Michael has taught college-level mathematics and sociology; high school math, history, science, and speech/drama; and has a doctorate in education.
How to Solve Trigonometric Equations: Practice Problems
Table of Contents
ShowIf you've been working in trigonometry, you've probably seen sines, cosines, tangents, and angle relationships until it's nearly driven you nuts. Before you lock yourself in a psych ward somewhere, let's take a look at some of them, and see if we can't make them just a bit easier.
A trigonometric equation is an equation that involves trigonometric functions, ratios of the sides of a right triangle. Trig functions help us solve many kinds of problems. In this lesson, we'll be looking for the angle(s) that make an equation true.
The table and diagram below are designed to help you understand and remember the key ratios that make up basic trig functions. The basic trig functions (or identities) are ratios of two of the three sides of a right triangle, and each function has a reciprocal function (where the numerator and denominator are reversed):
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| Function | Ratio | Reciprocal |
|---|---|---|
| Sine A | a / h | Cosecant A |
| Cosine A | b / h | Secant A |
| Tangent A | a / b | Cotangent A |
| Secant A | h / b | Cosign A |
| Cosecant A | h / a | Sine A |
| Cotangent A | b / a | Tangent A |
| Function | Ratio | Reciprocal |
|---|---|---|
| Sine B | b / h | Cosecant B |
| Cosine B | a / h | Secant B |
| Tangent B | b / a | Cotangent B |
| Secant B | h / a | Cosign B |
| Cosecant B | h / b | Sine B |
| Cotangent B | a / b | Tangent B |
When you're solving trig equations, you're looking for x (or some other value), using trig identities and the rules of algebra. You simplify the equation down to a single identity, such sin(x) = .5, then use your calculator or a table to find out the value of angle x. The following is a simple example:
- 3 sin(x) = 1.5
- sin(x) = .5 (divide both sides by 3)
- x = 30° (look up .5 on a sine table, or pull the arc-sign from your calculator)
30° is just one of the solutions. An angle is a measure of rotation, and if you keep rotating you'll get more angles with the same sine. The same ratio of sides happens at 150°, 390°, 510°, etc. Not only that, you can rotate backward (clockwise), and get negative angles. -210° and -330° also have a .5 sine.
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Because of this, most trig problems include a domain limitation (upper and lower limits for possible inputs). For example, domains might be limited to 0° to 180°, 0° to 360°, -180° to 180°, etc.
As a side note, remember that some trig problems work in radians instead of degrees. It's just another way to express angles. Radians use the π properties of a circle, and instead of expressing angles as part of a 360° circle, they express angles as a part of a 2π circle. Since a circle with a radius of 1 will have a circumference of 2π, any angle can be expressed as part of that 2π rotation. We'll stick to degrees in these exercises. A table of values has been included below to help you with your trig adventures:
| Angle | Radians | Sine Ratio | Cosine Ratio | Tangent Ratio |
|---|---|---|---|---|
| 30° | π/6 | 1/2 | √3/2 | √3/3 |
| 45° | π/4 | √2/2 | √2/2 | 1 |
| 60° | π/3 | √3/2 | 1/2 | √3 |
| 90° | π/2 | 1 | 0 | 1/0 = undefined |
| 120° | 2π/3 | √3/2 | -1/2 | -√3 |
| 135° | 3π/4 | √2/2 | -√2/2 | -1 |
| 150° | 5π/6 | 1/2 | -√3/2 | -√3/3 |
| 180° | π | 0 | -1 | 0/1=0 |
| 210° | 7π/6 | -1/2 | -√3/2 | √3/3 |
| 225° | 5π/4 | -√2/2 | -√2/2 | 1 |
| 240° | 4π/3 | -√3/2 | -1/2 | √3 |
| 270° | 3π/2 | -1 | 0 | 1/0 = undefined |
| 300° | 5π/3 | -√3/2 | 1/2 | -√3 |
| 315° | 7π/4 | -√2/2 | √2/2 | -1 |
| 330° | 11π/6 | -1/2 | √3/2 | -√3/3 |
All right, let's try one. It looks pretty scary, but remember, treat the sin(x) terms like variables and use algebraic rules. This example is a quadratic equation, so let's treat it like one.
- 2 sin²(x) - sin(x) = 1
- 2 sin²(x) - sin(x) - 1 = 0 (subtract 1 from both sides to form a trinomial)
- (2 sin(x) + 1) (sin(x) - 1) = 0 (factor the trinomial)
- 2 sin(x) + 1 = 0 OR sin(x) - 1 = 0
Solving for sin(x), we have the following:
- 2 sin(x) + 1 = 0
- 2 sin(x) = -1 (subtract 1 from both sides)
- sin(x) = -1/2 (divide both sides by 2)
For the second solution, we have these steps:
- sin(x) - 1 = 0
- sin(x) = 1 (add 1 to both sides)
Glancing at our table, we can see that our two values for x will show up at 90° (x = 1), 210° (x = -1/2), and 330° (x = -1/2). Remember, there are also an infinite number of other positions (in other rotations) where x also appears.
Okay, let's try one where the domain is pre-defined for us. Once again, factoring will help us. Remember, you can treat cot(x), cos(x), and other trig functions like algebraic variables.
- cot(x) cos²(x) - 2cot(x) = 0, where x is between -180° and 180°
- (cot(x)) (cos²(x) - 2) = 0 (factor out the cot(x) term)
- cot(x) = 0 OR cos²(x) - 2 = 0
The first solution, cot(x) = 0, is already simplified, but we have to do a little more with the second one:
- cos²(x) - 2 = 0
- cos²(x) = 2 (add 2 to both sides)
- cos(x) = √2 or -√2 (take the square root of both sides)
Since cot(x) is the reciprocal of tan(x), it will be 0 every time tan(x) is undefined (0/1 instead of 1/0). Within the given domain (-180° to 180°), the 90° angle and -90° angle meet that condition, so that's two of our answers.
So what about our two cos(x) solutions? Well, remember that the cosine is the ratio of the adjacent side to the hypotenuse in a triangle, so sine and cosine can never be larger than one or smaller than -1. In this case, the cosine solutions have no associated angles.
Trigonometric equations, equations that involve trigonometric functions (ratios of the sides of a right triangle), can be solved using algebraic steps, trig rules, and conversions. A domain limitation establishes upper and lower limits for possible input angles, otherwise you have to allow for an infinite number of solutions in positive and negative rotations.
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