Trigonometric functions

Recall the three trigonometric relationships in right-angled triangles (SOHCAHTOA):

sin(α)=OHcos(α)=AHtan(α)=OA\begin{array}{lll} \sin(\alpha)=\frac{O}{H}\\ \cos(\alpha)=\frac{A}{H}\\ \tan(\alpha)=\frac{O}{A} \end{array}

See the figure below (left). Here, HH, OO and AA denote the hypotenuse, opposite side and adjacent side. Note that 0<α<900^\circ<\alpha<90^\circ for the relationships to make sense. At the moment it is not clear what sin(140)\sin(140^\circ) means, as there is no right angled triangle with such an angle. But ultimately we want to be able to calculate sin(α),cos(α)\sin(\alpha), \cos(\alpha) and tan(α)\tan(\alpha) for α=140\alpha=140^\circ, or for any other angle of any size. This will lead to the trigonometric functions.

First, let's start by considering again right-angled triangles, but this time with hypotenuse H=1H=1 (figure above, middle). We then have

sin(α)=O1=O\sin(\alpha)=\frac{O}{1}=O

and

cos(α)=A1=A\cos(\alpha)=\frac{A}{1}=A

Thus, if H=1H=1, the sine is the length of the vertical side, and the cosine is length of the horizontal side. By the way, we can also represent tan(α)\tan(\alpha) as a side length. To do so, we have to extend the triangle so that the horizontal side has length 11 (figure above, right). The vertical side uu of the new triangle is then tan(α)\tan(\alpha), because

tan(α)=u1=u\tan(\alpha)=\frac{u}{1}=u

Extending the definition of the sine and cosine

To extend the sin(α)\sin(\alpha) and cos(α)\cos(\alpha) to arbitrary values of α\alpha, we focus on the fact we have observed above, namely that for H=1H=1 the sin(α)\sin(\alpha) is the vertical side length and the cos(α)\cos(\alpha) is the horizontal side length. So let us introduce a new definition of the sine and cosine. It involves the unit circle.

  1. Draw a unit circle, and indicate a vertical and a horizontal axis, both passing through the centre of the circle (see figure below). Each arbitrary angle α\alpha defines the position of a point PP on the unit circle. From now on we use the arc length xx rather than the angle α\alpha to describe the position of this point PP.

  2. For a given value of xx (the input to the machine) determine the corresponding point PP on the unit circle, and define the following output yy:

    y=sin(x)=distance between P and horizontal axis ("rise")y=cos(x)=distance between P and the vertical axis ("run")\boxed{\begin{array}{lll} y=\sin(x)&=&\text{distance between $P$ and horizontal axis ("rise")}\\ y=\cos(x)&=&\text{distance between $P$ and the vertical axis ("run")} \end{array}}

    In addition, "rise" is negative if PP is below the horizontal axis, and "run" is negative if PP is on the left side of the vertical axis (figure below, right).

It should be clear that for xx between 00 and π2\frac{\pi}{2}, or for α\alpha between 00^\circ and 9090^\circ, these definitions are simply the original trigonometric relationships SOHCAHTOA where H=1H=1:

The function tan(x)\tan(x) can be defined by the length of the vertical line of the extended triangles (see top of section). Or we can simply set

tan(x)=riserun=sin(x)cos(x)\boxed{\tan(x)=\frac{\text{rise}}{\text{run}}=\frac{\sin(x)}{\cos(x)}}

As a summary, we represent the three trigonometric functions sin(x),cos(x)\sin(x), \cos(x) and tan(x)\tan(x) as machines. The input xx is interpreted as the arc length (radians):

xxxsinrisecosruntanriserun=sin(x)cos(x)yyy\begin{array}{crcrcr} x & & x & & x &\\ \huge \downarrow & & \huge \downarrow & & \huge \downarrow &\\ \boxed{\large \sin}&\text{rise}\quad & \boxed{\large \cos} & \text{run}\quad & \boxed{\large \tan} & \frac{\text{rise}}{\text{run}}=\frac{\sin(x)}{\cos(x)}\\ \huge\downarrow & & \huge\downarrow& & \huge\downarrow&\\ y & & y & & y &\\ \end{array}
Exercise 1
  1. No calculator. Determine sin(x)\sin(x), cos(x)\cos(x), and tan(x)\tan(x) for the following xx-values: 00, π2\frac{\pi}{2}, π\pi, 3π2\frac{3\pi}{2}, 2π2\pi, 5π2\frac{5\pi}{2}, π2-\frac{\pi}{2}.

  2. Without using the trigonometric function keys sin, cos, and tan on the calculator, determine sin(x)\sin(x), cos(x)\cos(x), and tan(x)\tan(x) for the following xx-values: π4\frac{\pi}{4}, π6\frac{\pi}{6}, 7π4\frac{7\pi}{4}, 5π6\frac{-5\pi}{6}. Verify your results using the calculator.

  3. Without using trigonometric function keys sin^-1 and cos^-1 on the calculator, give an estimate of xx such that sin(x)=0.75\sin(x)=0.75. Also, if sin(x)=0.75\sin(x)=0.75, what is the exact value of sin(x+π)\sin(x+\pi)?

  4. No calculator: sketch the graph of the three trigonometric functions. Use the unit circle and indicate the points on the graph at x=0,π2,3π2x=0, \frac{\pi}{2},\frac{3\pi}{2} and 2π2\pi. Check the graphs using Geogebra. An animation illustrating the relationship between the unit circle and the graph of the sine and cosine functions can be found here: animation. Note: recall that 01=01=0\frac{0}{1}=\frac{0}{-1}=0 and 10=\frac{1}{0}=\infty and 10=\frac{-1}{0}=-\infty.

  5. No calculator. Express the functions f(x)=sin(x+π2)f(x)=\sin(x+\frac{\pi}{2}) and g(x)=sin(xπ2)g(x)=\sin(x-\frac{\pi}{2}) using cos(x)\cos(x).

  6. No calculator. What is sin2(x)+cos2(x)\sin^2(x)+\cos^2(x) for every xx? Note: the notation sin2(x)\sin^2(x) is a short form for sin(x)sin(x)=(sin(x))2\sin(x)\cdot \sin(x)=(\sin(x))^2.

  7. With calculator. Find an xx-intercept of the functions ff and gg by solving the equations f(x)=0f(x)=0 and g(x)=0g(x)=0.

    1. f(x)=2cos(2x+1)+1f(x)=2\cos(2x+1)+1
    2. g(x)=0.5tan(3x)10g(x)=0.5 \tan(3x)-10
  8. With calculator. Find an intersection point between the functions ff and gg

    1. f(x)=sin(x)f(x)=\sin(x), g(x)=0.4g(x)=0.4
    2. f(x)=3cos(2x0.5)f(x)=3\cos(2x-0.5), g(x)=0.1g(x)=0.1
    3. f(x)=sin(3x)f(x)=\sin(3x), g(x)=2cos(3x)g(x)=2\cdot \cos(3x)
  9. Given sin(x)=0.6\sin(x)=0.6, we need the arc sine to find xx:

    x=arcsin(0.6)=sin1(0.6)x=\arcsin(0.6)=\sin^{-1}(0.6)

    Why is there the name "arc" in the name? Argue with the unit circle.

  10. Without calculator. Find the solutions of

    1. cos(2x)=1\cos(2x)=-1

    2. 2sin(3x1)=02\sin(3x-1)=0

Solution
10
  1. Find xx with 2x=π2x=\pi, or 2x=5π2x=5\pi, and so on, so x=π2x=\frac{\pi}{2}, or x=5π2x=\frac{5\pi}{2}, ...
  2. Find xx with 3x1=03x-1=0 or 3x1=π3x-1=\pi, ..., so x=13x=\frac{1}{3}, or x=π+13x=\frac{\pi+1}{3}, ... .