Finding Exact Trigonometric Values Using Special Triangles and the Unit Circle (Step-by-Step Guide)

When your teacher asks for the exact value of a trigonometric function, they are not looking for a decimal from your calculator.

They want answers like:

• √3/2

• −√2/2

• 0, 1, or −1

In Canadian high school math, you usually get three kinds of angles where exact values are expected (in degrees):

1. Angles that come from the two special right triangles

2. Quadrantal angles (0°, 90°, 180°, 270°, 360° and coterminals)

3. Angles built from these using identities (sum and difference, double-angle, etc.)

   
   

This post will focus on the first two, and then explain what happens with angles that do not fit those patterns.

1. Using the two special triangles

The two special right triangles are:

The one on the left in the above image is called the 30°–60°–90° triangle.

The one on the right in the above image is called the 45°–45°–90° triangle.

We use a consistent three-step strategy:

Step 1: Figure out the quadrant

Use the angle to decide whether you are in:

• Quadrant I: 0° to 90°

• Quadrant II: 90° to 180°

• Quadrant III: 180° to 270°

• Quadrant IV: 270° to 360°

  
  

Then use the CAST rule to remember signs:

• Quadrant I: all trig functions are positive

• Quadrant II: sine is positive only

• Quadrant III: tangent is positive only

• Quadrant IV: cosine is positive only

  
  

In other words: C A S T if you go around the circle clockwise from Quadrant IV.

Step 2: Find the reference angle

The reference angle is the acute angle (between 0° and 90°) that your terminal arm makes with the x-axis.

For angles between 0° and 360°:

• QI: reference angle = θ

• QII: reference angle = 180° − θ

• QIII: reference angle = θ − 180°

• QIV: reference angle = 360° − θ

  
  

That reference angle will be 30°, 45°, or 60° in the typical exact-value questions where special triangles are used.

Step 3: Use the special triangle for the magnitude, and the quadrant for the sign

1. Use the reference angle and the correct triangle to get the magnitude (size) of the value.

2. Use the quadrant to decide if the final answer is positive or negative.

Worked Examples with Special Triangles

Example 1 (Quadrant I)

e.g. 1) Find the exact value of cos(30°)

Step 1: Quadrant

30° is in Quadrant I. All trig functions are positive here.

Step 2: Reference angle

Because the angle is already between 0° and 90°, the reference angle is 30°.

Step 3: Special triangle and sign

From the 30°–60°–90° triangle,

cos(30°) = adjacent / hypotenuse = √3/2

Quadrant I gives a positive value, which matches what we already have; therefore,

cos(30°) = √3/2

Here, step 3 is “redundant” in the sense that you already knew the sign would be positive from Quadrant I.

Example 2 (Quadrant II, III, or IV)

e.g. 2) Find the exact value of sin(150°)

Step 1: Quadrant

150° is between 90° and 180°, so it is in Quadrant II.

In Quadrant II, sine is positive, cosine and tangent are negative.

Step 2: Reference angle

Reference angle = 180° − 150° = 30°

Step 3: Special triangle and sign

The reference angle is 30°. From the 30°–60°–90° triangle,

sin(30°) = 1/2

Since 150° is in Quadrant II and sine is positive there,

sin(150°) = +1/2

You could do the same idea for an angle like 225° (Quadrant III, based on 45°) or 330° (Quadrant IV, based on 30°) using the same steps.

Example 3 (Angle bigger than 360°, use a coterminal

When an angle is larger than 360°, you first bring it back into the interval from 0° to 360° by subtracting 360° until it lands in that range.

e.g. 3) Find the exact value of sin(405°)

Step 0: Reduce the angle using coterminals

Subtract 360° and we get:

405° − 360° = 45°

So 405° is coterminal with 45°, which means sin(405°) = sin(45°)

Now proceed with 45°

Step 1: Quadrant

45° is in Quadrant I. All trig functions are positive.

Step 2: Reference angle

Reference angle is 45°.

Step 3: Special triangle and sign

From the 45°–45°–90° triangle,

sin(45°) = √2/2

Quadrant I keeps the value positive.

So sin(405°) = √2/2

Example 4 (Negative angle, use coterminals again)

For negative angles, you add 360° until the angle is between 0° and 360°. Then you follow the same three steps.

e.g. 4) Find the exact value of cos(-210°)

Step 0: Increase the angle using coterminals

Add 360° to make the angle positive:

-210° + 360° = 150°

so cos(-210°) = cos(150°)

Now proceed with 150°

Step 1: Quadrant

150° is in Quadrant II. In Quadrant II, cosine is negative.

Step 2: Reference angle

Reference angle = 180° − 150° = 30°

Step 3: Special triangle and sign

From the 30°–60°–90° triangle,

cos(30°) = √3/2

Since cosine is negative in Quadrant II,

cos(150°) = −√3/2

So

cos(-210°) = −√3/2

2. Quadrantal angles and the unit circle

The second major category is quadrantal angles. These are angles whose terminal arm lands exactly on one of the axes:

• 0°

• 90°

• 180°

• 270°

• 360°

• Any coterminal angles that differ from these by multiples of 360° (for example 450°, -90°, 720°, etc.).

On the unit circle (a circle with radius 1, centered at the origin), the key points are:

• At 0° (and 360°): (1, 0)

• At 90°: (0, 1)

• At 180°: (−1, 0)

• At 270°: (0, −1)

  
  

On the unit circle:

• cosθ is the x-coordinate

• sinθ is the y-coordinate

• tanθ = sinθ/cosθ

From this, we get:

sin(0°) = 0, cos(0°) = 1, tan(0°) = 0

sin(90°) = 1, cos(90°) = 0, tan(90°) is undefined

sin(180°) = 0, cos(180°) = −1, tan(180°) = 0

sin(270°) = −1, cos(270°) = 0, tan(270°) is undefined

You can also handle coterminal quadrantal angles the same way as in the earlier examples by adding or subtracting 360° until you land on one of these.

3. What about “weird” angles in Canadian high school courses?

In a standard Canadian high school curriculum (for example Pre-Calculus 11 and 12), exact values are normally expected for:

• special triangle angles (multiples of 30° or 45°)

• quadrantal angles (0°, 90°, 180°, 270°, 360°, and coterminals)

  
  

If the angle is not one of those, your course might either:

• let you use a calculator for a decimal approximation, or

• ask you to use identities to rewrite the angle in terms of special ones.

Common identities include:

• Sum and difference identities

  e.g.) sin(x + y) = sin(x)cos(y) + cos(x)sin(y)

• Double-angle identities

  e.g.) cos(2θ) = cos²(θ) − sin²(θ)

So an angle like 75° might be handled as:

• sin(75°) = sin(45° + 30°)

then you apply the sine addition formula and use the exact values of 45° and 30° from the special triangles.

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