Back to Use special triangles and unit circle

Exercises: Exact Trigonometric Values from Special Triangles

Use special triangles and unit circle symmetry. Express all answers in exact form (no decimals).

Grade 9·22 problems·~30 min·Common Core Math - HS Functions·standard·hsf-tf-a-3

Work through problems with immediate feedback

Warm-Up: Review What You Know

These problems review skills you have already learned.

A right triangle has legs of length 3 and 4. What is the length of its hypotenuse?

$\sqrt{7}$

$\sqrt{25}$

$Unit circle with a first-quadrant point P at (a, b), showing horizontal distance a and vertical distance b$

A point $P$ lies on the unit circle at angle $\theta$ . Its coordinates are $(a, b)$ . Which statements are correct?

$\cos\theta = a$ and $\sin\theta = b$

$\cos\theta = b$ and $\sin\theta = a$

$\cos\theta = \frac{a}{b}$ and $\sin\theta = \frac{b}{a}$

$\cos\theta = \frac{1}{a}$ and $\sin\theta = \frac{1}{b}$

The terminal side of angle $\frac{5\pi}{4}$ lies in the third quadrant. What is its reference angle? Express your answer in the form $\frac{a\pi}{b}$ (e.g., pi/4).

Fluency Practice

Find the exact value. Show the special triangle you used.

$45-45-90 triangle inscribed in the unit circle with legs sqrt(2)/2 and hypotenuse 1$

A 45-45-90 triangle is placed in the unit circle with its hypotenuse along the radius to angle $\frac{\pi}{4}$ . Each leg has length $\frac{\sqrt{2}}{2}$ . What is $\cos\!\left(\frac{\pi}{4}\right)$ ?

$30-60-90 triangle inscribed in the unit circle with the 30-degree angle at origin, vertical leg 1/2, horizontal leg sqrt(3)/2$

An equilateral triangle with side length 2 is bisected to form a 30-60-90 triangle with sides 1, $\sqrt{3}$ , and 2. When scaled to hypotenuse 1 and placed in the unit circle at angle $\frac{\pi}{6}$ , what is $\sin\!\left(\frac{\pi}{6}\right)$ ?

Using the 30-60-90 triangle on the unit circle, find the exact value of $\cos\!\left(\frac{\pi}{3}\right)$ .

Find the exact value of $\tan\!\left(\frac{\pi}{6}\right)$ . Express your answer as a fraction with a rational denominator.

Complete the unit circle coordinates for the three first-quadrant special angles:

Angle $\frac{\pi}{6}$ : point $= \left(\_\_\_, \, \_\_\_\right)$
Angle $\frac{\pi}{4}$ : point $= \left(\_\_\_, \, \_\_\_\right)$
Angle $\frac{\pi}{3}$ : point $= \left(\_\_\_, \, \_\_\_\right)$

cos(pi/6):

sin(pi/6):

cos(pi/4):

sin(pi/4):

cos(pi/3):

sin(pi/3):

Varied Practice

Which of the following gives the correct exact value of $\sin\!\left(\frac{\pi}{3}\right)$ ?

$\frac{1}{2}$

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

$\frac{\sqrt{2}}{2}$

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

A student claims that $\tan\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ . Is this correct?

No — $\tan\!\left(\frac{\pi}{4}\right) = 1$

Yes — $\tan\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

No — $\tan\!\left(\frac{\pi}{4}\right) = \sqrt{2}$

No — $\tan\!\left(\frac{\pi}{4}\right) = 2$

$Unit circle showing the reflection of pi/4 across the y-axis to give 3pi/4, with x-coordinates sqrt(2)/2 and -sqrt(2)/2$

Using the symmetry relation $\cos(\pi - x) = -\cos(x)$ , find the exact value of $\cos\!\left(\frac{3\pi}{4}\right)$ .

Find the exact value of $\sin\!\left(\frac{7\pi}{6}\right)$ by identifying its reference angle and quadrant.

A point at angle $x$ in the first quadrant has coordinates $(a, b)$ . A classmate says: "The point at angle $\pi + x$ has coordinates $(-a, b)$ because we negate the x-coordinate when moving to a new quadrant." Is the classmate correct? Explain using the unit circle.

Word Problems

$Unit circle showing all three first-quadrant special angle coordinates at pi/6, pi/4, and pi/3$

A surveyor uses a clinometer to measure angles. At angle $\frac{\pi}{3}$ from the horizontal, the instrument's unit-radius dial reads coordinates directly from the unit circle.

At angle $\frac{\pi}{3}$ , what exact y-coordinate (sine value) does the dial read?

The surveyor then rotates to angle $\frac{\pi}{6}$ . What exact x-coordinate (cosine value) does the dial read?

A ramp rises at an angle of $\frac{\pi}{4}$ from the horizontal. The ramp's length (hypotenuse) is 10 meters.

What is the exact vertical rise of the ramp in meters?

$Unit circle showing four positions at pi/6, 5pi/6, 7pi/6, and 11pi/6 with their coordinates$

A Ferris wheel has radius 1 unit and a passenger's position is modeled by the unit circle. The passenger starts at angle $\frac{\pi}{6}$ and the wheel rotates to new positions.

After the wheel rotates to angle $\frac{5\pi}{6}$ , what is the passenger's exact height (y-coordinate)?

At angle $\frac{11\pi}{6}$ , what is the passenger's exact height (y-coordinate)?

Error Analysis

Each problem shows a student's work that contains an error. Identify the mistake.

$Two-column card comparing Taylor's swapped values for sin(pi/6) and sin(pi/3) with the correct values$

Taylor evaluated $\sin\!\left(\frac{\pi}{6}\right)$ and $\sin\!\left(\frac{\pi}{3}\right)$ :

$\sin\!\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin\!\left(\frac{\pi}{3}\right) = \frac{1}{2}$

What mistake did Taylor make?

Taylor swapped the sine values for $\frac{\pi}{6}$ and $\frac{\pi}{3}$

Taylor confused sine with cosine

Taylor used $\sqrt{3}$ instead of $\frac{\sqrt{3}}{2}$

Taylor used degrees instead of radians

$Two-column card showing Jordan's error of negating sin when using the pi-minus symmetry relation$

Jordan evaluated $\sin\!\left(\frac{5\pi}{6}\right)$ using the symmetry relation:

$\frac{5\pi}{6} = \pi - \frac{\pi}{6}$
$\sin\!\left(\frac{5\pi}{6}\right) = -\sin\!\left(\frac{\pi}{6}\right) = -\frac{1}{2}$

What error did Jordan make when applying the symmetry relation $\sin(\pi - x) = {?}$

Jordan negated $\sin\!\left(\frac{\pi}{6}\right)$ incorrectly; the correct relation is $\sin(\pi - x) = +\sin(x)$

Jordan used the wrong reference angle — it should be $\frac{\pi}{3}$ , not $\frac{\pi}{6}$

Jordan applied the relation for $\cos(\pi - x)$ instead of $\sin(\pi - x)$

Jordan's calculation is correct

Challenge / Extension

These problems extend the core ideas. Try them if you've finished the rest.

Find the exact value of $\tan\!\left(\frac{11\pi}{6}\right)$ . Express your answer as a fraction with a rational denominator (or as an integer if it simplifies).

Without using the $\sqrt{n}/2$ pattern, derive the exact value of $\sin\!\left(\frac{\pi}{4}\right)$ from scratch using only the Pythagorean theorem and the definition of the unit circle. Explain each step.

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