Trigonometry Problems with Solutions: Easy to Hard

These trigonometry problems are organized from foundational right-triangle ratios and the unit circle up through trigonometric identities, solving trigonometric equations, graphing sine and cosine functions, and the Law of Sines and Law of Cosines for oblique triangles. Each trigonometry exercise includes a guided hint and a complete step-by-step solution using SOH-CAH-TOA, Pythagorean identities, double-angle formulas, sum and difference formulas, and real-world angle of elevation and depression applications. Work through the sections in order for a pedagogically progressive experience, or jump directly to the topic you need.

Contents6 sections

Right Triangle Trigonometry and the SOH-CAH-TOA Ratios

Right triangle trigonometry is the entry point for all further study in this subject. In a right triangle, the three primary trigonometric ratios — sine, cosine, and tangent — relate each acute angle to the ratio of two sides. The acronym SOH-CAH-TOA summarizes these ratios: $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ . Mastering these ratios, along with the Pythagorean theorem, is essential before moving on to the unit circle and identities.

Problem 1: Finding a Missing Side Using Sine

Easy

In a right triangle $ABC$ , the right angle is at $C$ . The hypotenuse $AB = 15$ cm and $\angle A = 32°$ .

Find the length of side $BC$ (opposite to $\angle A$ ).
Find the length of side $AC$ (adjacent to $\angle A$ ).

Hint

Identify which sides are opposite, adjacent, and the hypotenuse with respect to

\angle A

. Then choose the correct ratio: sine uses opposite and hypotenuse; cosine uses adjacent and hypotenuse.

View Solution

Solution to question 1:

Side $BC$ is opposite $\angle A$ and $AB$ is the hypotenuse, so use the sine ratio:

\sin(32°) = \frac{BC}{AB} = \frac{BC}{15}

BC = 15 \times \sin(32°) \approx 15 \times 0.5299 \approx 7.95 \text{ cm}

Solution to question 2:

Side $AC$ is adjacent to $\angle A$ , so use the cosine ratio:

\cos(32°) = \frac{AC}{AB} = \frac{AC}{15}

AC = 15 \times \cos(32°) \approx 15 \times 0.8480 \approx 12.72 \text{ cm}

Problem 2: Area and Sides of a Right Triangle

Medium

A right triangle has an acute angle of $40°$ at vertex $A$ and hypotenuse $c = 20$ cm.

Find both legs $a$ (opposite) and $b$ (adjacent).
Calculate the area of the triangle using the two legs.
Verify the result by computing the area using the formula $\text{Area} = \frac{1}{2}ab\sin C$ where $C = 90°$ .

Hint

Find

a

and

b

using sine and cosine of

40°

. For the verification, recall that

\sin(90°) = 1

View Solution

Solution to question 1:

a = 20\sin(40°) \approx 20 \times 0.6428 \approx 12.86 \text{ cm}

b = 20\cos(40°) \approx 20 \times 0.7660 \approx 15.32 \text{ cm}

Solution to question 2:

\text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} \times 12.86 \times 15.32 \approx 98.5 \text{ cm}^2

Solution to question 3:

\text{Area} = \frac{1}{2}ab\sin(90°) = \frac{1}{2} \times 12.86 \times 15.32 \times 1 \approx 98.5 \text{ cm}^2

Both methods agree, confirming the calculation.

The Unit Circle, Reference Angles, and Exact Values

The unit circle is a circle of radius 1 centered at the origin and is one of the most important tools in trigonometry. Every point $(\cos\theta,\, \sin\theta)$ on the unit circle encodes the exact values of cosine and sine for angle $\theta$ . Knowing the special angles — $30°$ , $45°$ , $60°$ , and their equivalents in radians — allows you to find exact values without a calculator. Reference angles, coterminal angles, and the signs of trig functions in each quadrant (summarized by the rule All Students Take Calculus) round out this foundational skill.

Problem 3: Exact Values from the Unit Circle

Easy

Without using a calculator, find the exact value of each expression.

$\sin\!\left(\dfrac{5\pi}{6}\right)$
$\cos\!\left(-\dfrac{4\pi}{3}\right)$
$\tan(315°)$

Hint

Find the reference angle for each angle, determine its quadrant, and apply the correct sign. Remember: cosine is an even function, so

\cos(-\theta) = \cos\theta

View Solution

Solution to question 1:

$\frac{5\pi}{6}$ is in Quadrant II. Its reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$ . Sine is positive in Q II.

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

Solution to question 2:

Since cosine is even: $\cos\!\left(-\frac{4\pi}{3}\right) = \cos\!\left(\frac{4\pi}{3}\right)$ . The angle $\frac{4\pi}{3}$ is in Q III; its reference angle is $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$ . Cosine is negative in Q III.

\cos\!\left(-\frac{4\pi}{3}\right) = -\cos\!\left(\frac{\pi}{3}\right) = -\frac{1}{2}

Solution to question 3:

$315°$ is in Q IV; its reference angle is $360° - 315° = 45°$ . Tangent is negative in Q IV.

\tan(315°) = -\tan(45°) = -1

Problem 4: Coterminal Angles and Reciprocal Functions

Medium

Consider the angle $\theta = -\dfrac{11\pi}{4}$ .

Find a positive coterminal angle $\alpha$ with $0 \leq \alpha < 2\pi$ .
Use $\alpha$ to find the exact values of $\csc\theta$ , $\sec\theta$ , and $\cot\theta$ .

Hint

Add multiples of

2\pi

to the given angle until the result is in

[0, 2\pi)

. Recall the reciprocal identities:

\csc\theta = \frac{1}{\sin\theta}

\sec\theta = \frac{1}{\cos\theta}

\cot\theta = \frac{1}{\tan\theta}

View Solution

Solution to question 1:

-\frac{11\pi}{4} + 2(2\pi) = -\frac{11\pi}{4} + \frac{16\pi}{4} = \frac{5\pi}{4}

The positive coterminal angle is $\alpha = \dfrac{5\pi}{4}$ , which lies in Q III with reference angle $\dfrac{\pi}{4}$ .

Solution to question 2:

In Q III, sine and cosine are both negative:

\sin\!\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}, \quad \cos\!\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}, \quad \tan\!\left(\frac{5\pi}{4}\right) = 1

\csc\theta = \frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2}, \quad \sec\theta = -\sqrt{2}, \quad \cot\theta = \frac{1}{1} = 1

Problem 5: Using the Pythagorean Identity to Find All Trig Values

Hard

Suppose $\sin\theta = \dfrac{5}{13}$ and $\theta$ is in Quadrant II.

Use the Pythagorean identity to find $\cos\theta$ .
Find the exact values of all six trigonometric functions of $\theta$ .
If $\phi$ is the reference angle of $\theta$ , state the exact value of $\tan\phi$ .

Hint

Start with

\sin^2\theta + \cos^2\theta = 1

and substitute the known value of

\sin\theta

. The sign of

\cos\theta

is determined by the quadrant. Build the remaining four functions from sine and cosine using ratio and reciprocal identities.

View Solution

Solution to question 1:

\left(\frac{5}{13}\right)^2 + \cos^2\theta = 1 \implies \cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}

Since $\theta$ is in Q II, $\cos\theta < 0$ :

\cos\theta = -\frac{12}{13}

Solution to question 2:

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{5/13}{-12/13} = -\frac{5}{12}

\csc\theta = \frac{13}{5}, \quad \sec\theta = -\frac{13}{12}, \quad \cot\theta = -\frac{12}{5}

Solution to question 3:

The reference angle $\phi$ is the acute angle in Q I with the same sine magnitude, so $\sin\phi = \frac{5}{13}$ and $\cos\phi = \frac{12}{13}$ :

\tan\phi = \frac{5}{12}

Trigonometry Identities: Pythagorean, Reciprocal, and Ratio

A trigonometry identity is an equation that is true for all values of the variable in its domain. The Pythagorean identities ( $\sin^2\theta + \cos^2\theta = 1$ and its derived forms), the reciprocal identities, and the ratio identities form the core toolkit for simplifying expressions and verifying more complex identities. The standard strategy is to work on one side only, converting everything to sines and cosines when stuck.

Problem 6: Simplifying with Reciprocal and Ratio Identities

Easy

Simplify each expression to a single trigonometric function or a constant.

$\dfrac{\tan\theta}{\sec\theta}$
$\sin\theta \cdot \cot\theta + \cos\theta \cdot \tan\theta$

Hint

Rewrite every function in terms of

\sin\theta

and

\cos\theta

first. Most simplifications become straightforward after this substitution.

View Solution

Solution to question 1:

\frac{\tan\theta}{\sec\theta} = \frac{\sin\theta/\cos\theta}{1/\cos\theta} = \frac{\sin\theta}{\cos\theta} \cdot \cos\theta = \sin\theta

Solution to question 2:

\sin\theta \cdot \frac{\cos\theta}{\sin\theta} + \cos\theta \cdot \frac{\sin\theta}{\cos\theta} = \cos\theta + \sin\theta

Problem 7: Proving a Pythagorean Identity

Medium

Verify the following identities by transforming one side into the other.

Prove that $\dfrac{1 - \cos^2\theta}{\sin\theta} = \sin\theta$ .
Prove that $\tan^2\theta + 1 = \sec^2\theta$ .

Hint

For (1), use the Pythagorean identity to replace the numerator. For (2), start from

\sin^2\theta + \cos^2\theta = 1

and divide every term by

\cos^2\theta

View Solution

Solution to question 1:

Working on the left side, apply $\sin^2\theta + \cos^2\theta = 1 \Rightarrow 1 - \cos^2\theta = \sin^2\theta$ :

\frac{1 - \cos^2\theta}{\sin\theta} = \frac{\sin^2\theta}{\sin\theta} = \sin\theta \quad \checkmark

Solution to question 2:

Begin with the Pythagorean identity and divide every term by $\cos^2\theta$ :

\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}

\tan^2\theta + 1 = \sec^2\theta \quad \checkmark

Trigonometry Formulas: Sum, Difference, Double-Angle, and Half-Angle

The trigonometry sum and difference formulas allow you to evaluate trigonometric functions at angles that are not standard unit-circle angles by expressing them as sums or differences of known angles. The double-angle formulas (e.g., $\cos 2\theta = \cos^2\theta - \sin^2\theta$ ) and half-angle formulas extend this toolkit, enabling exact computation of values like $\sin(15°)$ , $\cos(75°)$ , and more complex expressions that appear frequently in calculus and physics.

Problem 8: Exact Value Using Sum and Difference Formulas

Medium

Find the exact value of each expression without using a calculator.

$\sin(75°)$
$\cos\!\left(\dfrac{7\pi}{12}\right)$

Hint

Express each angle as a sum or difference of angles whose exact values you know:

75° = 45° + 30°

and

\frac{7\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3}

View Solution

Solution to question 1:

\sin(75°) = \sin(45° + 30°) = \sin45°\cos30° + \cos45°\sin30°

= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4}

Solution to question 2:

\cos\!\left(\frac{7\pi}{12}\right) = \cos\!\left(\frac{\pi}{4}+\frac{\pi}{3}\right) = \cos\frac{\pi}{4}\cos\frac{\pi}{3} - \sin\frac{\pi}{4}\sin\frac{\pi}{3}

= \frac{\sqrt{2}}{2}\cdot\frac{1}{2} - \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2}-\sqrt{6}}{4}

Problem 9: Applying the Double-Angle Formula

Medium

Let $\cos\theta = -\dfrac{3}{5}$ with $\theta$ in Quadrant III.

Find $\sin\theta$ and then compute $\sin(2\theta)$ and $\cos(2\theta)$ .
Determine which quadrant $2\theta$ lies in and verify the signs of your answers.

Hint

Use the Pythagorean identity to find

\sin\theta

, keeping the correct sign for Q III. Then apply

\sin 2\theta = 2\sin\theta\cos\theta

and

\cos 2\theta = \cos^2\theta - \sin^2\theta

View Solution

Solution to question 1:

From the Pythagorean identity with $\cos\theta = -\frac{3}{5}$ :

\sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25} \implies \sin\theta = -\frac{4}{5} \quad (\text{negative in Q III})

\sin(2\theta) = 2\sin\theta\cos\theta = 2\left(-\frac{4}{5}\right)\!\left(-\frac{3}{5}\right) = \frac{24}{25}

\cos(2\theta) = \cos^2\theta - \sin^2\theta = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}

Solution to question 2:

Since $\sin(2\theta) > 0$ and $\cos(2\theta) < 0$ , the angle $2\theta$ lies in Quadrant II. This is consistent with $\theta \in \left(\pi, \frac{3\pi}{2}\right)$ , giving $2\theta \in \left(2\pi, 3\pi\right)$ , i.e., effectively in Q II after reducing modulo $2\pi$ .

Solving Trigonometry Equations

Solving a trigonometry equation means finding all values of the variable (usually in a specified interval such as $[0, 2\pi)$ ) that satisfy the equation. Key techniques include algebraic factoring, using the Pythagorean identity to reduce an equation to one function, applying the zero-product property, and using inverse trigonometric functions. Always remember that most trigonometry equations have multiple solutions within one period, corresponding to multiple quadrant positions.

Problem 10: Linear Trigonometric Equation

Easy

Find all solutions to the equation on the interval $[0, 2\pi)$ .

$2\sin\theta - \sqrt{3} = 0$
$\sqrt{2}\cos\theta + 1 = 0$

Hint

Isolate the trig function first. Then identify the reference angle from the unit circle, and find all angles in

[0, 2\pi)

where the function takes that value, paying attention to the sign.

View Solution

Solution to question 1:

2\sin\theta = \sqrt{3} \implies \sin\theta = \frac{\sqrt{3}}{2}

The reference angle is $\frac{\pi}{3}$ . Sine is positive in Q I and Q II:

\theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}

Solution to question 2:

\cos\theta = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}

The reference angle is $\frac{\pi}{4}$ . Cosine is negative in Q II and Q III:

\theta = \frac{3\pi}{4} \quad \text{and} \quad \theta = \frac{5\pi}{4}

Problem 11: Solving by Factoring

Medium

Find all solutions on $[0, 2\pi)$ .

$2\sin^2\theta - \sin\theta - 1 = 0$
$\cos\theta\tan\theta - \cos\theta = 0$

Hint

For (1), treat

\sin\theta

as the variable and factor as you would a quadratic. For (2), factor out the common factor

\cos\theta

before applying the zero-product property.

View Solution

Solution to question 1:

Let $u = \sin\theta$ . The equation becomes $2u^2 - u - 1 = 0$ :

(2u + 1)(u - 1) = 0 \implies u = -\frac{1}{2} \text{ or } u = 1

$\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}$ .

$\sin\theta = -\dfrac{1}{2}$ : reference angle $\dfrac{\pi}{6}$ , negative in Q III and Q IV:

\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \text{and} \quad \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}

Solutions: $\dfrac{\pi}{2},\, \dfrac{7\pi}{6},\, \dfrac{11\pi}{6}$ .

Solution to question 2:

\cos\theta(\tan\theta - 1) = 0

$\cos\theta = 0 \Rightarrow \theta = \dfrac{\pi}{2},\, \dfrac{3\pi}{2}$ .

$\tan\theta = 1 \Rightarrow \theta = \dfrac{\pi}{4},\, \dfrac{5\pi}{4}$ .

Solutions: $\dfrac{\pi}{4},\, \dfrac{\pi}{2},\, \dfrac{5\pi}{4},\, \dfrac{3\pi}{2}$ .

Problem 12: Solving with Identities and the Quadratic Formula

Hard

Solve on $[0, 2\pi)$ .

$2\cos^2\theta + 3\sin\theta - 3 = 0$
$\sin(2\theta) = \cos\theta$

Hint

For (1), replace

\cos^2\theta

with

1 - \sin^2\theta

to obtain a quadratic in

\sin\theta

. For (2), expand

\sin(2\theta)

using the double-angle formula, then factor.

View Solution

Solution to question 1:

Substitute $\cos^2\theta = 1 - \sin^2\theta$ :

2(1 - \sin^2\theta) + 3\sin\theta - 3 = 0 \implies -2\sin^2\theta + 3\sin\theta - 1 = 0

2\sin^2\theta - 3\sin\theta + 1 = 0 \implies (2\sin\theta - 1)(\sin\theta - 1) = 0

$\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}$ .

$\sin\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{6},\, \dfrac{5\pi}{6}$ .

Solutions: $\dfrac{\pi}{6},\, \dfrac{\pi}{2},\, \dfrac{5\pi}{6}$ .

Solution to question 2:

Apply $\sin(2\theta) = 2\sin\theta\cos\theta$ :

2\sin\theta\cos\theta = \cos\theta \implies \cos\theta(2\sin\theta - 1) = 0

$\cos\theta = 0 \Rightarrow \theta = \dfrac{\pi}{2},\, \dfrac{3\pi}{2}$ .

$\sin\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{6},\, \dfrac{5\pi}{6}$ .

Solutions: $\dfrac{\pi}{6},\, \dfrac{\pi}{2},\, \dfrac{5\pi}{6},\, \dfrac{3\pi}{2}$ .

Trigonometry Laws: Law of Sines and Law of Cosines for Oblique Triangles

When a triangle does not contain a right angle, neither SOH-CAH-TOA nor the Pythagorean theorem applies directly. Two of the most powerful tools in trigonometry for these cases are the Law of Sines ( $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ ) and the Law of Cosines ( $c^2 = a^2 + b^2 - 2ab\cos C$ ). Recognizing which law to use — based on whether you are given AAS, ASA, SSA, SAS, or SSS information — is itself an important problem-solving skill. The SSA case requires special attention because it can produce the ambiguous case with two possible triangles.

Problem 13: Law of Sines (AAS Case)

Easy

In triangle $ABC$ , $\angle A = 42°$ , $\angle B = 71°$ , and side $a = 10$ cm.

Find $\angle C$ .
Use the Law of Sines to find sides $b$ and $c$ .

Hint

The angles of any triangle sum to

180°

. Once you have all three angles and one side, the Law of Sines

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

gives you the remaining sides.

View Solution

Solution to question 1:

\angle C = 180° - 42° - 71° = 67°

Solution to question 2:

\frac{b}{\sin 71°} = \frac{10}{\sin 42°} \implies b = \frac{10\sin 71°}{\sin 42°} \approx \frac{10 \times 0.9455}{0.6691} \approx 14.13 \text{ cm}

\frac{c}{\sin 67°} = \frac{10}{\sin 42°} \implies c = \frac{10\sin 67°}{\sin 42°} \approx \frac{10 \times 0.9205}{0.6691} \approx 13.76 \text{ cm}

Problem 14: Law of Cosines (SAS Case)

Medium

In triangle $PQR$ , $p = 8$ , $q = 11$ , and $\angle R = 55°$ (the angle between sides $p$ and $q$ ).

Find side $r$ using the Law of Cosines.
Use the Law of Sines to find $\angle P$ and $\angle Q$ .

Hint

Apply the Law of Cosines in the form

r^2 = p^2 + q^2 - 2pq\cos R

. Once

r

is known, switch to the Law of Sines to find the remaining angles, and verify that all three angles sum to

180°

View Solution

Solution to question 1:

r^2 = 8^2 + 11^2 - 2(8)(11)\cos(55°) = 64 + 121 - 176\cos(55°)

r^2 = 185 - 176 \times 0.5736 \approx 185 - 100.95 \approx 84.05

r \approx \sqrt{84.05} \approx 9.17

Solution to question 2:

\sin P = \frac{p\sin R}{r} = \frac{8\sin(55°)}{9.17} \approx \frac{8 \times 0.8192}{9.17} \approx \frac{6.554}{9.17} \approx 0.7147

\angle P \approx \sin^{-1}(0.7147) \approx 45.6°

\angle Q = 180° - 55° - 45.6° = 79.4°