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📐 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: 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: 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: 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: 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: 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: 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: 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: 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:

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

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q87

MSTE - Geometry and Trigonometry / Trigonometry / Engr. Janclyde Espinosa (Clidez)

If sin 40° + sin 20° = sinθ, find the value of θ.

Answer:

  1. 80º
  2. 20º
  3. 120º
  4. 60º
Use the sum-to-product identity:
$\sin40^\circ+\sin20^\circ=2\sin30^\circ\cos10^\circ$
$=\cos10^\circ$
Since $\cos10^\circ=\sin80^\circ$:
$\boxed{\theta=80^\circ}$

Question Bank: t328

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

Solve for $x$ from the following equation: $\cos 6x = 1 / \csc(3x+9)$. Note: all units are in degrees.

  1. $9^\circ$
  2. $8^\circ$
  3. $18^\circ$
  4. $12^\circ$
$\dfrac{1}{\csc(3x+9)}=\sin(3x+9)$, so the equation becomes $\cos 6x=\sin(3x+9)$.
Using $\cos\alpha=\sin(90^\circ-\alpha)$: $\sin(90^\circ-6x)=\sin(3x+9)$.
$90-6x=3x+9\Rightarrow 81=9x\Rightarrow x=9^\circ$.
$\boxed{9^\circ}$

Question Bank: t330

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

Find the value of $\phi$ in the equation $\cosh^2 \phi - \sinh^2 \phi = 3\sin \phi$.

  1. $19.47^\circ$
  2. $25.63^\circ$
  3. $36.21^\circ$
  4. $12.78^\circ$
The identity $\cosh^2\phi-\sinh^2\phi=1$ holds for all $\phi$.
So the equation reduces to $1=3\sin\phi$, giving $\sin\phi=\dfrac{1}{3}$.
$\phi=\sin^{-1}\left(\tfrac{1}{3}\right)=19.47^\circ$.
$\boxed{19.47^\circ}$

Question Bank: t336

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

Which of the following is equivalent to the given trigonometric expressions?

$(\csc x)(\cot x)$.

  1. $\cos x / (\sin^2 x - 1)$
  2. $\sin x / (\sin^2 x - 1)$
  3. $\cos x / (1 - \cos^2 x)$
  4. $\sin x / (\cos^2 x - 1)$

$(1/\sin x)(\cot x)$.

  1. $\csc^2 x / \sec x$
  2. $\csc x / \sec^2 x$
  3. $\csc^2 x / \sec^2 x$
  4. $\csc^2 x / \sec^2 x$

Which of the following is equivalent to $\sec^2 x - 1$?

  1. $\csc^2 x$
  2. $\cos^2 x$
  3. $\cot^2 x$
  4. $\tan^2 x$

Part 1.

$(\csc x)(\cot x)=\dfrac{1}{\sin x}\cdot\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^2 x}$.
Since $\sin^2 x=1-\cos^2 x$, this equals $\dfrac{\cos x}{1-\cos^2 x}$.
$\boxed{\cos x/(1-\cos^2 x)}$

Part 2.

$\dfrac{1}{\sin x}\cdot\cot x=\dfrac{1}{\sin x}\cdot\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^2 x}$.
Rewrite: $\dfrac{1}{\sin^2 x}\cdot\cos x=\csc^2 x\cdot\dfrac{1}{\sec x}=\dfrac{\csc^2 x}{\sec x}$.
$\boxed{\csc^2 x/\sec x}$

Part 3.

From the Pythagorean identity $\sec^2 x=1+\tan^2 x$.
Therefore $\sec^2 x-1=\tan^2 x$.
$\boxed{\tan^2 x}$

Question Bank: t1699

MSTE - Geometry and Trigonometry / Trigonometry / BEMz

Determine the period and amplitude of the function $y=3$ cos x.

  1. $2\pi, 3$
  2. $\pi/2, 3$
  3. 3/2, 3
  4. $\pi, 3$
For $y=A\cos bx$, the amplitude is $|A|$ and the period is
$$T=\frac{2\pi}{|b|}$$
Here $A=3$ and $b=1$, so
$$T=2\pi,\qquad \text{amplitude}=3$$
Therefore, the answer is $\boxed{2\pi,\ 3}$.