Year11-MATH-3-1-5

Chapter 5: Trigonometric Functions

In previous years, trigonometry was about triangles. In Mathematical Methods, we transition to Circular Functions. We treat sine and cosine as waves that repeat infinitely, which allows us to model periodic phenomena like tides, sound waves, and seasonal temperature shifts.


5.1 Radian Measure

Before performing calculus on trigonometric functions, we must use radians instead of degrees. Radians measure the arc length along a unit circle.

  • Conversion: π radians=180\pi \text{ radians} = 180^\circ
  • To Radians: Multiply by π180\frac{\pi}{180}
  • To Degrees: Multiply by 180π\frac{180}{\pi}

Note: Always ensure your calculator is in RAD mode when working with calculus in this subject.


5.2 The Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin (0,0)(0,0). For any point (x,y)(x,y) on the circle at an angle θ\theta:

  • x=cos(θ)x = \cos(\theta)
  • y=sin(θ)y = \sin(\theta)
  • tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

5.3 Graphs of Sine and Cosine

The standard functions y=sin(x)y = \sin(x) and y=cos(x)y = \cos(x) produce periodic waves. We often study transformed versions:

y=Asin(B(xC))+Dy = A\sin(B(x – C)) + D

  • AA (Amplitude): The height of the wave from the center.
  • BB (Period Factor): Used to find the Period (PP), which is the distance for one full cycle. P=2πBP = \frac{2\pi}{B}.
  • CC (Phase Shift): Horizontal translation.
  • DD (Mean Height): Vertical translation (the new center line).

5.4 Derivatives of Trigonometric Functions

One of the most remarkable patterns in calculus is how sine and cosine relate to each other’s gradients.

  • The Derivative of Sine:ddx(sin(kx))=kcos(kx)\frac{d}{dx}(\sin(kx)) = k\cos(kx)
  • The Derivative of Cosine:ddx(cos(kx))=ksin(kx)\frac{d}{dx}(\cos(kx)) = -k\sin(kx)

Worked Example 1: Differentiating

Differentiate f(x)=4sin(2x)+3cos(x)f(x) = 4\sin(2x) + 3\cos(x).

  1. Differentiate 4sin(2x)4\sin(2x): The derivative of the inside (2x2x) is 2. So, 2×4cos(2x)=8cos(2x)2 \times 4\cos(2x) = 8\cos(2x).
  2. Differentiate 3cos(x)3\cos(x): The derivative of cos\cos is sin-\sin. So, 3sin(x)-3\sin(x).
  3. Final Answer: f(x)=8cos(2x)3sin(x)f'(x) = 8\cos(2x) – 3\sin(x).

5.5 Practice Problems

Part A: Radians and Exact Values

  1. Convert 6060^\circ and 225225^\circ to radians (leave in terms of π\pi).
  2. Using the unit circle, find the exact value of sin(π2)\sin(\frac{\pi}{2}) and cos(π)\cos(\pi).

Part B: Graphing Features

  1. For the function y=5sin(2x)+1y = 5\sin(2x) + 1:
    • State the Amplitude.
    • Calculate the Period.
    • State the range of the function.
  2. Find the value of BB if the function y=cos(Bx)y = \cos(Bx) has a period of π\pi.

Part C: Calculus

  1. Find the derivative of y=sin(4x)y = \sin(4x).
  2. Find the derivative of f(x)=cos(5x)+x2f(x) = \cos(5x) + x^2.
  3. Challenge: Find the gradient of the curve y=sin(x)y = \sin(x) at the point where x=πx = \pi.

Solutions (Summary)

  • 1. π3\frac{\pi}{3}, 5π4\frac{5\pi}{4}
  • 2. sin(π2)=1\sin(\frac{\pi}{2}) = 1, cos(π)=1\cos(\pi) = -1
  • 3. Amp = 55; Period = 2π2=π\frac{2\pi}{2} = \pi; Range = [4,6][-4, 6] (Center is 1, goes up/down by 5).
  • 4. B=2B = 2 (since 2π2=π\frac{2\pi}{2} = \pi).
  • 5. 4cos(4x)4\cos(4x)
  • 6. 5sin(5x)+2x-5\sin(5x) + 2x
  • 7. y=cos(x)y’ = \cos(x). At x=πx = \pi, cos(π)=1\cos(\pi) = -1. The gradient is 1-1.


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