Exact Values and Special Triangles
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.
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?
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 $60^\circ$.
45-45-90 triangle: sides $1:1:\sqrt{2}$. $\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$, $\tan 45° = 1$.; 30-60-90 triangle: sides $1:\sqrt{3}:2$. $\sin 30° = \frac{1}{2}$, $\cos 30° = \frac{\sqrt{3}}{2}$, $\tan 30° = \frac{\sqrt{3}}{3}$.
Pause — copy the exact values from both special triangles: 45-45-90 ($\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$, $\tan 45° = 1$) and 30-60-90 ($\sin 30° = \frac{1}{2}$, $\cos 60° = \frac{1}{2}$, $\tan 30° = \frac{\sqrt{3}}{3}$) into your book.
Quick check: Which of the following is the correct exact value of $\tan 60°$?
Key facts
- Exact values of $\sin$, $\cos$, and $\tan$ at $30°$, $45°$, and $60°$
- How the two special triangles produce these values
- Rationalising denominators
Concepts
- Why $\sin 45° = \cos 45°$ and why $\sin 30° = \cos 60°$
- How exact values extend to other quadrants using ASTC
- Why rationalised form is standard
Skills
- Find exact trig values in any quadrant
- Simplify exact-value expressions
- Solve problems requiring exact answers without a calculator
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.
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 $\cos 150°$.
$\cos 150°$: 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 $\sin 150°$?
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 $\tan \frac{4\pi}{3}$.
Convert radians to degrees: $\frac{4\pi}{3} = 240°$ (QIII).; Reference angle $= 240° - 180° = 60°$.
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.
In quadrant II, ___ is ___, while ___ is ___.
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 $\sin^2 30° + \cos^2 30° + \tan 45°$.
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?
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.
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 $\sin 60° > \sin 30°$.
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.
Work these through step-by-step. Use exact values only — no calculators.
Find the exact value of $\sin 120°$.
Reference angle = $60°$, QII where sine is positive. $\sin 120° = +\sin 60° = \frac{\sqrt{3}}{2}$.
Find the exact value of $\cos \frac{5\pi}{4}$.
$\frac{5\pi}{4} = 225°$, reference angle = $45°$, QIII where cosine is negative. $\cos 225° = -\cos 45° = -\frac{\sqrt{2}}{2}$.
Find the exact value of $\tan 300°$.
Reference angle = $60°$, QIV where tangent is negative. $\tan 300° = -\tan 60° = -\sqrt{3}$.
Simplify $\sin^2 45° + \cos^2 45°$.
By the Pythagorean identity, $\sin^2\theta + \cos^2\theta = 1$ for any angle. Alternatively, $\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac12 + \frac12 = 1$.
Find the exact value of $\sin 135° + \cos 135°$.
$\sin 135° = +\sin 45° = \frac{\sqrt{2}}{2}$ (QII, sine positive). $\cos 135° = -\cos 45° = -\frac{\sqrt{2}}{2}$ (QII, cosine negative). Sum = $\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$.
Return to your original answer from Section 01. A square with side length 2 cm has a diagonal forming two $45°$-$45°$-$90°$ triangles. By Pythagoras:
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.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Exact value in quadrant IV
Find the exact value of $\cos 330°$.
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Working:
$330°$ is in quadrant IV. The reference angle is:
$\cos 30° = \frac{\sqrt{3}}{2}$. In quadrant IV, cosine is positive.
Answer: $\cos 330° = \mathbf{\frac{\sqrt{3}}{2}}$
Simplify an exact-value expression
Find the exact value of $2\sin 60° \cos 60°$.
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Working:
Substitute exact values: $\sin 60° = \frac{\sqrt{3}}{2}$ and $\cos 60° = \frac12$
Answer: $\mathbf{\frac{\sqrt{3}}{2}}$
Notice: this is exactly $\sin 60°$. In fact, this is a special case of the double-angle formula $\sin 2\theta = 2\sin\theta\cos\theta$.
Proving a relationship
Without using a calculator, show that $\tan 60° - \tan 30° = \frac{2\sqrt{3}}{3}$.
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Working:
Substitute exact values: $\tan 60° = \sqrt{3}$ and $\tan 30° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
Common denominator:
Hence shown: $\tan 60° - \tan 30° = \mathbf{\frac{2\sqrt{3}}{3}}$
Key step: rationalise $\frac{1}{\sqrt{3}}$ to $\frac{\sqrt{3}}{3}$ before subtracting. This makes the common denominator obvious.
Five timed questions on exact values and special triangles. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
Enter the arenaJump up the platform while answering Module 2 questions. Quick recall, lighter than the boss.
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