Surveyors, physicists, and engineers work with exact values like $\frac{\sqrt{3}}{2}$ every day, not decimal approximations. The $45^\circ$-$45^\circ$-$90^\circ$ and $30^\circ$-$60^\circ$-$90^\circ$ triangles give exact trig values that form the backbone of higher mathematics.
Today's hook, What is $\sin 75^\circ$? You can express it exactly using the values from the special triangles and a trig identity. Can you find it? (Hint: $75^\circ = 45^\circ + 30^\circ$)
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Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
A square has side length 2 cm. A diagonal is drawn, forming two 45°-45°-90° triangles. Without using a calculator, what is the exact length of the diagonal?
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Exact values from special triangles
Work through the core explanation before applying it.
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Exact values from special triangles
+5 XP to read
Two triangles produce every exact trig value you will ever need for HSC: the $45^\circ$-$45^\circ$-$90^\circ$ triangle and the $30^\circ$-$60^\circ$-$90^\circ$ triangle. Memorise them once and you will never need a calculator for exact-value questions again.
For the 45°-45°-90° triangle with legs of length 1, the hypotenuse is $\sqrt{2}$. For the 30°-60°-90° triangle, the sides are in ratio $1 : \sqrt{3} : 2$. These give exact values for $\sin\theta$, $\cos\theta$, and $\tan\theta$ at $30^\circ$, $45^\circ$, and $60^\circ$.
How exact values extend to other quadrants using ASTC
Why rationalised form is standard
Can do
Skills
Find exact trig values in any quadrant
Simplify exact-value expressions
Solve problems requiring exact answers without a calculator
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Key terms
45°-45°-90° triangleAn isosceles right triangle with sides in ratio $1:1:\sqrt{2}$. Gives exact values at $45°$.
30°-60°-90° triangleA right triangle with angles $30°$ and $60°$ and sides in ratio $1:\sqrt{3}:2$.
Exact valueA value expressed using radicals and fractions, not a decimal approximation.
Rationalising the denominatorMultiplying top and bottom by the surd so no radicals remain in the denominator.
Complementary anglesAngles that add to $90°$. Key property: $\sin\theta = \cos(90°-\theta)$.
Reference angleThe acute angle between the terminal arm and the $x$-axis. Used with ASTC for exact values in any quadrant.
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From special triangles to any quadrant
Work through the core explanation before applying it.
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From special triangles to any quadrant
We just saw that the 30-60-90 and 45-45-90 triangles give us exact values for angles in the first quadrant only. That raises a question: how do we find exact values like $\cos 150°$ or $\tan 240°$, angles outside the first quadrant? This card answers it → use the reference angle formula to reduce to a first-quadrant angle, then apply the ASTC sign rule.
The exact values from the two special triangles apply only to the first quadrant ($0°$ to $90°$). To find exact values in other quadrants, you need two things: the reference angle and the ASTC sign rule.
For any angle $\theta$, the reference angle $\alpha$ is the acute angle between the terminal arm and the $x$-axis:
Quadrant II: $\alpha = 180° - \theta$
Quadrant III: $\alpha = \theta - 180°$
Quadrant IV: $\alpha = 360° - \theta$
Then: find the exact value at the reference angle, and apply the correct sign from ASTC. This lets you find exact values for $120°$, $225°$, $300°$, and even $-30°$ without ever touching a calculator.
Also note the complementary relationship: $\sin\theta = \cos(90°-\theta)$. This is why $\sin 30° = \cos 60° = \frac12$ and $\sin 60° = \cos 30° = \frac{\sqrt{3}}{2}$.
Reference angle rule: QII: $\alpha = 180° - \theta$; QIII: $\alpha = \theta - 180°$; QIV: $\alpha = 360° - \theta$.; ASTC signs: All positive (QI), Sine positive (QII), Tan positive (QIII), Cos positive (QIV).
Pause, copy the three reference angle rules (QII: $180°-\theta$; QIII: $\theta-180°$; QIV: $360°-\theta$) and the ASTC signs for each quadrant into your book.
True or false: In quadrant III, the cosine of the reference angle is positive, so $\cos 210°$ is positive.
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Worked Example, Exact value in quadrant II
+5 XP for trying first
We just saw that reference angles and ASTC let us evaluate any angle. That raises a question: what does the procedure actually look like, step by step, for a QII angle like $150°$? This card answers it → reference angle $= 180° - 150° = 30°$; cosine is negative in QII; so $\cos 150° = -\cos 30° = -\frac{\sqrt{3}}{2}$.
Find the exact value of $\cos 150°$.
Your turn first. Try it yourself before viewing the solution.
SOLUTION
2 marks · HSC band 4+
Step 1: Find the reference angle.
$150°$ is in quadrant II. The reference angle is:
$$\alpha = 180° - 150° = 30°$$
Step 2: Find the exact value at the reference angle.
From the $30°$-$60°$-$90°$ triangle: $\cos 30° = \frac{\sqrt{3}}{2}$
Step 3: Apply the ASTC sign.
In quadrant II, cosine is negative (only sine positive).
$\cos 150°$: QII, reference angle $= 30°$, cosine is negative in QII.; $\cos 150° = -\cos 30° = -\dfrac{\sqrt{3}}{2}$.
Pause, copy the QII worked result: $\cos 150° = -\cos 30° = -\dfrac{\sqrt{3}}{2}$, with the three-step method (identify quadrant → find reference angle → attach ASTC sign) into your book.
Follow-up: What is the exact value of $\sin 150°$?
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Worked Example, Exact value in quadrant III
+5 XP for trying first
We just saw the QII method with a degree angle. That raises a question: what if the angle is given in radians, does the same method still work, or do we need to convert first? This card answers it → convert radians to degrees first ($\frac{4\pi}{3} = 240°$), then apply the same reference angle and ASTC procedure.
Find the exact value of $\tan \frac{4\pi}{3}$.
Your turn first. Try it yourself before viewing the solution.
Pause, copy the QIII worked result: $\tan\frac{4\pi}{3} = \sqrt{3}$, with the conversion step ($\frac{4\pi}{3} = 240°$) and reference angle derivation ($240° - 180° = 60°$) into your book.
Fill the blanks: drag each token to the correct blank.
positivenegativesinecosine
In quadrant II, ___ is ___,
while ___ is ___.
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Worked Example, Simplifying an exact-value expression
+5 XP for trying first
We just saw how to find exact trig values in any quadrant. That raises a question: what if a question gives a combination of exact values, like $\sin^2 30° + \cos^2 30°$, do we evaluate each one or is there a shortcut? This card answers it → recognise the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to collapse the expression instantly.
Find the exact value of $\sin^2 30° + \cos^2 30° + \tan 45°$.
Your turn first. Try it yourself before viewing the solution.
Notice that $\sin^2 30° + \cos^2 30° = 1$ by the Pythagorean identity, so the expression equals $1 + 1 = 2$ immediately.
Answer: $\mathbf{2}$
This is a typical HSC-style question: it tests both exact values and the Pythagorean identity in one expression.
Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$ for any angle $\theta$.; Recognising the identity first saves time: $\sin^2 30° + \cos^2 30° = 1$.
Pause, copy the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ and the strategy tip (look for the identity pattern before substituting individual values) into your book.
Odd one out: Three of these equal 1 by the Pythagorean identity. Which one does NOT?
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Common traps
Meet the mistakes that cost marks, then do it yourself.
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Common traps
We just saw that recognising the Pythagorean identity saves time on simplification problems. That raises a question: what ordering and rationalisation errors do students most often make with exact values? This card answers it → two traps: confusing $\sin 30°$ with $\sin 60°$ (sin increases from 0° to 90°), and leaving surd denominators unrationalised.
Trap 1, Swapping $\sin 30°$ and $\sin 60°$
These are the two most commonly confused exact values. Remember: $\sin 30° = \frac12$ (the smaller angle gets the smaller value) and $\sin 60° = \frac{\sqrt{3}}{2}$ (the larger angle gets the larger value). A quick check: $\sin$ increases from $0°$ to $90°$, so $\sin 60° > \sin 30°$.
Trap 2, Leaving the denominator un-rationalised
HSC marking guidelines almost always require rationalised denominators. Write $\frac{\sqrt{3}}{3}$ instead of $\frac{1}{\sqrt{3}}$, and $\frac{\sqrt{2}}{2}$ instead of $\frac{1}{\sqrt{2}}$. If you're unsure, rationalise: multiply top and bottom by the surd in the denominator.
The exact values from the special triangles are only for acute angles. When finding $\sin 150°$ or $\cos 240°$, you must find the reference angle and apply the correct sign from ASTC. $\sin 150° = +\frac12$ (QII, sine positive), but $\cos 150° = -\frac{\sqrt{3}}{2}$ (QII, cosine negative).
Sin increases $0° \to 90°$, so $\sin 30° = \tfrac{1}{2} < \sin 60° = \tfrac{\sqrt{3}}{2}$.; Always rationalise: $\tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3}$ and $\tfrac{1}{\sqrt{2}} = \tfrac{\sqrt{2}}{2}$.
Pause, copy the two trap reminders: sin increases from 0° to 90° ($\sin 30° < \sin 60°$), and always rationalise surds ($\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$, $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$) into your book.
True or false: $\dfrac{1}{\sqrt{3}}$ and $\dfrac{\sqrt{3}}{3}$ represent the same value.
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Drill, build fluency
Work through the core explanation before applying it.
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Drill, build fluency
+5 XP for 5 correct
Work these through step-by-step. Use exact values only, no calculators.
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Find the exact value of $\sin 120°$.
Show answer
$\frac{\sqrt{3}}{2}$ Reference angle = $60°$, QII where sine is positive. $\sin 120° = +\sin 60° = \frac{\sqrt{3}}{2}$.
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Find the exact value of $\cos \frac{5\pi}{4}$.
Show answer
$-\frac{\sqrt{2}}{2}$ $\frac{5\pi}{4} = 225°$, reference angle = $45°$, QIII where cosine is negative. $\cos 225° = -\cos 45° = -\frac{\sqrt{2}}{2}$.
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Find the exact value of $\tan 300°$.
Show answer
$-\sqrt{3}$ Reference angle = $60°$, QIV where tangent is negative. $\tan 300° = -\tan 60° = -\sqrt{3}$.
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Simplify $\sin^2 45° + \cos^2 45°$.
Show answer
1 By the Pythagorean identity, $\sin^2\theta + \cos^2\theta = 1$ for any angle. Alternatively, $\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac12 + \frac12 = 1$.
Alternatively, using the $45°$-$45°$-$90°$ triangle ratio $1:1:\sqrt{2}$, scaling by 2 gives $2:2:2\sqrt{2}$.
Answer: $2\sqrt{2}$ cm (approximately 2.83 cm).
Did you guess $2\sqrt{2}$ or something close? Many students guess 2 or 3 cm, but the actual diagonal is longer than the side, as expected, since it's the hypotenuse of a right triangle.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 4–5
Exact value in quadrant IV
Find the exact value of $\cos 330°$.
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View comprehensive answer
Working:
$330°$ is in quadrant IV. The reference angle is:
$$\alpha = 360° - 330° = 30°$$
$\cos 30° = \frac{\sqrt{3}}{2}$. In quadrant IV, cosine is positive.
