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Trigonometry - Trigonometric Graphs and Identities

Grade 12A Level

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

🔑Concepts

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The Sine function graph y=sin⁡(x)y = \sin(x) is a periodic wave with a period of 360∘360^\circ (or 2π2\pi) and an amplitude of 11. It starts at the origin (0,0)(0,0), reaches its maximum at 90∘90^\circ, and crosses the x-axis at 180∘180^\circ.

Graph of the sine function from 0 to 360 degrees.
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The Cosine function graph y=cos⁡(x)y = \cos(x) is a periodic wave identical in shape to the sine wave but shifted (phase-shifted) to the left by 90∘90^\circ. It starts at its maximum value (0,1)(0,1) and crosses the x-axis at 90∘90^\circ and 270∘270^\circ.

Graph of the cosine function from 0 to 360 degrees.
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For the general trigonometric function y=asin⁡(bx)+cy = a \sin(bx) + c or y=acos⁡(bx)+cy = a \cos(bx) + c: The value ∣a∣|a| determines the amplitude (vertical stretch), bb determines the frequency (horizontal stretch/compression), and cc determines the vertical shift (the principal axis).

Graph showing amplitude vertical stretch.
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The Tangent function y=tan⁡(x)y = \tan(x) differs from sine and cosine as it has a period of 180∘180^\circ and contains vertical asymptotes where cos⁡(x)=0\cos(x) = 0 (at x=90∘,270∘x = 90^\circ, 270^\circ, etc.).

Graph of the tangent function showing the asymptote at 90 degrees.

📐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).

Problem 4:

Determine the equation of the trigonometric function shown in the graph, which has a maximum value of 55, a minimum value of −1-1, and a period of 180∘180^\circ.

Cosine wave shifted vertically to center at y=2 with amplitude 3.

Solution:

  1. Find the amplitude ∣a∣|a|: a=max−min2=5−(−1)2=3a = \frac{\text{max} - \text{min}}{2} = \frac{5 - (-1)}{2} = 3
  2. Find the vertical shift cc: c=max+min2=5+(−1)2=2c = \frac{\text{max} + \text{min}}{2} = \frac{5 + (-1)}{2} = 2
  3. Find bb using the period: Period=360∘b  ⟹  180=360b  ⟹  b=2\text{Period} = \frac{360^\circ}{b} \implies 180 = \frac{360}{b} \implies b = 2
  4. Since the graph starts at the maximum at x=0x=0, we use a cosine function: y=3cos⁡(2x)+2y = 3\cos(2x) + 2

Explanation:

To identify the equation from a graph, calculate the amplitude (half the distance between max and min), the principal axis (the average of max and min), and the coefficient bb based on how many full cycles occur within 360∘360^\circ.

Problem 5:

Solve the identity 1cos⁡2θ−1=tan⁡2θ\frac{1}{\cos^2 \theta} - 1 = \tan^2 \theta visually using a right-angled triangle or algebra.

Right triangle with sides sin, cos and hypotenuse 1.

Solution:

  1. Start with the Pythagorean identity: sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1
  2. Divide every term by cos⁡2θ\cos^2 \theta: sin⁡2θcos⁡2θ+cos⁡2θcos⁡2θ=1cos⁡2θ\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}
  3. Simplify the terms: tan⁡2θ+1=1cos⁡2θ\tan^2 \theta + 1 = \frac{1}{\cos^2 \theta}
  4. Rearrange to match the target expression: 1cos⁡2θ−1=tan⁡2θ\frac{1}{\cos^2 \theta} - 1 = \tan^2 \theta

Explanation:

This identity is a variation of the fundamental Pythagorean identity. By dividing sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 by cos⁡2θ\cos^2\theta, we derive the relationship between tangent and secant (reciprocal of cosine).