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 confidencethat 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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