D1.1 Determine exact trigonometric ratios for special angles

Ontario Grade 11 Mathematics

Trigonometric Functions

Finding sin, cos, and tan without a calculator — MCR3U Unit D

When you press sin(30°) on a calculator, you get 0.5. That decimal is exact, but for many other angles the calculator gives a rounded decimal — not the true value. In this lesson you will learn to write trigonometric ratios as exact fractions or expressions involving square roots. These exact values appear throughout MCR3U whenever you work with functions, equations, or identities, so building fluency now saves a lot of effort later. No calculator is needed once you know the two special triangles.

What you will learn

Identify the two special triangles (30–60–90 and 45–45–90) and explain why their side lengths are exact.
State the exact values of sine, cosine, and tangent for 30°, 45°, and 60°.
Apply exact trigonometric ratios to evaluate expressions and solve problems without a calculator.
Recognize when an angle of 0° or 90° appears and state the corresponding exact ratio from the unit definition.

Prerequisite Bridge: Right Triangles and the Primary Trig Ratios

From Grade 10 you know that in any right triangle, the three primary trigonometric ratios connect an acute angle to the sides of the triangle. Label the sides relative to angle

\theta

: the side directly across from

\theta

is the opposite side, the side next to

\theta

(not the hypotenuse) is the adjacent side, and the longest side (across from the right angle) is the hypotenuse.

The three ratios are:

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}

,

\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}

, and

\tan\theta = \frac{\text{opposite}}{\text{adjacent}}

. The memory device SOH-CAH-TOA still applies.

You also need the Pythagorean theorem: in a right triangle with legs

a

and

b

and hypotenuse

c

, we have

a^2 + b^2 = c^2

. This theorem is what lets us find exact side lengths — and therefore exact trig ratios — for specific triangles.

\sin\theta = \frac{\text{opp}}{\text{hyp}}, \cos\theta = \frac{\text{adj}}{\text{hyp}}, \tan\theta = \frac{\text{opp}}{\text{adj}}

SOH-CAH-TOA defines the three primary ratios relative to a chosen acute angle.
The Pythagorean theorem connects all three sides of a right triangle.
Exact values come from exact side lengths, not decimal approximations.

The 45–45–90 Triangle

Start with a square that has side length 1. Draw one diagonal. The diagonal cuts the square into two right triangles. Each triangle has two legs of length 1 and two equal angles of 45°. Using the Pythagorean theorem, the hypotenuse has length

\sqrt{1^2 + 1^2} = \sqrt{2}

.

Fix one of the 45° angles as

\theta

. The side opposite

\theta

has length 1, the side adjacent to

\theta

has length 1, and the hypotenuse has length

\sqrt{2}

. Substituting into SOH-CAH-TOA gives the three exact ratios for 45°.

Notice that

\sin 45° = \cos 45°

because the triangle is isosceles — the opposite and adjacent sides are equal. Also,

\tan 45° = 1

because opposite equals adjacent, so their ratio is exactly 1. These are clean, memorable results.

\sin 45° = \frac{\sqrt{2}}{2}, \cos 45° = \frac{\sqrt{2}}{2}, \tan 45° = 1

The 45–45–90 triangle comes from halving a unit square along its diagonal.
Side lengths are in the ratio $1 : 1 : \sqrt{2}$ .
$\sin 45° = \cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ and $\tan 45° = 1$ .
Rationalizing the denominator: $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ — both forms are correct and exact.

The 30–60–90 Triangle

Start with an equilateral triangle — all three sides equal 2, all three angles equal 60°. Draw the perpendicular from one vertex to the opposite side. This line is both an altitude and a line of symmetry, so it cuts the equilateral triangle into two identical right triangles. Each right triangle has angles of 30°, 60°, and 90°.

In each right triangle, the hypotenuse is 2 (a full side of the equilateral triangle) and the shortest leg is 1 (half of the base, which was 2). The Pythagorean theorem gives the remaining leg:

1^2 + b^2 = 2^2

, so

b^2 = 3

and

b = \sqrt{3}

. The side lengths are in the ratio

1 : \sqrt{3} : 2

.

For the 30° angle: opposite = 1, adjacent =

\sqrt{3}

, hypotenuse = 2. For the 60° angle: opposite =

\sqrt{3}

, adjacent = 1, hypotenuse = 2. Applying SOH-CAH-TOA to each angle gives the six exact values shown in the table below. Notice that

\sin 30° = \cos 60°

and

\sin 60° = \cos 30°

— the two angles are complementary (they add to 90°), so this swap always happens.

\sin 60° = \frac{\sqrt{3}}{2}, \cos 60° = \frac{1}{2}, \tan 60° = \sqrt{3}

The 30–60–90 triangle comes from halving an equilateral triangle.
Side lengths are in the ratio $1 : \sqrt{3} : 2$ .
$\sin 30° = \cos 60° = \frac{1}{2}$ and $\sin 60° = \cos 30° = \frac{\sqrt{3}}{2}$ .
$\tan 30° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ and $\tan 60° = \sqrt{3}$ .
Complementary angles (summing to 90°) always have their sine and cosine swapped.

The Boundary Angles: 0° and 90°

The angles 0° and 90° are not inside a triangle in the usual sense — a triangle cannot have an angle of 0° or 90° while still being a valid right triangle with three distinct sides. Instead, think of what happens to the opposite and adjacent sides as the angle shrinks toward 0° or grows toward 90°.

As

\theta

approaches 0°, the opposite side shrinks to 0 while the hypotenuse stays fixed. So

\sin 0° = 0

and

\cos 0° = 1

. Since opposite = 0,

\tan 0° = 0

as well.

As

\theta

approaches 90°, the opposite side grows until it equals the hypotenuse, giving

\sin 90° = 1

and

\cos 90° = 0

. Because the adjacent side becomes 0,

\tan 90°

is undefined — you cannot divide by zero. These four boundary values appear often in graphing and identities, so memorize them alongside the triangle values.

\sin 0° = 0, \cos 0° = 1, \sin 90° = 1, \cos 90° = 0

$\sin 0° = 0$ , $\cos 0° = 1$ , $\tan 0° = 0$ .
$\sin 90° = 1$ , $\cos 90° = 0$ , $\tan 90°$ is undefined.
Boundary values follow from what happens to the sides at the extremes, not from a triangle construction.

Using Exact Ratios in Expressions and Equations

Once you know the exact values, you can evaluate trigonometric expressions by substitution and then simplify. The key skill is recognizing which special angle appears, recalling the correct ratio, and then carrying out exact arithmetic — no rounding at any stage.

For example, to evaluate

2\sin 60° - \cos 30°

, substitute

\sin 60° = \frac{\sqrt{3}}{2}

and

\cos 30° = \frac{\sqrt{3}}{2}

, giving

2 \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}

. The answer is exact.

A slightly harder task is verifying or using a trigonometric identity at a special angle. Because the ratios are exact, you can check both sides of an identity numerically without any rounding error. This technique appears later in the course when you study identities and transformations of trig functions.

Substitute the exact fraction or surd, then simplify using ordinary fraction arithmetic.
Never round intermediate steps — the whole point is to keep the answer exact.
Recognizing the special angle quickly is a skill that improves with practice.

Exact Trigonometric Ratios for Special Angles

Angle	sin	cos	tan
0°	0	1	0
30°	1/2	√3/2	√3/3
45°	√2/2	√2/2	1
60°	√3/2	1/2	√3
90°	1	0	undefined

Worked example

Evaluating an Expression Using Special Angle Ratios

Without a calculator, find the exact value of

E = \tan 60° \cdot \cos 45° + \sin 30°

.

Identify each special angle and recall its ratio
List the three ratios needed. From the 30–60–90 triangle, $\tan 60° = \sqrt{3}$ . From the 45–45–90 triangle, $\cos 45° = \frac{\sqrt{2}}{2}$ . From the 30–60–90 triangle, $\sin 30° = \frac{1}{2}$ .
$\tan 60° = \sqrt{3}, \cos 45° = \frac{\sqrt{2}}{2}, \sin 30° = \frac{1}{2}$
Substitute the exact values into the expression
Replace each trigonometric ratio in $E$ with its exact value.
$E = \sqrt{3} · \frac{\sqrt{2}}{2} + \frac{1}{2}$
Multiply the first term
Multiply $\sqrt{3}$ by $\frac{\sqrt{2}}{2}$ . Because $\sqrt{3} \cdot \sqrt{2} = \sqrt{6}$ , the product is $\frac{\sqrt{6}}{2}$ .
$\sqrt{3} · \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{2}$
Add the two fractions
Both terms already share the denominator 2, so add the numerators directly.
$E = \frac{\sqrt{6}}{2} + \frac{1}{2} = \frac{\sqrt{6} + 1}{2}$

Answer:

E = \frac{\sqrt{6} + 1}{2}

Check: Approximate numerically:

\sqrt{6} \approx 2.449

, so

\frac{2.449 + 1}{2} \approx 1.725

. Using a calculator directly:

\tan 60° \times \cos 45° + \sin 30° \approx 1.7321 \times 0.7071 + 0.5 \approx 1.2247 + 0.5 = 1.7247

. The values match (rounding accounts for the small difference), confirming the exact answer is correct.

Worked example

Finding a Missing Side Using an Exact Ratio

A ramp makes an angle of 30° with the ground. The ramp is 8 m long. Find the exact height the ramp rises above the ground.

Draw and label the triangle
The ramp is the hypotenuse (length 8 m). The height is the side opposite the 30° angle. The ground is the adjacent side. Sine connects the opposite side to the hypotenuse, so use $\sin 30°$ .
$\sin 30° = \frac{\text{height}}{8}$
Recall the exact value of sin 30°
From the 30–60–90 triangle, $\sin 30° = \frac{1}{2}$ .
$\frac{1}{2} = \frac{\text{height}}{8}$
Solve for the height
Multiply both sides by 8 to isolate the height.
$\text{height} = 8 × \frac{1}{2} = 4$

Answer: The ramp rises exactly 4 m above the ground.

Check: Verify with the Pythagorean theorem: the adjacent side should be

\sqrt{8^2 - 4^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}

. Check

\cos 30°

:

\frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}

, which matches the known exact value. Everything is consistent.

Common mistakes and how to avoid them

Swapping sin and cos for 30° and 60° — writing sin 60° = 1/2 instead of √3/2.

Correction: Remember the larger angle (60°) has the larger sine value. Since √3/2 ≈ 0.866 > 1/2 = 0.5, sin 60° must be the bigger number, √3/2.

Leaving an irrational number in the denominator without rationalizing, e.g. writing tan 30° = 1/√3 as a final answer when the rationalized form √3/3 is expected.

Correction: Multiply numerator and denominator by √3: 1/√3 × √3/√3 = √3/3. Both are mathematically equal, but rationalized form is standard.

Stating that tan 90° = 0 or that it equals a very large number.

Correction: tan 90° is undefined because it requires dividing by cos 90° = 0, and division by zero is never defined.

Using a rounded decimal (e.g. 0.866) instead of an exact surd (√3/2) when the question asks for an exact value.

Correction: An exact value must be written as a fraction, integer, or expression with surds. A decimal rounded to any number of places is not exact.

Confusing which side is opposite and which is adjacent when the triangle is drawn in an unfamiliar orientation.

Correction: Always label opposite, adjacent, and hypotenuse relative to the angle you are working with, not relative to the page orientation.

Lesson summary

The 45–45–90 triangle, built by halving a unit square, has sides in the ratio 1 : 1 : √2, giving sin 45° = cos 45° = √2/2 and tan 45° = 1.
The 30–60–90 triangle, built by halving an equilateral triangle of side 2, has sides in the ratio 1 : √3 : 2, giving all six exact ratios for 30° and 60°.
The boundary values sin 0° = 0, cos 0° = 1, sin 90° = 1, cos 90° = 0, and tan 90° undefined follow from the geometry of the sides at the extremes.
Exact values are fractions or surds — never rounded decimals. Rationalize denominators when a surd appears in the denominator.
To evaluate an expression, identify each special angle, substitute the exact ratio, and simplify with fraction arithmetic.
Recognizing complementary pairs (30° and 60°, which add to 90°) explains why sin and cos swap between those two angles.

Check your understanding

Question 1

What is the exact value of

\cos 30°

?

$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$
$\frac{\sqrt{2}}{2}$
$\sqrt{3}$

Show answer and explanation

\frac{\sqrt{3}}{2}

In the 30–60–90 triangle with sides 1, √3, and 2, the adjacent side to 30° is √3 and the hypotenuse is 2, so cos 30° = √3/2. The value 1/2 is cos 60°, not cos 30°.

Question 2

Which expression equals

\tan 45°

?

$\frac{\sqrt{3}}{3}$
$\frac{\sqrt{2}}{2}$
$\sqrt{3}$
$1$

Show answer and explanation

1

In the 45–45–90 triangle the opposite and adjacent sides are both 1, so tan 45° = 1/1 = 1. The other options are tan 30°, cos 45°, and tan 60° respectively.

Question 3

A ladder leans against a wall at 60° to the ground. The ladder is 6 m long. What is the exact vertical height it reaches up the wall?

$2\sqrt{3}$ m
$3$ m
$3\sqrt{3}$ m
$6\sqrt{3}$ m

Show answer and explanation

3\sqrt{3}

m

The height is opposite the 60° angle and the ladder is the hypotenuse, so height = 6 × sin 60° = 6 × (√3/2) = 3√3 m. Option B (3 m) uses sin 30° by mistake; option A uses half of 2√3 incorrectly; option D multiplies incorrectly.

Question 4

Which of the following is undefined?

$\sin 90°$
$\cos 0°$
$\tan 90°$
$\tan 0°$

Show answer and explanation

\tan 90°

tan 90° = sin 90° / cos 90° = 1/0, and division by zero is undefined. The other three values are all defined: sin 90° = 1, cos 0° = 1, and tan 0° = 0.

Key terms

Exact value: A value written as an integer, fraction, or expression with square roots — with no rounding at any stage.
Special angle: One of the angles 0°, 30°, 45°, 60°, or 90°, whose trigonometric ratios can be written as exact values.
Hypotenuse: The longest side of a right triangle, always opposite the 90° angle.
Opposite side: The side of a right triangle that is directly across from the angle being considered.
Adjacent side: The side of a right triangle that is next to the angle being considered, but is not the hypotenuse.
Surd: An exact expression that contains an unresolved square root, such as √2 or √3.
Rationalize the denominator: Rewrite a fraction so that no square root appears in the denominator, by multiplying top and bottom by the same surd.
Complementary angles: Two angles that add to exactly 90°. For complementary angles, the sine of one equals the cosine of the other.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation D1.1. It is a study resource, not an official curriculum publication.

Official curriculum reference

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D1.1 · Determine exact trigonometric ratios for special angles

Ontario Grade 11 Mathematics

Finding sin, cos, and tan without a calculator — MCR3U Unit D

What you will learn

Prerequisite Bridge: Right Triangles and the Primary Trig Ratios

The 45–45–90 Triangle

The 30–60–90 Triangle

The Boundary Angles: 0° and 90°

Using Exact Ratios in Expressions and Equations

Exact Trigonometric Ratios for Special Angles

Evaluating an Expression Using Special Angle Ratios

Finding a Missing Side Using an Exact Ratio

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

About this lesson