Skip to main content
ExamExplained

Trigonometric identities (QCE Math Methods Unit 2)

Syllabus dot point

“State and apply the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, and use it together with related identities to simplify expressions and solve equations”

7 min read

Quick answer

The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 comes from the unit-circle definition, together with tan⁡θ=sin⁡θ/cos⁡θ\tan\theta = \sin\theta / \cos\theta it lets you express any trig expression in one function, and quadrant signs determine which root to take when computing the missing trig value from one given trig value.

Jump to a section
  1. What this dot point is asking
  2. The Pythagorean identity
  3. The tangent identity
  4. Why the identity matters for solving equations
  5. Standard manipulations
  6. Given one trig value, find the others
  7. Solving trig equations using identities
  8. Exam-style questions
Key fact

In the 2025 syllabus, trigonometric functions are introduced in Unit 1 (Topic 4: Trigonometric functions). The unit that the 2025 syllabus uses for trigonometric identities is not confirmed, so check your teacher's unit plan. This page stays in Unit 2 so its link does not change.

What this dot point is asking

QCAA wants you to know the Pythagorean identity and to apply it together with the quadrant rules (from the unit circle) to find missing trig values, simplify trig expressions, and solve trig equations.

The Pythagorean identity

For any angle θ\theta:

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

This comes directly from the unit-circle definition: a point at angle θ\theta on the unit circle has coordinates (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), and these satisfy x2+y2=1x^2 + y^2 = 1.

The tangent identity

tan⁡θ=sin⁡θcos⁡θ,cos⁡θ≠0\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cos\theta \neq 0

Dividing the Pythagorean identity by cos⁡2θ\cos^2 \theta gives:

tan⁡2θ+1=sec⁡2θ\tan^2 \theta + 1 = \sec^2 \theta

(QCAA Methods uses this less than the basic form, but it appears occasionally.)

Why the identity matters for solving equations

An equation mixing sin⁡θ\sin\theta and cos⁡θ\cos\theta (or their squares) usually cannot be solved directly, because two different functions of the same angle appear. The Pythagorean identity lets you replace cos⁡2θ\cos^2\theta with 1−sin⁡2θ1 - \sin^2\theta (or vice versa) so that only one function remains, turning the equation into a quadratic in that single function. Factoring or the quadratic formula then gives the values of that function, and the unit circle converts each value into all the angles in the required interval. This reduce-to-one-function strategy is the most common identity application in Methods.

Standard manipulations

Express in one function
sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta, and cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta. Useful for rewriting an expression in a single trig function before solving.
Difference of squares
1−sin⁡2θ=(1−sin⁡θ)(1+sin⁡θ)=cos⁡2θ1 - \sin^2 \theta = (1-\sin\theta)(1+\sin\theta) = \cos^2\theta.
Combine over a common denominator
1cos⁡θ−cos⁡θ=1−cos⁡2θcos⁡θ=sin⁡2θcos⁡θ=sin⁡θtan⁡θ\frac{1}{\cos\theta} - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta} = \sin\theta \tan\theta.

Given one trig value, find the others

Use the Pythagorean identity to compute the magnitude, then use the quadrant to fix the sign.

Example: cos⁡θ=−7/25\cos\theta = -7/25 and θ\theta is in Q3.

sin⁡2θ=1−49/625=576/625\sin^2 \theta = 1 - 49/625 = 576/625.

sin⁡θ=±24/25\sin\theta = \pm 24/25. In Q3, sine is negative, so sin⁡θ=−24/25\sin\theta = -24/25.

tan⁡θ=(−24/25)/(−7/25)=24/7\tan\theta = (-24/25) / (-7/25) = 24/7 (positive, as expected in Q3).

Solving trig equations using identities

If an equation contains both sin⁡\sin and cos⁡\cos (or sin⁡2\sin^2 and cos⁡2\cos^2), use the identity to reduce it to one function.

Example: solve 2cos⁡2θ−sin⁡θ−1=02 \cos^2 \theta - \sin\theta - 1 = 0 for θ∈[0,2π]\theta \in [0, 2\pi].

Replace cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta:

2(1−sin⁡2θ)−sin⁡θ−1=02(1 - \sin^2 \theta) - \sin\theta - 1 = 0

1−2sin⁡2θ−sin⁡θ=01 - 2 \sin^2 \theta - \sin\theta = 0

2sin⁡2θ+sin⁡θ−1=02 \sin^2 \theta + \sin\theta - 1 = 0

Factor: (2sin⁡θ−1)(sin⁡θ+1)=0(2\sin\theta - 1)(\sin\theta + 1) = 0.

sin⁡θ=1/2  ⟹  θ=π/6\sin\theta = 1/2 \implies \theta = \pi/6 or 5π/65\pi/6.

sin⁡θ=−1  ⟹  θ=3π/2\sin\theta = -1 \implies \theta = 3\pi/2.

Three solutions in the stated interval.

Given one ratio, find the others

Apply the Pythagorean identity

Given cos⁡θ=−725\cos\theta = -\dfrac{7}{25} with θ\theta in the third quadrant, find sin⁡θ\sin\theta and tan⁡θ\tan\theta. From sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta: sin⁡2θ=1−49625=576625\sin^2\theta = 1 - \dfrac{49}{625} = \dfrac{576}{625}, so sin⁡θ=±2425\sin\theta = \pm\dfrac{24}{25}.

Fix the sign from the quadrant

In the third quadrant sine is negative, so sin⁡θ=−2425\sin\theta = -\dfrac{24}{25}.

Compute the tangent

tan⁡θ=sin⁡θcos⁡θ=−24/25−7/25=247,\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-24/25}{-7/25} = \frac{24}{7},

positive, as expected for the third quadrant. Markers reward the identity, the quadrant sign, and the tangent ratio.

Common traps
Forgetting the sign
sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 gives a positive value for each square. The sign of the trig function itself comes from the quadrant.
Treating sin⁡2θ\sin^2 \theta as sin⁡θ2\sin \theta^2
sin⁡2θ\sin^2 \theta means (sin⁡θ)2(\sin\theta)^2. The squared notation is conventional; do not confuse it with sin⁡(θ2)\sin(\theta^2).
Losing solutions during factoring
Always set each factor to zero and find all solutions in the stated interval before deciding which to accept.
Dividing by cos⁡θ\cos\theta without checking cos⁡θ=0\cos\theta = 0
When cos⁡θ=0\cos\theta = 0, dividing loses solutions. Always check the endpoint solutions separately.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Paper 1 (technique). Given sin⁡θ=35\sin\theta = \dfrac{3}{5} with θ\theta in the second quadrant, determine the exact values of cos⁡θ\cos\theta and tan⁡θ\tan\theta.
Show worked answer →

Use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1: cos⁡2θ=1−925=1625\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}, so cos⁡θ=±45\cos\theta = \pm\dfrac{4}{5}.

In the second quadrant cosine is negative, so cos⁡θ=−45\cos\theta = -\dfrac{4}{5}. Then tan⁡θ=sin⁡θcos⁡θ=3/5−4/5=−34\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}.

Markers reward the Pythagorean identity, the quadrant sign, and the tangent as a ratio.

Practice4 marks
Paper 2 (complex familiar). Solve 2cos⁡2θ−sin⁡θ−1=02\cos^2\theta - \sin\theta - 1 = 0 for θ∈[0,2π]\theta \in [0, 2\pi].
Show worked answer →

Replace cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta: 2(1−sin⁡2θ)−sin⁡θ−1=02(1 - \sin^2\theta) - \sin\theta - 1 = 0, which rearranges to 2sin⁡2θ+sin⁡θ−1=02\sin^2\theta + \sin\theta - 1 = 0.

Factor: (2sin⁡θ−1)(sin⁡θ+1)=0(2\sin\theta - 1)(\sin\theta + 1) = 0. So sin⁡θ=12\sin\theta = \dfrac{1}{2} giving θ=π6,5π6\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, or sin⁡θ=−1\sin\theta = -1 giving θ=3π2\theta = \dfrac{3\pi}{2}.

Markers reward using the identity to reduce to one function, factoring the quadratic in sin⁡θ\sin\theta, and all solutions in the interval.

Practise this

Sources & how we know this

Based on the official QCAA syllabus and published QCAA past papers and marking guidelines. Not yet individually reviewed by a person; always check QCAA and your school's materials for current rules.

ExamExplained