A sine wave starts at zero when $\theta=0$, rises to its maximum, drops through zero to its minimum and then returns to zero. This cycle is repeated again and again. The vertical distance from zero to the peak value is called the **amplitude**. The number of peaks per unit time is called the **frequency**. Frequency is generally given the symbol $\nu$ (called nu). The horizontal distance between two peaks is called the **wavelength**. Wavelength is usually given the symbol $\lambda$ (called lambda).

Figure 8.1: A sine wave $y=sin \theta$

A cosine wave is similar to a sine wave except it starts at its maximum, drops through zero to its minimum and then returns back through zero to its maximum again. This cycle, just like the sine wave, is repeated again and again.

Figure 8.2: A cosine wave $y=cos \theta$

Imagine a cosine wave and a sine wave adding together to create a combined wave. What would the combined wave be like? To make the question a bit more general let the sine wave have an amplitude of $a$ and the cosine wave have an amplitude of $b$.

Figure 8.3: A combined wave $y=asin \theta+bcos \theta$

We know from section 7.3 that | ||

$ \sin(\theta + \phi)$ | $=$ | $\sin \theta \cos \phi + \cos \theta \sin \phi $. |

so we can write | ||

$ r\sin(\theta + \phi)$ | $=$ | $r\sin \theta \cos \phi + r\cos \theta \sin \phi $. |

For the combined wave we have | ||

$y$ | $=$ | $a\sin \theta + b\cos \theta $. |

let $a$ | $=$ | $r\cos \phi $ |

and $b$ | $=$ | $r\sin \phi $ |

then $a^2 + b^2$ | $=$ | $r^2 cos^2 \phi + r^2\sin^2 \phi $ |

$=$ | $r^2 (cos^2 \phi + sin^2 \phi) $ | |

we know | ||

$cos^2 \phi + sin^2 \phi$ | $=$ | $ 1 $ |

which means | ||

$a^2 + b^2$ | $=$ | $ r^2 $ |

so $r$ | $=$ | $\sqrt{a^2 + b^2} $. |

If we divide $b$ by $a$ we get | ||

$\frac{b}{a}$ | $=$ | $ \frac{r sin \phi}{r cos \phi} $ |

$=$ | $ tan \phi $ | |

so $\phi$ | $=$ | $tan^{-1}(b/a)$. |

When:

$a > 0, b > 0 $ means $ 0 < \phi < \pi/2$

$a < 0, b > 0 $ means $ \pi/2 < \phi < \pi$

$a < 0, b < 0 $ means $ \pi < \phi < 3\pi/2$

$a > 0, b < 0 $ means $ 3\pi/2 < \phi < 2\pi$

Example 8.3b: Find the amplitude and phase of the combined wave $y=-4sin \theta+3cos \theta$ | ||

We know | ||

$ r\sin(\theta + \phi)$ | $=$ | $r\sin \theta \cos \phi + r\cos \theta \sin \phi $ |

and we are given $y$ | $=$ | $-4\sin \theta + 3\cos \theta$ |

Comparing the equations we can see: | ||

$rcos\phi$ | $=$ | $-4$ |

$rsin\phi$ | $=$ | $3$ |

$r^2 cos^2\phi + r^2 sin^2 \phi$ | $=$ | $(-4)^2+3^2$ |

$=$ | $5$ | |

If we divide $rsin\phi$ by $rcos\phi$ we get | ||

$tan\phi$ | $=$ | $3/-4$ |

so $\phi$ | $=$ | $tan^{-1}(3/-4)$ |

Your calculator gives an answer of $-0.64$ radians (-36.87°) | ||

so we need to add $\pi$ radians (180°) to get | ||

$\phi$ | $=$ | $2.50$ radians (143.13°) |

$y=-4sin \theta+3cos \theta$ | $=$ | $5sin(\theta+2.50)$ |

Exercise 8.3

- Transform the following expressions to the form $r sin(\theta + \phi)$
- a. $ sin \theta - cos \theta $
- b. $ 3sin \theta + 4cos \theta $
- c. $ 5sin \theta + 12cos \theta $
- d. $ sin \theta + \sqrt{3} cos \theta $

There is a set of problems that require us to solve equations of the form | ||

$a sin\theta+b cos\theta$ | $=$ | $c$ |

To solve these equations we transform $a sin\theta+b cos\theta$ into $rsin(\theta+\phi)$ and set the result equal to $c$. | ||

$rsin(\theta+\phi)$ | $=$ | $c$ |

$sin(\theta+\phi)$ | $=$ | $c/r$ |

$\theta+\phi$ | $=$ | $sin^{-1}(c/r)$ |

$\theta$ | $=$ | $sin^{-1}(c/r)-\phi$ |

If the result is negative you normally add $2\pi$ to get it into the range $0$ to $2\pi$. | ||

Look at the diagram below. You will see there is another solution. To find the other solution subtract $\theta$ and $\phi$ from $\pi$. | ||

$\theta_1 = sin^{-1}(c/r)-\phi$
As before, add $2\pi$ if the result is negative.
$\theta_2 = \pi-sin^{-1}(c/r)-\phi$ |

Example 8.4: Solve the following equation $4sin \theta+3cos \theta = 2$ for values of $\theta$ between $0$ and $2\pi$ | ||

$r$ | $=$ | $\sqrt{3^2 + 4^2} $. |

$=$ | $5$ | |

and $tan \phi$ | $=$ | $3/4$ |

$\phi$ | $=$ | $0.64$ radians (36.87°) |

$5sin(\theta+0.64)$ | $=$ | $2$ |

$sin(\theta+0.64)$ | $=$ | $2/5$ |

$\theta+0.64$ | $=$ | $sin^{-1}(2/5)$ |

$\theta+0.64$ | $=$ | $0.41$ |

$\theta$ | $=$ | $0.41-0.64$ |

$\theta$ | $=$ | $-0.23$ |

Adding $2\pi$ we get $\theta_1$ | $=$ | $6.05$ radians (346.71°) |

$\theta_2$ | $=$ | $\pi-sin^{-1}(2/5)-\phi$ |

$=$ | $\pi-0.41-0.64$ | |

$=$ | $2.09$ radians (119.6°) | |

Sanity Check | ||

Look at the diagram above. You can see the smaller intersection of the purple sine wave and the green line is a bit after $\pi/2$ which is about $2$ so we can be confident with $2.09$. | ||

The second intersection is between $3\pi/2$ and $2\pi$ which is about $6$ so we can be confident with $6.05$. |

Exercise 8.4

- Solve the following equations for $ \theta $ in the range $ 0 < \theta < \pi $
- a. $ sin \theta - cos \theta =-1$
- b. $ 2sin \theta - cos \theta =2 $
- c. $ 3sin \theta - 4cos \theta =-1$
- d. $ \sqrt{3}sin \theta + cos \theta=1 $

A combined wave is made from two or more sine and/or cosine waves. Once a combined wave has been calculated we would like to know the position and magnitude of the turning points, that is the points at which the combined wave is at a maximum or a minimum.

Consider $ y= a sin \theta + b cos \theta + c $. Sines and cosines vary between $+1$ and $-1$ so the maximum value is equal to $\sqrt{a^2+b^2}+c$ and the minimum value is equal to $-\sqrt{a^2+b^2}+c$.

Maximum = $\sqrt{a^2+b^2}+c$

Minimum = $-\sqrt{a^2+b^2}+c$

Minimum = $-\sqrt{a^2+b^2}+c$

When the total angle equals $\pi / 2$ the sine is at its maximum value of $1$. This means a maximum is at $\theta+\phi = \pi / 2$ so $\theta=\pi / 2 - \phi$

The wave then drops through zero to its minimum value of $-1$ when the angle equals $3 \pi / 2$. This means a minimum is at $\theta+\phi = 3\pi / 2$ so $\theta=3\pi / 2-\phi$

There is an infinite number of maxima and minima. It is normal to give values in the range $0 ≤ \theta ≤ 2\pi$.

Maxima are at $\theta=2n\pi+\phi$

Minima are at $\theta=(2n+1)\pi+\phi$

It is normal to give values of $\theta$ in the range $0 ≤ \theta ≤ 2\pi$.

Minima are at $\theta=(2n+1)\pi+\phi$

It is normal to give values of $\theta$ in the range $0 ≤ \theta ≤ 2\pi$.

Example 8.5: Find the value and position of the first maximum and minimum of $y= 4sin\theta+3cos \theta + 3$ | ||

$r$ | $=$ | $\sqrt{3^2 + 4^2}=5 $. |

and $\phi$ | $=$ | $tan^{-1}(3/4) = 0.64$ radians |

Maximum | $=$ | $5 + 3=8$ |

Minimum | $=$ | $-5 + 3=-2$ |

Maximum $=8$ and is at | ||

$\theta+0.64$ | $=$ | $\pi/2$ |

so $\theta_{max}$ | $=$ | $\pi/2 - 0.64=0.93$ radians |

Minimum $=-2$ and is at | ||

$\theta+0.64$ | $=$ | $3 \pi/2$ |

so $\theta_{min}$ | $=$ | $3 \pi/2-0.64=4.07$ radians |

Exercise 8.5

- Find the position and magnitude of the maximum value for the following equations. Give your answer in the range $ 0 < \theta < 2\pi $
- a. $ y= 3 sin \theta + 4cos \theta + 4 $
- b. $ y= sin \theta - 2cos \theta -3 $
- c. $ y= 5sin \theta + 12 cos \theta +6$
- d. $ y= \sqrt{3}sin \theta + cos \theta +2 $