Trigonometric Identities & Equations

One identity comes straight from Pythagoras, and it unlocks nearly every trig equation.

Drop a right triangle inside the unit circle.

Its horizontal leg is cosθ, its vertical leg is sinθ, and the hypotenuse is the radius — which is 1.

the oldest theorem, hiding in a circle 1 cos θ sin θ θ legs cos and sin, hypotenuse 1 — Pythagoras does the rest

Pythagoras has something to say about every right triangle, and this one is no exception.

So (cosθ)2 + (sinθ)2 = 12, whatever angle θ happens to be.

That is the Pythagorean identity, sin2θ + cos2θ = 1. It is not a new fact — it is the oldest theorem in the book, read off a circle of radius one.

the three ratios all live in the same triangle θ adjacent opposite hypotenuse sin θ = opp / hyp cos θ = adj / hyp tan θ = opp / adj "opposite" and "adjacent" are named from θ, so they swap if you move to the other acute angle the hypotenuse never moves — it is always opposite the right angle

The identities worth knowing

The Pythagorean one, sin2θ + cos2θ = 1, is the one you will reach for most.

The quotient identity tanθ = sinθcosθ lets you turn tangents into sines and cosines.

Between them, almost any expression can be pushed into sines and cosines alone, where it is easier to handle.

When stuck, convert everything

Rewriting every term in sines and cosines is the standard opening move. It rarely makes things worse and usually reveals the cancellation.

Proving an identity

Work on one side only and transform it until it matches the other. You are not solving an equation.

Start with the messier side — there is more to simplify and more to work with.

Common moves: convert to sines and cosines, find a common denominator, factor, or swap 1 for sin2 + cos2.

Do not move terms across

For a clear first proof, transform one side into the other. A chain of reversible equivalences can also prove an identity, but simply assuming the starting equality and manipulating it is not a proof.

Solving a trig equation

Solve as you would any equation until you reach something like sinθ = 12.

Then use the reference angle to find the first solution, and the quadrants to find the others in one turn.

sinθ = 12 gives θ = π6 and 5π6, because sine is positive in quadrants I and II.

The calculator gives one answer

sin-1(0.5) returns 30° and stops. The second solution is yours to find, and questions almost always want both.

Infinitely many answers

Going round again lands on the same place, so every solution repeats every 2π.

Write the general solution as θ = π6 + 2π n and θ = 5π6 + 2π n, with n any integer.

If the question restricts to 0 ≤ θ < 2π, list only the solutions inside that window.

Read the interval first

The interval given in the question decides how many answers to report. Checking it before you start saves listing solutions nobody wanted.

Quadratic-looking equations

2sin2θ - sinθ - 1 = 0 is a quadratic in disguise — let s = sinθ and it becomes 2s2 - s - 1 = 0.

Factor to (2s+1)(s-1) = 0, so sinθ = -12 or sinθ = 1.

Then solve each piece separately. Each value of the sine may give two angles, so count carefully.

Substitute to see it

Writing s for sinθ makes the structure obvious and stops you trying to divide by sinθ — which would lose solutions.

Worked examples

Using the identity

If sinθ = 35 and θ is in quadrant II, find cosθ.

  1. sin2θ + cos2θ = 1, so cos2θ = 1 - 925 = 1625. Substitute and rearrange.
  2. So cosθ = ±45. Squaring lost the sign, so both are still candidates.
  3. Quadrant II has x negative, so cosθ = -45. The quadrant settles which one.

-45

Proving one

Prove that sinθcosθ + cosθsinθ = 1sinθcosθ.

  1. Take the left side and use the common denominator sinθcosθ. Start with the messier side.
  2. It becomes sin2θ + cos2θsinθcosθ. Add the two fractions.
  3. The top is 1 by the Pythagorean identity, giving the right side. One side transformed into the other — the proof is complete.

Proved

Solving in an interval

Solve 2cosθ + 1 = 0 for 0 ≤ θ < 2π.

  1. Rearrange: cosθ = -12. Ordinary algebra first.
  2. The reference angle is π3, since cosπ3 = 12. Size from the reference angle.
  3. Cosine is negative in quadrants II and III: θ = 2π3 and 4π3. Two answers inside one turn.

2π3, 4π3

Words to know

Identity
An equation true for every value of the variable, not just some.
Pythagorean identity
sin2θ + cos2θ = 1.
Reciprocal identities
csc = 1sin, sec = 1cos, cot = 1tan.
Quotient identity
tanθ = sinθcosθ.
General solution
All the angles satisfying an equation, usually written with +2π n for the repeats.

Mistakes to watch for

Reporting only the calculator’s answer. cos-1(-0.5) gives 2π3 and stops, but 4π3 works too. Every trig equation has a second solution in the turn, and usually infinitely many beyond it.

Dividing both sides by sinθ. In sinθcosθ = sinθ, dividing throws away every solution where sinθ = 0. Factor instead and set each piece to zero.

Losing the sign when taking a square root. cos2θ = 1625 gives cosθ = ±45, and only the quadrant decides which. Taking just the positive root silently assumes quadrant I.

Working on both sides while proving an identity. Moving terms across assumes the result you are trying to establish. Transform one side alone until it looks like the other.

Reading sin2θ as sin(θ2). The square applies to the whole sine value, so sin2 30° = (0.5)2 = 0.25, while sin(900°) = 1 — completely different quantities.

The identity holds for every angle, including ones past 360° and negative ones, because the triangle argument is really about coordinates on a circle.

Ptolemy was proving results equivalent to the addition formulas around 150 AD, to build tables of chords for astronomy.

Sound engineers use these identities constantly: two tones played together beat against each other, and the formula for the beat is an identity rearranged.

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