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Trigonometry - Trigonometric Graphs and Identities

Grade 12IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Unit Circle: Understanding coordinates (cos θ, sin θ) and the CAST diagram (quadrants where functions are positive).

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Fundamental Identities: Using the relationship between sine, cosine, and tangent to simplify expressions.

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Graph Transformations: Identifying amplitude, period, and phase shifts for y = a sin(bx + c) + d.

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Solving Trigonometric Equations: Finding all solutions within a specific domain (e.g., 0 to 360 degrees) using reference angles.

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Periodicity: Recognizing that sin and cos repeat every 360° (2π rad) while tan repeats every 180° (π rad).

📐Formulae

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

tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}

Amplitude=∣a∣ for y=asin⁡(bx)\text{Amplitude} = |a| \text{ for } y = a \sin(bx)

Period=360∘b or 2πb\text{Period} = \frac{360^\circ}{b} \text{ or } \frac{2\pi}{b}

sin⁡(180∘−θ)=sin⁡θ\sin(180^\circ - \theta) = \sin \theta

cos⁡(360∘−θ)=cos⁡θ\cos(360^\circ - \theta) = \cos \theta

💡Examples

Problem 1:

Solve the equation 2sin⁡(x)=32\sin(x) = \sqrt{3} for 0∘≤x≤360∘0^\circ \le x \le 360^\circ.

Solution:

  1. Isolate the sine function: sin⁡(x)=32\sin(x) = \frac{\sqrt{3}}{2}.
  2. Find the reference angle: x=arcsin⁡(32)=60∘x = \arcsin(\frac{\sqrt{3}}{2}) = 60^\circ.
  3. Identify quadrants where sine is positive: Quadrant 1 and Quadrant 2.
  4. Calculate angles: x1=60∘x_1 = 60^\circ, x2=180∘−60∘=120∘x_2 = 180^\circ - 60^\circ = 120^\circ. Final Answer: x=60∘,120∘x = 60^\circ, 120^\circ.

Explanation:

To solve trigonometric equations, first isolate the function, find the principal value (reference angle), and then use the CAST rule to find other values within the specified range.

Problem 2:

Simplify the expression: (cos⁡θ)(tan⁡θ)+sin⁡θ(\cos \theta)(\tan \theta) + \sin \theta.

Solution:

  1. Substitute tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}.
  2. Expression becomes: cos⁡θ⋅sin⁡θcos⁡θ+sin⁡θ\cos \theta \cdot \frac{\sin \theta}{\cos \theta} + \sin \theta.
  3. Cancel cos⁡θ\cos \theta: sin⁡θ+sin⁡θ\sin \theta + \sin \theta.
  4. Result: 2sin⁡θ2\sin \theta.

Explanation:

Simplification often involves converting all terms to sine and cosine and using algebraic cancellation.

Problem 3:

State the amplitude and period of the function y=3cos⁡(2x)+1y = 3\cos(2x) + 1.

Solution:

  1. Amplitude: The coefficient 'a' is 3, so ∣3∣=3|3| = 3.
  2. Period: The coefficient 'b' is 2. Period=360∘2=180∘\text{Period} = \frac{360^\circ}{2} = 180^\circ.
  3. Vertical Shift: The graph is shifted up by 1 unit.

Explanation:

In the general form y=acos⁡(bx)+dy = a \cos(bx) + d, 'a' determines the vertical stretch (amplitude) and 'b' determines the horizontal compression (affecting the period).