − t) = 1/(t(1 − t)) = −1/(t(t − 1)). Sum = (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 2. (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A. Divide numerator and denominator by sin A: (cot A − 1 + cosec A)/(cot A + 1 − cosec A). In the numerator replace 1 by cosec²A − cot²A: cot A + cosec A − (cosec A − cot A)(cosec A + cot A) = (cosec A + cot A)(1 − cosec A + cot A) = (cosec A + cot A)(cot A + 1 − cosec A). The second factor is exactly the denominator, so the fraction equals cosec A + cot A.

Proof 3. 1/(sec θ − tan θ) − 1/cos θ = 1/cos θ − 1/(sec θ + tan θ). Since (sec θ − tan θ)(sec θ + tan θ) = 1, 1/(sec θ − tan θ) = sec θ + tan θ and 1/(sec θ + tan θ) = sec θ − tan θ. LHS = sec θ + tan θ − sec θ = tan θ. RHS = sec θ − (sec θ − tan θ) = tan θ. Equal.

Proof 4. (1 + sec A)/sec A = sin²A/(1 − cos A). LHS = (1 + 1/cos A)/(1/cos A) = cos A + 1. RHS = (1 − cos²A)/(1 − cos A) = (1 − cos A)(1 + cos A)/(1 − cos A) = 1 + cos A. Equal.

Proof 5. sin θ/(1 + cos θ) + (1 + cos θ)/sin θ = 2 cosec θ. Common denominator: (sin²θ + (1 + cos θ)²)/((1 + cos θ) sin θ) = (sin²θ + 1 + 2 cos θ + cos²θ)/((1 + cos θ) sin θ) = (2 + 2 cos θ)/((1 + cos θ) sin θ) = 2/sin θ = 2 cosec θ.

Proof 6. 1/(1 + sin θ) + 1/(1 − sin θ) = 2 sec²θ. Common denominator: ((1 − sin θ) + (1 + sin θ))/(1 − sin²θ) = 2/cos²θ = 2 sec²θ.

Proof 7. (tan θ + sec θ − 1)/(tan θ − sec θ + 1) = (1 + sin θ)/cos θ. Replace the 1 in the numerator by sec²θ − tan²θ: tan θ + sec θ − (sec θ − tan θ)(sec θ + tan θ) = (sec θ + tan θ)(1 − sec θ + tan θ), and the second factor is the denominator. So the fraction equals sec θ + tan θ = (1 + sin θ)/cos θ.

Proofs 2 and 7 use the same trick, replacing a lone 1 by a difference of squares so that the denominator appears as a factor of the numerator; it is worth practising until it is recognised at sight.

When an identity has a square root of a fraction such as (1 + sin A)/(1 − sin A), the method is to multiply the numerator and denominator inside the root by the conjugate of the denominator, so that the denominator becomes a perfect square by an identity and the root can be taken.

Proof 1. √((1 + sin A)/(1 − sin A)) = sec A + tan A. Multiply inside by (1 + sin A)/(1 + sin A): √((1 + sin A)²/(1 − sin²A)) = √((1 + sin A)²/cos²A) = (1 + sin A)/cos A = 1/cos A + sin A/cos A = sec A + tan A. (For acute A all quantities are positive, so the positive root is correct.)

Proof 2. √((1 − cos θ)/(1 + cos θ)) = cosec θ − cot θ. Multiply inside by (1 − cos θ)/(1 − cos θ): √((1 − cos θ)²/(1 − cos²θ)) = (1 − cos θ)/sin θ = cosec θ − cot θ.

Proof 3. √((sec θ − 1)/(sec θ + 1)) + √((sec θ + 1)/(sec θ − 1)) = 2 cosec θ. Write both over the common root: the sum is ((sec θ − 1) + (sec θ + 1))/√((sec θ − 1)(sec θ + 1)) = 2 sec θ/√(sec²θ − 1) = 2 sec θ/tan θ = 2 (1/cos θ)(cos θ/sin θ) = 2/sin θ = 2 cosec θ.

Proof 4. √((1 + cos A)/(1 − cos A)) = (1 + cos A)/sin A. Multiply inside by (1 + cos A)/(1 + cos A): √((1 + cos A)²/(1 − cos²A)) = (1 + cos A)/sin A.

Proof 5. (sec A − tan A)² = (1 − sin A)/(1 + sin A). LHS = (1/cos A − sin A/cos A)² = (1 − sin A)²/cos²A = (1 − sin A)²/((1 − sin A)(1 + sin A)) = (1 − sin A)/(1 + sin A). This is the square of Proof 1's reciprocal, and shows the same conjugate at work without a root.

Proof 6. (cosec A − cot A)² = (1 − cos A)/(1 + cos A), proved in the main chapter the same way; and its companion (cosec A + cot A)² = (1 + cos A)/(1 − cos A).

A note on signs. For an acute angle every ratio is positive, so the square root of a square of a positive quantity is that quantity. In the Intermediate course, where angles beyond 90° are allowed, the sign must be checked; here it never needs to be.

Worked numerical check. At A = 30°: √((1 + 1/2)/(1 − 1/2)) = √3 ≈ 1.732, and sec 30° + tan 30° = 2/√3 + 1/√3 = 3/√3 = √3. Proof 1 checks.

Some identities are awkward to prove by transforming one side into the other, because both sides are complicated. The accepted alternative is to simplify each side separately until they reach the same expression, and then to state that the two sides are equal. This is a valid proof, provided the two chains of reasoning are kept apart and no step assumes what is to be proved.

Proof 1. tan θ/(sec θ − 1) + tan θ/(sec θ + 1) = 2 cosec θ. LHS with a common denominator: tan θ((sec θ + 1) + (sec θ − 1))/(sec²θ − 1) = 2 sec θ tan θ/tan²θ = 2 sec θ/tan θ = 2 (1/cos θ)(cos θ/sin θ) = 2/sin θ = 2 cosec θ. Here one side sufficed, but notice the factor sec²θ − 1 = tan²θ appearing from the conjugate denominators.

Proof 2. (sin θ + cos θ)/(sin θ − cos θ) + (sin θ − cos θ)/(sin θ + cos θ) = 2/(1 − 2 cos²θ). LHS = ((sin θ + cos θ)² + (sin θ − cos θ)²)/(sin²θ − cos²θ) = (2 sin²θ + 2 cos²θ)/(sin²θ − cos²θ) = 2/(sin²θ − cos²θ). RHS: 1 − 2 cos²θ = sin²θ + cos²θ − 2 cos²θ = sin²θ − cos²θ, so RHS = 2/(sin²θ − cos²θ). Both sides equal 2/(sin²θ − cos²θ); hence LHS = RHS.

Proof 3. sin θ/(cot θ + cosec θ) − sin θ/(cot θ − cosec θ) = 2. Since cot θ + cosec θ = (cos θ + 1)/sin θ, the first term is sin²θ/(1 + cos θ) = (1 − cos²θ)/(1 + cos θ) = 1 − cos θ. Since cot θ − cosec θ = (cos θ − 1)/sin θ, the second term is sin²θ/(cos θ − 1) = −(1 − cos²θ)/(1 − cos θ) = −(1 + cos θ). LHS = (1 − cos θ) − (−(1 + cos θ)) = 2 = RHS.

Proof 4. (sin A − sin B)/(cos A + cos B) + (cos A − cos B)/(sin A + sin B) = 0. Common denominator: ((sin A − sin B)(sin A + sin B) + (cos A − cos B)(cos A + cos B))/((cos A + cos B)(sin A + sin B)) = (sin²A − sin²B + cos²A − cos²B)/(…) = ((sin²A + cos²A) − (sin²B + cos²B))/(…) = (1 − 1)/(…) = 0. This is an identity in two angles, and it is proved with nothing more than sin²θ + cos²θ = 1 applied twice.

Proof 5. (cot A + cosec A − 1)/(cot A − cosec A + 1) = (1 + cos A)/sin A. Replace the 1 in the numerator by cosec²A − cot²A: numerator = (cosec A + cot A) − (cosec A − cot A)(cosec A + cot A) = (cosec A + cot A)(1 − cosec A + cot A) = (cosec A + cot A)(cot A − cosec A + 1). The second factor is the denominator, so LHS = cosec A + cot A = 1/sin A + cos A/sin A = (1 + cos A)/sin A = RHS.

Layout for a two-sided proof. Write 'LHS = … = E' as one block, then 'RHS = … = E' as a second block, then the single line 'Since LHS = E = RHS, the identity holds.' Examiners award full marks for this form; they do not award marks for a proof that writes LHS = RHS at the top and works downwards on both sides at once, because that form assumes the conclusion.

Numerical check. Proof 2 at θ = 30°: LHS = ((1/2 + √3/2)/(1/2 − √3/2)) + ((1/2 − √3/2)/(1/2 + √3/2)) = (1 + √3)/(1 − √3) + (1 − √3)/(1 + √3) = ((1 + √3)² + (1 − √3)²)/(1 − 3) = 8/(−2) = −4; RHS = 2/(1 − 2 × 3/4) = 2/(−1/2) = −4. They agree.

Higher powers of the ratios yield to factorisation. The useful factorisations are a³ ± b³ = (a ± b)(a² ∓ ab + b²), a⁴ − b⁴ = (a² − b²)(a² + b²), and taking out a common square factor.

Proof 1. (sin θ − 2 sin³θ)/(2 cos³θ − cos θ) = tan θ. Factorise: sin θ(1 − 2 sin²θ)/(cos θ(2 cos²θ − 1)). Now 1 − 2 sin²θ = 1 − 2(1 − cos²θ) = 2 cos²θ − 1, so the brackets cancel and the fraction is sin θ/cos θ = tan θ.

Proof 2. sin³θ + cos³θ = (sin θ + cos θ)(1 − sin θ cos θ). Using a³ + b³: (sin θ + cos θ)(sin²θ − sin θ cos θ + cos²θ) = (sin θ + cos θ)(1 − sin θ cos θ). Hence (sin³θ + cos³θ)/(sin θ + cos θ) + sin θ cos θ = 1.

Proof 3. sec⁴θ − sec²θ = tan⁴θ + tan²θ. LHS = sec²θ(sec²θ − 1) = sec²θ tan²θ = (1 + tan²θ) tan²θ = tan²θ + tan⁴θ.

Proof 4. sin⁴θ − cos⁴θ = sin²θ − cos²θ = 1 − 2 cos²θ = 2 sin²θ − 1. The first step uses a⁴ − b⁴ = (a² − b²)(a² + b²) with a² + b² = 1; the others use sin²θ = 1 − cos²θ.

Proof 5. sin⁶θ + cos⁶θ = 1 − 3 sin²θ cos²θ. Write as (sin²θ)³ + (cos²θ)³ = (sin²θ + cos²θ)(sin⁴θ − sin²θ cos²θ + cos⁴θ) = sin⁴θ + cos⁴θ − sin²θ cos²θ. And sin⁴θ + cos⁴θ = (sin²θ + cos²θ)² − 2 sin²θ cos²θ = 1 − 2 sin²θ cos²θ. So the total is 1 − 3 sin²θ cos²θ.

Proof 6. cosec⁴θ − cosec²θ = cot⁴θ + cot²θ, by the same steps as Proof 3 with cosec and cot.

Proof 7. (sin A + cos A)² + (sin A − cos A)² = 2. Expand: (1 + 2 sin A cos A) + (1 − 2 sin A cos A) = 2. The companion (sin A + cos A)² − (sin A − cos A)² = 4 sin A cos A follows the same way.

Worked numerical check. At θ = 30° for Proof 5: sin⁶30° + cos⁶30° = 1/64 + 27/64 = 28/64 = 7/16; and 1 − 3(1/4)(3/4) = 1 − 9/16 = 7/16. The identity checks.

The habit to build is to look at the highest power and ask which factorisation reduces it: a common factor for a⁴ − a², the sum-of-cubes formula for a³ + b³, the difference of squares for a⁴ − b⁴.

A relation such as sec θ + tan θ = p is one equation in two unknown ratios, but the identity sec²θ − tan²θ = 1 supplies a second equation, and then everything can be found in terms of p.

Worked example 1. If sec θ + tan θ = p, find sin θ. Since (sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ = 1, sec θ − tan θ = 1/p. Adding: 2 sec θ = p + 1/p = (p² + 1)/p, so sec θ = (p² + 1)/(2p) and cos θ = 2p/(p² + 1). Subtracting: 2 tan θ = (p² − 1)/p, tan θ = (p² − 1)/(2p). Then sin θ = tan θ/sec θ = (p² − 1)/(p² + 1). Check with p = 2: sin θ = 3/5, cos θ = 4/5, and indeed sec θ + tan θ = 5/4 + 3/4 = 2.

Worked example 2. If cosec θ + cot θ = k, show that cos θ = (k² − 1)/(k² + 1). cosec θ − cot θ = 1/k. So cosec θ = (k² + 1)/(2k), cot θ = (k² − 1)/(2k), and cos θ = cot θ/cosec θ = (k² − 1)/(k² + 1). Also sin θ = 2k/(k² + 1).

Worked example 3. If cosec θ − cot θ = 1/3, find all six ratios. cosec θ + cot θ = 3. So cosec θ = 5/3, cot θ = 4/3, sin θ = 3/5, tan θ = 3/4, cos θ = 4/5, sec θ = 5/4.

Worked example 4. If sin θ + cos θ = p and sec θ + cosec θ = q, show that q(p² − 1) = 2p. q = 1/cos θ + 1/sin θ = (sin θ + cos θ)/(sin θ cos θ) = p/(sin θ cos θ). Squaring p: p² = 1 + 2 sin θ cos θ, so p² − 1 = 2 sin θ cos θ. Hence q(p² − 1) = (p/(sin θ cos θ)) × 2 sin θ cos θ = 2p.

Worked example 5. If tan θ + sin θ = m and tan θ − sin θ = n, show that m² − n² = 4√(mn). m² − n² = (m + n)(m − n) = (2 tan θ)(2 sin θ) = 4 tan θ sin θ. And mn = tan²θ − sin²θ = sin²θ/cos²θ − sin²θ = sin²θ(1 − cos²θ)/cos²θ = sin⁴θ/cos²θ, so √(mn) = sin²θ/cos θ = sin θ tan θ. Hence 4√(mn) = 4 sin θ tan θ = m² − n².

Worked example 6. If sin θ + cos θ = √2, find sin θ cos θ and tan θ + cot θ. Squaring: 1 + 2 sin θ cos θ = 2, sin θ cos θ = 1/2. tan θ + cot θ = 1/(sin θ cos θ) = 2. (θ is 45°.)

Worked example 7. If sec θ = x + 1/(4x), show that sec θ + tan θ = 2x or 1/(2x). tan²θ = sec²θ − 1 = x² + 1/2 + 1/(16x²) − 1 = x² − 1/2 + 1/(16x²) = (x − 1/(4x))², so tan θ = ±(x − 1/(4x)). Then sec θ + tan θ = 2x (with the plus sign) or 1/(2x) (with the minus sign).

The method is always: pair the given relation with the identity that links the same two ratios, solve the pair by adding and subtracting, then compute whatever is asked.

When two quantities x and y are each expressed in terms of θ, a relation between x and y that does not involve θ can often be found by squaring and adding or subtracting so that an identity absorbs θ. This is called eliminating θ, and it is the algebraic form of the fact that the identities hold for all θ.

Worked example 1. If x = a cos θ and y = b sin θ, then x/a = cos θ and y/b = sin θ, so x²/a² + y²/b² = cos²θ + sin²θ = 1.

Worked example 2. If x = a sec θ and y = b tan θ, then x²/a² − y²/b² = sec²θ − tan²θ = 1.

Worked example 3. If x = a sec θ + b tan θ and y = a tan θ + b sec θ, show that x² − y² = a² − b². x² = a² sec²θ + 2ab sec θ tan θ + b² tan²θ; y² = a² tan²θ + 2ab sec θ tan θ + b² sec²θ. Subtracting, the middle terms cancel: x² − y² = a²(sec²θ − tan²θ) − b²(sec²θ − tan²θ) = a² − b².

Worked example 4. If a cos θ + b sin θ = m and a sin θ − b cos θ = n, show that a² + b² = m² + n². m² + n² = a² cos²θ + 2ab sin θ cos θ + b² sin²θ + a² sin²θ − 2ab sin θ cos θ + b² cos²θ = a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ) = a² + b².

Worked example 5. If x = r sin θ cos φ, y = r sin θ sin φ and z = r cos θ, then x² + y² + z² = r² sin²θ(cos²φ + sin²φ) + r² cos²θ = r² sin²θ + r² cos²θ = r². This is the relation between spherical and rectangular coordinates, met later.

Worked example 6. If x = a cos³θ and y = b sin³θ, then (x/a)^(2/3) + (y/b)^(2/3) = cos²θ + sin²θ = 1.

Worked example 7. If x sin³θ + y cos³θ = sin θ cos θ and x sin θ = y cos θ, show that x² + y² = 1. From the second equation, x = y cos θ/sin θ. Substitute in the first: y cos θ sin²θ + y cos³θ = sin θ cos θ, so y cos θ(sin²θ + cos²θ) = sin θ cos θ, giving y = sin θ. Then x = sin θ cos θ/sin θ = cos θ. Hence x² + y² = cos²θ + sin²θ = 1.

Worked example 8. If tan θ + sec θ = x, then sec θ − tan θ = 1/x, so sec θ = (x + 1/x)/2 and tan θ = (x − 1/x)/2; here θ is eliminated by expressing each ratio in x, which is the previous topic seen from this angle.

The signals to watch for: a cos and a sin with the same coefficients suggest squaring and adding; a sec and a tan suggest squaring and subtracting; cubes suggest a two-thirds power. Always check the final relation at a convenient θ, such as θ = 0° or 45°.

Since the six ratios are linked by the reciprocal, quotient and Pythagorean relations, any one of them determines the other five for an acute angle. The exercise asks for these expressions, and the same skill is used whenever a problem gives a single ratio.

In terms of sin θ. cos θ = √(1 − sin²θ); tan θ = sin θ/√(1 − sin²θ); cot θ = √(1 − sin²θ)/sin θ; sec θ = 1/√(1 − sin²θ); cosec θ = 1/sin θ.

In terms of cos θ. sin θ = √(1 − cos²θ); tan θ = √(1 − cos²θ)/cos θ; cot θ = cos θ/√(1 − cos²θ); sec θ = 1/cos θ; cosec θ = 1/√(1 − cos²θ).

In terms of tan θ. sec θ = √(1 + tan²θ); cos θ = 1/√(1 + tan²θ); sin θ = tan θ/√(1 + tan²θ); cot θ = 1/tan θ; cosec θ = √(1 + tan²θ)/tan θ.

In terms of sec θ. cos θ = 1/sec θ; tan θ = √(sec²θ − 1); sin θ = √(sec²θ − 1)/sec θ; cot θ = 1/√(sec²θ − 1); cosec θ = sec θ/√(sec²θ − 1).

In terms of cot θ. tan θ = 1/cot θ; cosec θ = √(1 + cot²θ); sin θ = 1/√(1 + cot²θ); cos θ = cot θ/√(1 + cot²θ); sec θ = √(1 + cot²θ)/cot θ.

In terms of cosec θ. sin θ = 1/cosec θ; cot θ = √(cosec²θ − 1); cos θ = √(cosec²θ − 1)/cosec θ; tan θ = 1/√(cosec²θ − 1); sec θ = cosec θ/√(cosec²θ − 1).

There is no need to memorise these tables; each entry is obtained in one or two steps from the identities. The triangle method gives the same results: for 'in terms of tan θ' take opposite = tan θ, adjacent = 1, hypotenuse = √(1 + tan²θ), and read the ratios off.

Worked example 1. Write sin θ, tan θ and cosec θ in terms of cos θ, then check at θ = 60°: sin 60° = √(1 − 1/4) = √3/2 ✓; tan 60° = (√3/2)/(1/2) = √3 ✓; cosec 60° = 1/(√3/2) = 2/√3 ✓.

Worked example 2. If tan θ = 1/√7, find (cosec²θ − sec²θ)/(cosec²θ + sec²θ). cot²θ = 7, so cosec²θ = 1 + 7 = 8; tan²θ = 1/7, so sec²θ = 8/7. The expression = (8 − 8/7)/(8 + 8/7) = (48/7)/(64/7) = 3/4.

Worked example 3. If cos θ = 4/5, find (2 tan θ)/(1 + tan²θ) and compare with 2 sin θ cos θ. sin θ = 3/5, tan θ = 3/4; (2 × 3/4)/(1 + 9/16) = (3/2)/(25/16) = 24/25; and 2 sin θ cos θ = 2 × 3/5 × 4/5 = 24/25. Equal, which is the identity 2 tan θ/(1 + tan²θ) = sin 2θ met in the next course.

Worked example 4. Express sin θ in terms of cot θ and verify at 45°: sin θ = 1/√(1 + cot²θ) = 1/√2 ✓.

Some problems in the exercise give a condition on the angle or on a ratio and ask for a value, and some ask the student to verify a statement at a particular angle. Both use the standard values and the identities.

Worked example 1. If A, B and C are the angles of a triangle, show that sin((B + C)/2) = cos(A/2). Since A + B + C = 180°, B + C = 180° − A, so (B + C)/2 = 90° − A/2, and sin(90° − A/2) = cos(A/2). Similarly tan((B + C)/2) = cot(A/2).

Worked example 2. If cos(40° + A) = sin 30°, find A. sin 30° = cos 60°, so 40° + A = 60°, A = 20°.

Worked example 3. If sec 2A = cosec(A − 42°) with 2A acute, find A. cosec(A − 42°) = sec(90° − A + 42°) = sec(132° − A). So 2A = 132° − A, A = 44°.

Worked example 4. Verify cos 2θ = 2 cos²θ − 1 at θ = 30°. cos 60° = 1/2; 2 cos²30° − 1 = 2(3/4) − 1 = 1/2. Verified. And sin 3θ = 3 sin θ − 4 sin³θ at θ = 30°: sin 90° = 1; 3(1/2) − 4(1/8) = 3/2 − 1/2 = 1. Verified.

Worked example 5. Verify tan 2θ = 2 tan θ/(1 − tan²θ) at θ = 30°. tan 60° = √3; 2(1/√3)/(1 − 1/3) = (2/√3)/(2/3) = 3/√3 = √3. Verified.

Worked example 6. If θ is acute and 2 sin²θ − 1 = 0, find θ and hence tan θ + cot θ. sin²θ = 1/2, sin θ = 1/√2, θ = 45°, tan θ + cot θ = 2.

Worked example 7. If 4 cos²θ − 3 = 0 for acute θ, then cos θ = √3/2, θ = 30°, and sec θ + cosec θ = 2/√3 + 2 = (2 + 2√3)/√3.

Worked example 8. If sin θ = cos θ, find θ and evaluate 2 tan²θ + sin²θ − 1. tan θ = 1, θ = 45°; 2(1) + 1/2 − 1 = 3/2.

Worked example 9. Is the statement sin θ = 4/3 possible for some angle θ? No: sin θ is the opposite side divided by the hypotenuse, and the hypotenuse is the longest side, so sin θ ≤ 1. Similarly cos θ = 1.2 is impossible, but tan θ = 4/3 and sec θ = 4/3 are both possible.

Worked example 10. Is cos θ = (a² + b²)/(2ab) possible for a ≠ b? Since (a − b)² > 0, a² + b² > 2ab, so the fraction exceeds 1 and it is impossible. On the other hand, sec θ = (a² + b²)/(2ab) is possible because sec θ ≥ 1.

Verification at a standard angle is also the recommended way to check any identity before writing its proof, and the recommended way to check an answer obtained by eliminating θ.

Problem 1. Prove (cosec θ − sin θ)(sec θ − cos θ) = 1/(tan θ + cot θ). Both sides reduce to sin θ cos θ (see the topic on products).

Problem 2. Prove (1 + tan²A)/(1 + cot²A) = tan²A. LHS = sec²A/cosec²A = sin²A/cos²A = tan²A.

Problem 3. If cosec θ + cot θ = k, prove cos θ = (k² − 1)/(k² + 1). cosec θ − cot θ = 1/k; cosec θ = (k² + 1)/(2k), cot θ = (k² − 1)/(2k); cos θ = cot θ/cosec θ = (k² − 1)/(k² + 1).

Problem 4. If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove x² − y² = a² − b². Square, subtract, use sec²θ − tan²θ = 1.

Problem 5. Prove (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A. Expand and use all three identities.

Problem 6. Simplify (1 + tan²θ)(1 − sin θ)(1 + sin θ). = sec²θ(1 − sin²θ) = sec²θ cos²θ = 1.

Problem 7. Prove tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ. Use t = tan θ.

Problem 8. If sin θ + cos θ = √2, evaluate tan θ + cot θ. Squaring gives sin θ cos θ = 1/2, so tan θ + cot θ = 1/(sin θ cos θ) = 2.

Problem 9. Prove √((1 + sin A)/(1 − sin A)) = sec A + tan A. Multiply inside the root by (1 + sin A).

Problem 10. Prove (sin θ − 2 sin³θ)/(2 cos³θ − cos θ) = tan θ. Factorise and use 1 − 2 sin²θ = 2 cos²θ − 1.

Problem 11. If tan θ + sin θ = m and tan θ − sin θ = n, prove m² − n² = 4√(mn). m² − n² = 4 tan θ sin θ; mn = sin²θ tan²θ.

Problem 12. Prove (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A. Divide by sin A and replace 1 by cosec²A − cot²A.

Problem 13. If sec θ + tan θ = p, find cosec θ. tan θ = (p² − 1)/(2p), sec θ = (p² + 1)/(2p), sin θ = (p² − 1)/(p² + 1), so cosec θ = (p² + 1)/(p² − 1).

Problem 14. Prove sec⁴θ − sec²θ = tan⁴θ + tan²θ. Take out sec²θ and use sec²θ − 1 = tan²θ.

Problem 15. Evaluate (sec²θ − 1)(cosec²θ − 1). = tan²θ cot²θ = 1.

Problem 16. If 2 cos θ = √3 for acute θ, find sin 2θ using sin 2θ = 2 sin θ cos θ. cos θ = √3/2, θ = 30°, sin θ = 1/2, sin 2θ = 2 × (1/2) × (√3/2) = √3/2 = sin 60°, consistent.

In an examination each of these is a two- or four-mark question; the four-mark ones are the identities in Problems 1, 5, 7, 9, 10, 11 and 12 and the conditional Problems 3, 4 and 13.

Planning a proof. Read both sides. If one side is a single ratio or a simple expression, start from the other. Look for a conjugate, a difference of squares, or a common factor before converting everything to sine and cosine; if nothing is visible, convert. Keep the target in view: if the RHS is tan θ, aim for sin θ over cos θ; if it is 2 sec A, aim for 2 over cos A. After each step ask whether the expression is closer to the target.

Writing the proof. Begin 'LHS ='. One transformation per line. Name the identity used in brackets: (sin²θ + cos²θ = 1), (sec²θ − tan²θ = 1), (a² − b² = (a − b)(a + b)). End '= RHS'. If both sides are reduced separately, write 'LHS = ... = sin θ cos θ' and 'RHS = ... = sin θ cos θ', then 'Hence LHS = RHS'.

Common errors. Cross-multiplying the identity and proving a tautology. Cancelling a factor that is not common to the whole numerator and denominator. Writing √(a² + b²) as a + b. Forgetting that √(1 − sin²θ) is cos θ only because θ is acute. Replacing sin²θ + cos²θ by 1 correctly but then also replacing sin θ + cos θ by 1, which is wrong. Confusing sec θ − tan θ with sec θ − tan θ squared. Writing tan θ = sin θ/cos θ upside down. Using a standard value from the wrong row of the table. Assuming that sin(A + B) = sin A + sin B.

Checking. Substitute θ = 30° or 45° into both sides of any identity and any final relation. For a conditional problem, choose a specific value of the parameter (p = 2 in sec θ + tan θ = p gives the 3-4-5 triangle) and check that the derived ratios satisfy the identities.

What lies ahead. In the Intermediate course the identities are extended to all angles via the unit circle, and new ones are added: the compound-angle formulas sin(A ± B), cos(A ± B), tan(A ± B); the multiple-angle formulas sin 2θ = 2 sin θ cos θ, cos 2θ = 2 cos²θ − 1 and sin 3θ = 3 sin θ − 4 sin³θ, which were verified at 30° in this exercise; and the transformations of sums into products. The elimination problems of this exercise are the parametric equations of the ellipse (x = a cos θ, y = b sin θ) and the hyperbola (x = a sec θ, y = b tan θ), which appear in coordinate geometry. Physics uses the same identities for projectiles, waves and alternating currents. The habit of pairing a given relation with the identity that completes it, learned here, is the essential skill for all of that work.