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

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 $\theta$:

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

The tangent identity

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

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

Standard manipulations

Express in one function. $\sin^2 \theta = 1 - \cos^2 \theta$, and $\cos^2 \theta = 1 - \sin^2 \theta$. Useful for rewriting an expression in a single trig function before solving.

Difference of squares. $1 - \sin^2 \theta = (1-\sin\theta)(1+\sin\theta) = \cos^2\theta$.

Combine over a common denominator. $\frac{1}{\cos\theta} - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta} = \sin\theta \tan\theta$.

Given one trig value, find the others

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

Example: $\cos\theta = -7/25$ and $\theta$ is in Q3.

$\sin^2 \theta = 1 - 49/625 = 576/625$.

$\sin\theta = \pm 24/25$. In Q3, sine is negative, so $\sin\theta = -24/25$.

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

Solving trig equations using identities

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

Example: solve $2 \cos^2 \theta - \sin\theta - 1 = 0$ for $\theta \in [0, 2\pi]$.

Replace $\cos^2 \theta = 1 - \sin^2 \theta$:

$2(1 - \sin^2 \theta) - \sin\theta - 1 = 0$

$1 - 2 \sin^2 \theta - \sin\theta = 0$

$2 \sin^2 \theta + \sin\theta - 1 = 0$

Factor: $(2\sin\theta - 1)(\sin\theta + 1) = 0$.

$\sin\theta = 1/2 \implies \theta = \pi/6$ or $5\pi/6$.

$\sin\theta = -1 \implies \theta = 3\pi/2$.

Three solutions in the stated interval.

Common traps

Forgetting the sign. $\sin^2 \theta + \cos^2 \theta = 1$ gives a positive value for each square. The sign of the trig function itself comes from the quadrant.

Treating $\sin^2 \theta$ as $\sin \theta^2$. $\sin^2 \theta$ means $(\sin\theta)^2$. The squared notation is conventional; do not confuse it with $\sin(\theta^2)$.

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 $\cos\theta$ without checking $\cos\theta = 0$. When $\cos\theta = 0$, dividing loses solutions. Always check the endpoint solutions separately.

In one sentence

The Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ comes from the unit-circle definition, together with $\tan\theta = \sin\theta / \cos\theta$ 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.