Trigonometry, Part 7

Another way to analyse trig equations is to use the graph of the trig function involved instead of the unit circle. The following graphs of the basic trig functions were obtained from https://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html.

sine graph
cosine graph
tangent graph

Let’s talk about the sine function, the first graph, though the following discussion can be generalised to the other two.

There are key points on this graph: the peaks and where it crosses the x-axis. The values of x where the sine is zero are 0, ±𝜋, ±2𝜋, ±3𝜋, … . The positive and negative peaks occur at ±𝜋/2, ±3𝜋/2, ±5𝜋/2, … . Also note that the sine goes from -1 to 1, a total distance of 2. The amplitude is defined as half of this distance, so the amplitude of the basic sine wave is 1.

Another aspect to notice is that the curve repeats every 2𝜋 radians. That means that given any x value, adding or subtracting any multiple of 2𝜋 to x will result in the same sine value. This is the definition of a function’s period. The basic sine function has a period of 2𝜋. Mathematically,

sin(x +2n𝜋) = sin(x), where n is any integer.

So let’s look at the solution to sin x = 1/2, 0 ≤ x ≤ 2𝜋. Instead of the unit circle, the sine graph can be used instead:

Noticing that 𝜋/6 has a sine of 1/2, we can use the symmetry of the sine graph to find the other angle with the same sine, 𝜋 – 𝜋/6 = 5𝜋/6. These are the two answers for the given domain.

What if the given domain was -2𝜋 ≤ x ≤ 2𝜋? Because the period of the basic sine wave is 2𝜋, you can add or subtract multiples of 2𝜋 to the basic angles found to get the answers in the required interval. In this case, subtract 2𝜋 from each of the basic angles previously found to get:

𝜋/6 – 2𝜋 = -11𝜋/6
5𝜋/6 – 2𝜋 = -7𝜋/6

as the other two answers.

If the problem at hand is more complex like 2sin(2x +𝜋) -1 which I solved in my last post, you solve it as before except you use the sine graph instead of the unit circle to find the basic angles, then proceed as before.