Year 12 Methods (Unit 3 & 4) Trigonometry using the sine and cosine rules

Reviewing trigonometry

Q: How do you convert between degrees and radians?

Use 180 degrees equals pi radians. To change degrees to radians multiply by pi over 180; to change radians to degrees multiply by 180 over pi. For example 135 degrees equals three pi over four.

Q: What are the exact values of sine, cosine and tangent?

From the 30-60-90 triangle, sin 30 equals one half, cos 30 equals root three over two, sin 60 equals root three over two and tan 60 equals root three; from the 45-45-90 triangle, sin 45 equals cos 45 equals one over root two.

Q: What is ASTC on the unit circle?

The point at angle theta is cos theta, sin theta. Anticlockwise from quadrant one the positive ratios are All, Sine, Tangent, Cosine, so in quadrant four sine is negative and cosine is positive.

Q: What is the Pythagorean identity?

sin squared theta plus cos squared theta equals one; rearrange to find one ratio from the other, choosing the sign from the quadrant.

Q: What are the arc length and sector area formulas?

With theta in radians, arc length is l equals r theta and sector area is A equals one half r squared theta.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), reviewing trigonometry gathers the tools the sine and cosine rules rely on: right-angled-triangle ratios (SOH–CAH–TOA) and Pythagoras, radian measure, the exact values, the unit circle and the sign of each ratio in the four quadrants (ASTC), the Pythagorean identity, arc length and sector area, and solving trigonometric equations over a domain.

In a right-angled triangle, for an angle \(\theta\) the three ratios are \(\sin\theta=\dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta=\dfrac{\text{adj}}{\text{hyp}}\) and \(\tan\theta=\dfrac{\text{opp}}{\text{adj}}\) (remembered as SOH–CAH–TOA), and the sides satisfy Pythagoras, \(\text{opp}^{2}+\text{adj}^{2}=\text{hyp}^{2}\).

An angle can be measured in degrees or in radians, where one radian is the angle subtended by an arc equal in length to the radius, and \(180^\circ=\pi\) radians.

The unit circle (radius \(1\), centre the origin) extends these ideas to any angle: the point reached by rotating through \(\theta\) is \((\cos\theta,\sin\theta)\), so \(\cos\theta\) is its \(x\)-coordinate and \(\sin\theta\) its \(y\)-coordinate. The signs of the ratios in the four quadrants are captured by ASTC — All, Sine, Tangent, Cosine positive, quadrants one to four.

Key idea. On the unit circle the point at angle \(\theta\) is \((\cos\theta,\sin\theta)\); its coordinates give the ratios and their signs, and \(\sin^{2}\theta+\cos^{2}\theta=1\) always holds.

The \(30\text{--}60\text{--}90\) triangle: \(\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}\), \(\cos\dfrac{\pi}{3}=\dfrac{1}{2}\), \(\tan\dfrac{\pi}{3}=\sqrt{3}\)

Unit circle: the point is \((\cos\theta,\sin\theta)\); positive ratios are \(A,S,T,C\) in quadrants 1–4

Right-angled-triangle ratios and Pythagoras:

\[\sin\theta=\dfrac{\text{opp}}{\text{hyp}},\quad \cos\theta=\dfrac{\text{adj}}{\text{hyp}},\quad \tan\theta=\dfrac{\text{opp}}{\text{adj}},\quad \text{opp}^{2}+\text{adj}^{2}=\text{hyp}^{2}\]

\sin θ = \frac{opp}{hyp}

Radian–degree conversion (\(180^\circ=\pi\)):

\[\text{radians}=\text{degrees}\times\dfrac{\pi}{180^\circ},\qquad \text{degrees}=\text{radians}\times\dfrac{180^\circ}{\pi}\]

Exact values (integer multiples of \(\dfrac{\pi}{6}\) and \(\dfrac{\pi}{4}\)):

\[\sin\dfrac{\pi}{6}=\dfrac{1}{2},\ \cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2},\ \sin\dfrac{\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}},\ \sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2},\ \cos\dfrac{\pi}{3}=\dfrac{1}{2},\ \tan\dfrac{\pi}{3}=\sqrt{3}\]

Pythagorean identity, arc length and sector area (with \(\theta\) in radians):

\[\sin^{2}\theta+\cos^{2}\theta=1,\qquad l=r\theta,\qquad A=\dfrac{1}{2}r^{2}\theta\]

\sin^{2} θ + \cos^{2} θ = 1

ASTC. Reading anticlockwise from the first quadrant, the positive ratios are All, then Sine, then Tangent, then Cosine. So in the fourth quadrant only \(\cos\theta>0\) (with \(\sin\theta<0\)).

Working with trigonometry review

Right triangles. Label opposite, adjacent and hypotenuse relative to the angle, pick the ratio (SOH–CAH–TOA) that uses the known and unknown sides, and solve; use Pythagoras for a missing side.
Convert angles. Multiply by \(\dfrac{\pi}{180}\) to go from degrees to radians, or by \(\dfrac{180}{\pi}\) to go the other way.
Exact values. Read them from the \(30\text{--}60\text{--}90\) and \(45\text{--}45\text{--}90\) triangles, or from the unit circle.
Signs. Use ASTC to decide whether a ratio is positive or negative in a given quadrant.
Solve equations. Find the base angle, then use ASTC (or the graph) to list every solution in the required domain.

Arcs and sectors. With \(\theta\) in radians, \(l=r\theta\) gives the arc length and \(A=\dfrac{1}{2}r^{2}\theta\) the sector area — convert a degree angle to radians first.

Example 1 — Exact value

Find the exact value of \(\cos\dfrac{\pi}{3}\).

Solution

\(\dfrac{\pi}{3}=60^\circ\); read the ratio from the \(30\text{--}60\text{--}90\) triangle.

\(\cos 60^\circ\)	\(=\)	\(\dfrac{\text{adjacent}}{\text{hypotenuse}}\)
	\(=\)	\(\dfrac{1}{2}\)

\cos \frac{π}{3} = \frac{1}{2}

Example 2 — Radians

Write \(135^\circ\) in radians.

Solution

Multiply by \(\dfrac{\pi}{180}\) and simplify.

\(135^\circ\)	\(=\)	\(135\times\dfrac{\pi}{180}\)
	\(=\)	\(\dfrac{3\pi}{4}\)

135 ° = \frac{3 π}{4}

Example 3 — Pythagorean identity

Given \(\sin\theta=\dfrac{3}{5}\) with \(\theta\) acute, find \(\cos\theta\) and \(\tan\theta\).

Solution

Use \(\sin^{2}\theta+\cos^{2}\theta=1\), then \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\).

\(\cos^{2}\theta\)	\(=\)	\(1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\cos\theta\)	\(=\)	\(\dfrac{4}{5}\ (\theta\text{ acute})\)
\(\tan\theta\)	\(=\)	\(\dfrac{3}{4}\)

\cos θ = \frac{4}{5}

Example 4 — Solve over a domain

Solve \(\sin x=\dfrac{1}{2}\) for \(0\le x\le 2\pi\).

Solution

Base angle \(\dfrac{\pi}{6}\); sine is positive in quadrants 1 and 2.

\(x\)	\(=\)	\(\dfrac{\pi}{6}\ \text{ or }\ \pi-\dfrac{\pi}{6}\)
\(x\)	\(=\)	\(\dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

x = \frac{π}{6}, \frac{5 π}{6}

Common pitfalls

Use radians in \(l=r\theta\) and \(A=\tfrac12 r^{2}\theta\). These formulas fail if \(\theta\) is in degrees — convert first (e.g. \(60^\circ=\dfrac{\pi}{3}\)).

List every solution. Over a full turn, \(\sin x=k\) or \(\cos x=k\) usually has two solutions; find the base angle, then use ASTC to place the second one.

Do not swap neighbouring exact values. \(\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}\) but \(\cos\dfrac{\pi}{3}=\dfrac{1}{2}\); and a ratio is not its reciprocal.

Frequently asked questions

How do you convert between degrees and radians?

Use \(180^\circ=\pi\). Multiply degrees by \(\dfrac{\pi}{180}\) for radians, or radians by \(\dfrac{180}{\pi}\) for degrees. So \(135^\circ=\dfrac{3\pi}{4}\).

What are the exact values of sine, cosine and tangent?

From the special triangles: \(\sin\dfrac{\pi}{6}=\dfrac12\), \(\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}\), \(\sin\dfrac{\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}\), \(\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}\), \(\tan\dfrac{\pi}{3}=\sqrt3\).

What is ASTC on the unit circle?

The point at angle \(\theta\) is \((\cos\theta,\sin\theta)\). Anticlockwise from quadrant 1 the positive ratios are All, Sine, Tangent, Cosine, so in quadrant 4 \(\sin\theta<0\) and \(\cos\theta>0\).

What is the Pythagorean identity?

\(\sin^{2}\theta+\cos^{2}\theta=1\). Rearrange to find one ratio from the other, choosing the sign from the quadrant.

What are the arc length and sector area formulas?

With \(\theta\) in radians, \(l=r\theta\) and \(A=\dfrac12 r^{2}\theta\).

How do you solve a trigonometric equation over a domain?

Find the base angle from the exact values, then use ASTC to list every solution in the domain. For \(\sin x=\dfrac12\) on \([0,2\pi]\), \(x=\dfrac{\pi}{6},\dfrac{5\pi}{6}\).

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