MathematicsCore26 min read

Trigonometric Identities and Equations

Six functions, a handful of identities, and how to pick the right one

This topic appears in:

A Level Mathematics (9709) · ch.2A Level Mathematics (9709) · ch.3
01

Three more functions, each a reciprocal

Secant, cosecant and cotangent are not new ideas — each is simply one divided by one of the three functions you already know. The only genuine difficulty is that the names do not pair with the obvious partners.

The reliable trick is to look at the third letter: sec goes with cosine, cosec goes with sine. Matching the first letters instead is the mistake almost everybody makes at least once.

sec θ = 1 / cos θcosec θ = 1 / sin θcot θ = 1 / tan θ = cos θ / sin θthird letter of the name tells you which function it inverts
sec θ
reciprocal of cosineundefined wherever cos θ = 0
cosec θ
reciprocal of sineundefined wherever sin θ = 0
cot θ
reciprocal of tangentequals 0 where tan θ is undefined

Every one of the six functions is a length read off this one diagram. Drag the angle and watch sine and cosine change; the reciprocal functions blow up exactly where their partners pass through zero.

Reciprocals explain the asymptotes

Dividing by something approaching zero produces something growing without limit. So sec θ has vertical asymptotes wherever cos θ = 0 — at 90°, 270° and so on — and cosec θ has them wherever sin θ = 0. You never need to memorise where the asymptotes are; you can always work them out from the partner function.

02

The Pythagorean identities

One identity generates the other two. Starting from sin²θ + cos²θ = 1, dividing every term by cos²θ gives the second, and dividing by sin²θ gives the third. Deriving them takes ten seconds and is safer than trusting memory under pressure.

sin²θ + cos²θ = 1÷ cos²θ:tan²θ + 1 = sec²θ÷ sin²θ:1 + cot²θ = cosec²θderive the second and third rather than memorising them
sin²θ
means (sin θ)²not sin(θ²), which is something else entirely
sec²θ
the identity with tanappears constantly in integration too

An identity is not an equation

An identity holds for every value of θ; an equation holds only for particular values. That distinction decides what you are allowed to do. Proving an identity means transforming one side into the other — you may not move terms across the identity sign as though solving, because that assumes what you are trying to prove.

03

Compound and double angles

The compound angle formulae expand a function of a sum. They matter because sin(A + B) is emphatically not sin A + sin B — the sine function does not distribute over addition, and assuming it does is one of the costliest errors in the subject.

Putting B = A into the compound formulae produces the double angle formulae, which is why they never need separate memorising. The three versions of cos 2A all follow from the first by applying the Pythagorean identity, and having all three available is what makes many integrals and equations tractable.

sin(A ± B) = sin A cos B ± cos A sin Bcos(A ± B) = cos A cos B ∓ sin A sin Btan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)sin 2A = 2 sin A cos Acos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²Atan 2A = 2 tan A / (1 − tan²A)the cosine formulae flip the sign — that is what ∓ means
± ∓
signs taken in the same ordertop with top, bottom with bottom
cos 2A
three equivalent formschoose the one containing the function you want
2cos²A − 1
the form with cosine onlyrearranges to cos²A = (1 + cos 2A)/2 for integration

The cosine formula changes the sign

cos(A + B) = cos A cos B − sin A sin B. The minus with a plus inside catches people every year. A quick check: put A = B = 0. The left side is cos 0 = 1, and the right is 1 × 1 − 0 × 0 = 1 ✓. Try it with a plus sign instead and you would still get 1, so use A = B = 90° as well: left is cos 180° = −1, right is 0 − 1 = −1 ✓.

04

The R form: two waves become one

An expression such as 3 sin θ + 4 cos θ is awkward because θ appears twice. Writing it as a single sine or cosine wave with a shifted starting point removes that problem entirely, and immediately hands you the maximum and minimum values.

The method is always the same: expand the target form, compare coefficients, and read off R and α. R comes from Pythagoras and α from a tangent, which is worth remembering as "R by Pythagoras, alpha by tan".

a sin θ + b cos θ = R sin(θ + α)R = √(a² + b²)tan α = b / amaximum value R, minimum value −RR is always taken positive, and α is chosen in the first quadrant
R
the amplitude of the combined wave√(a² + b²), always positive
α
the phase shiftfrom tan α = b/a for this particular form
max, min
R and −Rbecause a sine can only reach ±1
Worked example

Express 3 sin θ + 4 cos θ in the form R sin(θ + α), and hence find the maximum value of the expression and the value of θ at which it occurs, for 0° ⩽ θ ⩽ 360°.

  1. Expand the target: R sin(θ + α) = R sin θ cos α + R cos θ sin α.Using the compound angle formula, so the two sides can be compared term by term.
  2. Compare coefficients: R cos α = 3 and R sin α = 4.Matching the sin θ terms and the cos θ terms separately.
  3. Square and add: R² = 3² + 4² = 25, so R = 5.Since cos²α + sin²α = 1, squaring and adding eliminates α entirely. R is taken positive.
  4. Divide: tan α = 4/3, so α = 53.1°.Dividing eliminates R. Both 3 and 4 are positive, so α is in the first quadrant.
  5. So the expression is 5 sin(θ + 53.1°).One wave instead of two, with θ appearing only once.
  6. Maximum is 5, when sin(θ + 53.1°) = 1, that is θ + 53.1° = 90°, so θ = 36.9°.The maximum of any sine is 1, so the maximum of the expression is R. Solving for θ needs the shift subtracted again.

5 sin(θ + 53.1°); maximum 5 at θ = 36.9°

05

Solving equations without losing solutions

A trigonometric equation over a stated interval nearly always has several solutions, and marks are lost by finding one and stopping. The safe procedure is to widen the interval first, then solve, then narrow back.

If the equation contains 2θ and θ runs from 0° to 360°, then 2θ runs from 0° to 720° — so solutions must be hunted across two full revolutions. Halving them at the end returns them to the original range.

Never divide by a trigonometric function

Solving sin θ cos θ = sin θ by cancelling sin θ destroys every solution where sin θ = 0. Factorise instead: sin θ(cos θ − 1) = 0 gives sin θ = 0 or cos θ = 1, keeping both families. This is the same rule as in ordinary algebra, and it is broken just as often.

The procedure that finds every solution

  1. Simplify to a single trigonometric function using the identities, if more than one appears.
  2. Adjust the interval to match the argument: for 2θ, double the interval; for θ + 30°, shift it.
  3. Find the principal value from the calculator, then use symmetry or the CAST diagram for the others in the widened interval.
  4. Add multiples of 360° (or 180° for tan) until you leave the interval.
  5. Convert back to θ, then discard any that fall outside the original range.
  6. Count your answers against the shape of the graph — a sine equation over a full revolution usually has two.

Practice questions

5 questions · 18 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 7 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Starting from sin²θ + cos²θ = 1, derive the identity tan²θ + 1 = sec²θ.
Model answer

Dividing every term of sin²θ + cos²θ = 1 by cos²θ gives sin²θ/cos²θ + 1 = 1/cos²θ. Since sin θ/cos θ = tan θ and 1/cos θ = sec θ, this is tan²θ + 1 = sec²θ.

Examiner tip. One mark for dividing by cos²θ, one for the correct simplification. Show the division explicitly.

SQ2[2 marks]
Explain why sec θ is undefined at θ = 90°, and state what happens to the graph there.
Model answer

sec θ = 1/cos θ, and cos 90° = 0. Division by zero is undefined, so sec 90° does not exist. The graph has a vertical asymptote at θ = 90°, with the curve tending to +∞ on one side and −∞ on the other.

Examiner tip. Name the reciprocal relationship and the asymptote — both are needed for the two marks.

SQ3[3 marks]
Prove the identity (1 − cos 2θ) / sin 2θ = tan θ.
Model answer

Using cos 2θ = 1 − 2sin²θ, the numerator becomes 1 − (1 − 2sin²θ) = 2sin²θ. Using sin 2θ = 2 sin θ cos θ, the denominator is 2 sin θ cos θ. So the fraction is 2sin²θ / (2 sin θ cos θ) = sin θ / cos θ = tan θ.

Examiner tip. Choosing the right form of cos 2θ is the whole question — the version with sin²θ is the one that cancels. Work on one side only.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
Solve 2sin²θ + 3cos θ = 3 for 0° ⩽ θ ⩽ 360°.
Full working
  1. Replace sin²θ using sin²θ = 1 − cos²θ: 2(1 − cos²θ) + 3cos θ = 3Reducing to a single trigonometric function is always the first move.[1]
  2. Rearrange: 2 − 2cos²θ + 3cos θ − 3 = 0, so 2cos²θ − 3cos θ + 1 = 0A quadratic in cos θ. Multiplying by −1 makes the leading coefficient positive.[1]
  3. Factorise: (2cos θ − 1)(cos θ − 1) = 0, so cos θ = ½ or cos θ = 1Both roots are valid since each lies between −1 and 1.[1]
  4. cos θ = ½ gives θ = 60° and 300°; cos θ = 1 gives θ = 0° and 360°Cosine is positive in the first and fourth quadrants, so the second solution is 360 − 60. Both endpoints are included by the stated interval.[1]

θ = 0°, 60°, 300°, 360°

Exam questions

1 · 7 marks

Multi-part questions with a full mark scheme.

Q1[7 marks]
(a) Express 5 sin θ − 12 cos θ in the form R sin(θ − α), where R > 0 and 0° < α < 90°.
(b) Hence solve 5 sin θ − 12 cos θ = 6.5 for 0° ⩽ θ ⩽ 360°.
(c) State the maximum value of 5 sin θ − 12 cos θ and the value of θ at which it occurs.
Mark scheme
  1. (a) R sin(θ − α) = R sin θ cos α − R cos θ sin α, so R cos α = 5 and R sin α = 12The minus in the target form matches the minus in the expression, so both comparison equations are positive.[1]
  2. R = √(25 + 144) = 13, and tan α = 12/5 so α = 67.4°A 5-12-13 triangle. α is in the first quadrant as required.[1]
  3. So the expression is 13 sin(θ − 67.4°)Worth stating explicitly before using it in part (b).[1]
  4. (b) 13 sin(θ − 67.4°) = 6.5, so sin(θ − 67.4°) = 0.5Dividing by R reduces it to a standard equation.[1]
  5. Interval for θ − 67.4° is −67.4° to 292.6°. Solutions: 30° and 150°Adjusting the interval before solving is what prevents solutions being missed.[1]
  6. θ = 97.4° and θ = 217.4°Adding 67.4° to each, then checking both lie in the original range.[1]
  7. (c) Maximum is 13, when sin(θ − 67.4°) = 1, so θ − 67.4° = 90° and θ = 157.4°The maximum of the combined wave is R, which is the main reason for using this form.[1]

(a) 13 sin(θ − 67.4°); (b) θ = 97.4°, 217.4°; (c) maximum 13 at θ = 157.4°