CE Board Exam Randomizer

⬅ Back to Trigonometry
📐 Key Concepts: Trigonometric Identities

Pythagorean Identities:
$\sin^2\theta + \cos^2\theta = 1$  |  $1 + \tan^2\theta = \sec^2\theta$  |  $1 + \cot^2\theta = \csc^2\theta$

Reciprocal Identities:
$\csc\theta = \dfrac{1}{\sin\theta}$  |  $\sec\theta = \dfrac{1}{\cos\theta}$  |  $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$

Sum and Difference Formulas:
$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$
$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$

Double Angle Formulas:
$\sin 2\theta = 2\sin\theta\cos\theta$
$\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$
$\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$

Half-Angle Formulas:
$\sin\tfrac{\theta}{2} = \pm\sqrt{\dfrac{1-\cos\theta}{2}}$  |  $\cos\tfrac{\theta}{2} = \pm\sqrt{\dfrac{1+\cos\theta}{2}}$
$\tan\tfrac{\theta}{2} = \pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \dfrac{\sin\theta}{1+\cos\theta} = \dfrac{1-\cos\theta}{\sin\theta}$

Sign (±) depends on the quadrant of θ/2.

Problem 1: Six Trig Functions from tan θ (QI)

Given that $\tan\theta = \dfrac{4}{3}$ and $\theta$ is in the first quadrant, find the exact values of all six trigonometric functions: $\sin\theta$, $\cos\theta$, $\tan\theta$, $\csc\theta$, $\sec\theta$, and $\cot\theta$.

Trig Identities – Problem 1 – Diagram Trig Identities – Problem 1 – Diagram Trig Identities – Problem 1 – Diagram
Solution Solution Solution Solution

Draw a right triangle with opposite = 4, adjacent = 3. Hypotenuse $= \sqrt{4^2+3^2} = 5$. All functions are positive in QI:

$$\sin\theta = \frac{4}{5},\quad \cos\theta = \frac{3}{5},\quad \tan\theta = \frac{4}{3}$$
$$\csc\theta = \frac{5}{4},\quad \sec\theta = \frac{5}{3},\quad \cot\theta = \frac{3}{4}$$

Problem 2: Six Trig Functions from cos θ (QII)

Given that $\cos\theta = -\dfrac{5}{13}$ and $\theta$ is in the second quadrant, find the exact values of all six trigonometric functions.

Trig Identities – Problem 2 – Diagram Trig Identities – Problem 2 – Diagram Trig Identities – Problem 2 – Diagram
Solution Solution Solution Solution

$\cos\theta = -5/13$, so adjacent $= -5$, hypotenuse $= 13$. In QII, sine is positive: opposite $= \sqrt{13^2 - 5^2} = \sqrt{144} = 12$.

$$\sin\theta = \frac{12}{13},\quad \cos\theta = -\frac{5}{13},\quad \tan\theta = -\frac{12}{5}$$
$$\csc\theta = \frac{13}{12},\quad \sec\theta = -\frac{13}{5},\quad \cot\theta = -\frac{5}{12}$$

Problem 3: Exact Value Using Sum Formula — sin(75°)

Find the exact value of $\sin(75°)$ by expressing it as $\sin(45° + 30°)$ and applying the sum formula.

Trig Identities – Problem 3 – Diagram Trig Identities – Problem 3 – Diagram
Solution Solution Solution
$$\sin75° = \sin(45°+30°) = \sin45°\cos30° + \cos45°\sin30°$$
$$= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \boxed{\frac{\sqrt{6}+\sqrt{2}}{4}}$$

Problem 4: Exact Value Using Difference Formula — cos(15°)

Find the exact value of $\cos(15°)$ by expressing it as $\cos(45° - 30°)$ and applying the difference formula.

Trig Identities – Problem 4 – Diagram Trig Identities – Problem 4 – Diagram
Solution Solution Solution
$$\cos15° = \cos(45°-30°) = \cos45°\cos30° + \sin45°\sin30°$$
$$= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \boxed{\frac{\sqrt{6}+\sqrt{2}}{4}}$$

Problem 5: Double Angle Formulas

If $\sin\alpha = \dfrac{5}{13}$ and $\alpha$ is in the first quadrant, find the exact values of $\sin 2\alpha$, $\cos 2\alpha$, and $\tan 2\alpha$.

Trig Identities – Problem 5 – Diagram Trig Identities – Problem 5 – Diagram Trig Identities – Problem 5 – Diagram
Solution Solution Solution Solution

$\sin\alpha = 5/13$, QI → $\cos\alpha = 12/13$, $\tan\alpha = 5/12$.

$$\sin2\alpha = 2\sin\alpha\cos\alpha = 2\!\left(\frac{5}{13}\right)\!\left(\frac{12}{13}\right) = \boxed{\frac{120}{169}}$$
$$\cos2\alpha = \cos^2\alpha - \sin^2\alpha = \frac{144}{169} - \frac{25}{169} = \boxed{\frac{119}{169}}$$
$$\tan2\alpha = \frac{\sin2\alpha}{\cos2\alpha} = \frac{120/169}{119/169} = \boxed{\frac{120}{119}}$$

Problem 6: Half-Angle Formula — tan(22.5°)

Find the exact value of $\tan(22.5°)$ using the half-angle formula. Express your answer in simplified radical form.

Trig Identities – Problem 6 – Diagram Trig Identities – Problem 6 – Diagram
Solution Solution Solution

Use $\tan\!\left(\tfrac{\theta}{2}\right) = \dfrac{1 - \cos\theta}{\sin\theta}$ with $\theta = 45°$:

$$\tan22.5° = \frac{1 - \cos45°}{\sin45°} = \frac{1 - \dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \frac{(2-\sqrt{2})\sqrt{2}}{2} = \frac{2\sqrt{2}-2}{2} = \boxed{\sqrt{2}-1}$$

Problem 7: Solving a Trigonometric Equation

Find all values of $\theta$ in $[0°, 360°)$ that satisfy the equation: $2\sin^2\theta - \sin\theta - 1 = 0$.

Trig Identities – Problem 7 – Diagram Trig Identities – Problem 7 – Diagram
Solution Solution Solution Solution

Let $u = \sin\theta$. The equation becomes:

$$2u^2 - u - 1 = 0 \Rightarrow (2u + 1)(u - 1) = 0$$

Case 1: $\sin\theta = 1 \Rightarrow \theta = 90°$

Case 2: $\sin\theta = -\tfrac{1}{2}$ → reference angle = 30°; negative sine in QIII and QIV:

$$\theta = 180° + 30° = 210° \quad \text{and} \quad \theta = 360° - 30° = 330°$$
$$\boxed{\theta = 90°,\; 210°,\; 330°}$$

Problem 8: Identity Simplification

Simplify the expression: $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2$

Trig Identities – Problem 8 – Diagram Trig Identities – Problem 8 – Diagram
Solution Solution Solution

Expand each square:

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

Adding the two expressions, the cross terms cancel:

$$= 2\sin^2\theta + 2\cos^2\theta = 2(\sin^2\theta + \cos^2\theta) = 2(1) = \boxed{2}$$