What Are the Trigonometric Ratios of Specific Angles?
The trigonometric ratios of specific angles are the exact values that the six ratios — sine, cosine, tangent, cosecant, secant, and cotangent — take at the standard angles $0°, 30°, 45°, 60°,$ and $90°$. They are called specific (or standard) angles because their ratios can be written as exact fractions and surds rather than approximate decimals, which is why they appear in nearly every textbook derivation and exam question.
If the words opposite, adjacent, and hypotenuse aren't yet second nature, the six ratios are built directly from them — the full SOH-CAH-TOA picture lives in trigonometric ratios. Here we take those definitions as given and pin down the actual numbers.
Here is the master table. Memorise it eventually — but the rest of this article shows you how to rebuild it, so you never have to trust your memory alone.
$\theta$ | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ |
|---|---|---|---|---|---|
$\sin\theta$ | $0$ | $\frac{1}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{\sqrt{3}}{2}$ | $1$ |
$\cos\theta$ | $1$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{2}$ | $0$ |
$\tan\theta$ | $0$ | $\frac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | undefined |
$\csc\theta$ | undefined | $2$ | $\sqrt{2}$ | $\frac{2}{\sqrt{3}}$ | $1$ |
$\sec\theta$ | $1$ | $\frac{2}{\sqrt{3}}$ | $\sqrt{2}$ | $2$ | undefined |
$\cot\theta$ | undefined | $\sqrt{3}$ | $1$ | $\frac{1}{\sqrt{3}}$ | $0$ |
The pattern that locks the sine row in. Read the sine row left to right and rewrite each value over $2$: $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$. The numerators count $\sqrt{0}$ through $\sqrt{4}$. The cosine row is the same five values in reverse, and tangent is just $\frac{\sin\theta}{\cos\theta}$. Reconstruct sine and cosine, and the other four ratios fall out for free.
Where do these exact values come from?
Two triangles and one circle generate the whole table. The $45°$ column comes from a right isosceles triangle; the $30°$ and $60°$ columns come from half an equilateral triangle; the $0°$ and $90°$ edge values come from the unit circle, where they are read off as coordinates. The next three sections build each in turn — define the reference triangle before trusting the number, and the table stops being something you memorise and starts being something you can derive.
How Are the Values at 45° Derived?
Start with a right-angled triangle whose two legs are equal — a $45$-$45$-$90$ triangle. If each leg has length $1$, the hypotenuse follows from the Pythagorean theorem:
$$\text{hypotenuse} = \sqrt{1^2 + 1^2} = \sqrt{2}$$
For the $45°$ angle, the opposite and adjacent sides are both $1$, and the hypotenuse is $\sqrt{2}$. So:
$$\sin 45° = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}}$$
$$\cos 45° = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}}$$
$$\tan 45° = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{1} = 1$$
Because the triangle is symmetric, $\sin 45° = \cos 45°$ — the only special angle where the two are equal. The reciprocals follow at once: $\csc 45° = \sqrt{2}$, $\sec 45° = \sqrt{2}$, and $\cot 45° = 1$.
How Are the Values at 30° and 60° Derived?
These two angles share one triangle. Take an equilateral triangle with each side of length $2$ and drop a perpendicular from the top vertex to the base. That perpendicular cuts the triangle into two identical right triangles, each a $30$-$60$-$90$ triangle, and it bisects both the top angle (into $30°$) and the base (into a segment of length $1$).
The new right triangle has:
hypotenuse $= 2$ (the original side),
base $= 1$ (half of the bisected side),
height $= \sqrt{2^2 - 1^2} = \sqrt{3}$ (Pythagorean theorem).
Now read the ratios for $\angle = 30°$ (at the top, where the opposite side is the base of length $1$):
$$\sin 30° = \frac{1}{2}, \quad \cos 30° = \frac{\sqrt{3}}{2}, \quad \tan 30° = \frac{1}{\sqrt{3}}$$
And for $\angle = 60°$ (at the bottom, where opposite and adjacent swap):
$$\sin 60° = \frac{\sqrt{3}}{2}, \quad \cos 60° = \frac{1}{2}, \quad \tan 60° = \sqrt{3}$$
Notice that $\sin 30° = \cos 60°$ and $\cos 30° = \sin 60°$. That is no accident — $30°$ and $60°$ are complementary (they sum to $90°$), and the sine of one always equals the cosine of the other. That relationship is its own topic; see trigonometric ratios of complementary angles.
What Are the Values at 0° and 90°, and Why Is tan 90° Undefined?
The edge angles $0°$ and $90°$ have no proper triangle — a triangle with a $0°$ angle is squashed flat. So we read them off the unit circle instead, where the point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$.
At $0°$ the point sits at $(1, 0)$: so $\cos 0° = 1$ and $\sin 0° = 0$.
At $90°$ the point sits at $(0, 1)$: so $\cos 90° = 0$ and $\sin 90° = 1$.
Tangent is the quotient $\tan\theta = \frac{\sin\theta}{\cos\theta}$, and this is exactly where the "undefined" entries come from. At $90°$:
$$\tan 90° = \frac{\sin 90°}{\cos 90°} = \frac{1}{0}$$
Division by zero has no value — so $\tan 90°$ is undefined, not infinite and not zero. The same logic explains $\cot 0° = \frac{\cos 0°}{\sin 0°} = \frac{1}{0}$, and the undefined cosecant and secant entries: $\csc 0° = \frac{1}{\sin 0°} = \frac{1}{0}$ and $\sec 90° = \frac{1}{\cos 90°} = \frac{1}{0}$. Wherever a denominator hits $0$, the ratio is undefined.
In radians these five angles are $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$ — the same values, different units. A radian is the angle that subtends an arc equal in length to the radius; once you're comfortable with trigonometric ratios in radians the table reads identically.
Examples of Trigonometric Ratios of Specific Angles
Example 1
Evaluate $\sin 30° + \cos 60°$.
Read both values from the table:
$$\sin 30° = \frac{1}{2}, \quad \cos 60° = \frac{1}{2}$$
$$\sin 30° + \cos 60° = \frac{1}{2} + \frac{1}{2} = 1$$
Final answer: $1$.
Example 2
Evaluate $\tan 45° - \tan 30°$ without a calculator.
A first instinct is to treat $\tan$ as something you can split or subtract angle-wise — to write $\tan 45° - \tan 30° = \tan(45° - 30°) = \tan 15°$. Test that against the actual numbers. The right-hand side $\tan 15°$ is about $0.268$. The left-hand side uses the exact table values:
$$\tan 45° - \tan 30° = 1 - \frac{1}{\sqrt{3}} \approx 1 - 0.577 = 0.423$$
The two results disagree ($0.423 \neq 0.268$), so the shortcut is wrong: tangent does not distribute over subtraction of angles. The correct method is simply to substitute the table values and simplify.
$$\tan 45° - \tan 30° = 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}$$
Rationalising the denominator:
$$\frac{\sqrt{3} - 1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3 - \sqrt{3}}{3}$$
Final answer: $\frac{3 - \sqrt{3}}{3} \approx 0.423$.
Example 3
Show that $\sin^2 60° + \cos^2 60° = 1$.
Substitute the table values for $60°$:
$$\sin 60° = \frac{\sqrt{3}}{2}, \quad \cos 60° = \frac{1}{2}$$
$$\sin^2 60° + \cos^2 60° = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1$$
Final answer: the identity holds. This is the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ checked at a specific angle.
Example 4
Evaluate $\dfrac{\tan 60° - \tan 30°}{1 + \tan 60° \cdot \tan 30°}$.
Substitute $\tan 60° = \sqrt{3}$ and $\tan 30° = \frac{1}{\sqrt{3}}$, one move per line.
Numerator:
$$\sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$
Denominator:
$$1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 + 1 = 2$$
Divide:
$$\frac{2/\sqrt{3}}{2} = \frac{1}{\sqrt{3}}$$
Final answer: $\frac{1}{\sqrt{3}}$. (This is $\tan 30°$ — and the expression is the tangent-subtraction formula for $\tan(60° - 30°) = \tan 30°$, a nice cross-check.)
Example 5
A ramp makes a $30°$ angle with the ground and rises to a height of $2$ metres. How long is the ramp?
The height is the side opposite the $30°$ angle, and the ramp itself is the hypotenuse. The ratio linking opposite and hypotenuse is sine:
$$\sin 30° = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\text{ramp}}$$
Since $\sin 30° = \frac{1}{2}$:
$$\frac{1}{2} = \frac{2}{\text{ramp}} \implies \text{ramp} = 4 \text{ m}$$
Final answer: the ramp is $4$ metres long. This is the same right-triangle reasoning that powers heights and distances problems.
Example 6
Evaluate $4(\sin^4 30° + \cos^4 60°) - 3(\cos^2 45° - \sin^2 90°)$.
Pull every value from the table first: $\sin 30° = \frac{1}{2}$, $\cos 60° = \frac{1}{2}$, $\cos 45° = \frac{1}{\sqrt{2}}$, $\sin 90° = 1$.
First bracket:
$$\sin^4 30° + \cos^4 60° = \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4 = \frac{1}{16} + \frac{1}{16} = \frac{1}{8}$$
Second bracket:
$$\cos^2 45° - \sin^2 90° = \left(\frac{1}{\sqrt{2}}\right)^2 - 1^2 = \frac{1}{2} - 1 = -\frac{1}{2}$$
Combine:
$$4 \cdot \frac{1}{8} - 3 \cdot \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{3}{2} = 2$$
Final answer: $2$.
Why the Specific Angles Earn Their Keep
The reason these five angles dominate trigonometry is exactness — you can compute with them by hand and the answers stay clean. That single property is what makes them the workhorses of the subject.
They make every other identity checkable. When you meet $\sin^2\theta + \cos^2\theta = 1$ or a sum-difference formula, the fastest way to trust it is to plug in $30°$ or $45°$ and watch both sides match — a derivation, then a spot-check.
They are the entry point to real measurement. Surveying, the angle of elevation to the top of a building, ramp gradients, roof pitches — the first version a student solves uses a $30°$ or $45°$ angle precisely because the arithmetic stays exact while the method generalises to any angle.
They anchor the unit circle. Plot the five first-quadrant points and the shape of the sine and cosine curves becomes visible — these angles are the reference posts the whole circle is read against.
The destination this points toward is the full trigonometric table: once the five reference angles are exact in your head, extending to $120°$, $135°$, or $210°$ is just a reference-angle-plus-sign step away.
Where Students Slip on Specific-Angle Values
Mistake 1: Mixing up which value belongs to 30° and which to 60°
Where it slips in: Recalling the table under time pressure, where $30°$ and $60°$ sit next to each other.
Don't do this: Writing $\sin 60° = \frac{1}{2}$ because "$\frac{1}{2}$ feels like the easy one."
The correct way: The sine of the smaller angle is the smaller value: $\sin 30° = \frac{1}{2}$ and $\sin 60° = \frac{\sqrt{3}}{2}$. The single most common first-instinct error here is swapping $\sin 30°$ and $\sin 60°$ — anchoring to the $\frac{\sqrt{n}}{2}$ pattern (sine increases from $0°$ to $90°$) settles it every time.
Mistake 2: Treating tan 90° as zero or infinity
Where it slips in: Any expression that lands a $90°$ inside a tangent or a $0°$ inside a cotangent.
Don't do this: Writing $\tan 90° = 0$ (confusing it with $\tan 0°$) or quietly using $\infty$ as if it were a number you can add and multiply.
The correct way: $\tan 90°$ is undefined because $\cos 90° = 0$ and you cannot divide by zero. If a problem forces $\tan 90°$ to appear, that usually signals a setup error worth re-checking — the habit of writing the ratio as $\frac{\sin}{\cos}$ and looking at the denominator first is what catches it.
Mistake 3: Leaving an answer in unrationalised form when the standard form is expected
Where it slips in: Final answers that contain a surd in the denominator, like $\frac{1}{\sqrt{3}}$ or $\frac{1}{\sqrt{2}}$.
Don't do this: Assuming $\frac{1}{\sqrt{3}}$ and $\frac{\sqrt{3}}{3}$ are interchangeable in every marking scheme — they are equal in value, but graders often want the rationalised form.
The correct way: Multiply top and bottom by the surd: $\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$. Both are correct values; rationalising the denominator is the conventional final form unless the question says otherwise.
Key Takeaways
The trigonometric ratios of specific angles are the exact sin, cos, tan, csc, sec, and cot values at $0°, 30°, 45°, 60°,$ and $90°$.
The $45°$ values come from a $45$-$45$-$90$ triangle; the $30°$ and $60°$ values from half an equilateral triangle; the $0°$ and $90°$ values are read as coordinates on the unit circle.
The sine row follows the $\frac{\sqrt{0}}{2}$ to $\frac{\sqrt{4}}{2}$ pattern, and the cosine row is its reverse — rebuild these two and the rest follow.
$\tan 90°$, $\cot 0°$, $\csc 0°$, and $\sec 90°$ are undefined because each has a zero in the denominator.
These five angles matter because their values are exact, which makes them the testing ground for every identity and the entry point to real measurement.
Practice Before Moving On
Work through these and verify each answer against the table.
Evaluate $\cos 0° + \sin 90°$.
Evaluate $\dfrac{\tan 30°}{1 - \tan^2 30°}$ and identify which standard ratio it equals.
Show that $2\sin 30° \cos 30° = \sin 60°$.
Answer to Question 1: $1 + 1 = 2$.
Answer to Question 2: substituting $\tan 30° = \frac{1}{\sqrt{3}}$ gives $\frac{1/\sqrt{3}}{1 - 1/3} = \frac{1/\sqrt{3}}{2/3} = \frac{\sqrt{3}}{2}$… which simplifies to $\frac{1}{\sqrt{3}} \cdot \frac{3}{2} = \frac{\sqrt{3}}{2}$; this equals $\tan 60° \div 2$, and is the half-angle route to $\tan 60°$ — re-check the algebra if you didn't land on $\frac{\sqrt{3}}{2}$.
Answer to Question 3: $2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} = \sin 60°$.
Once these five angles are exact in your head, the natural next step is extending them across all four quadrants. To build that fluency with a live teacher, explore Bhanzu's trigonometry tutor, the high school math tutor track, or flexible math classes online.
Want a Bhanzu trainer to walk through the derivation triangles with your child until the table becomes automatic? Book a free demo class.
Read More
Trigonometric functions — extending the special-angle values to any angle on the unit circle.
Sin cos tan — the three primary ratios and how they connect.
Trigonometry formulas — the full reference list of trigonometric formulas.
What is a radian — the angle unit behind the radian forms of these values.
Trigonometric identities — the identities the special-angle values verify.
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