5-16 Solving Basic Trigonometric Equations Solve the given equation. tanθ=-(1)/(3)

Name: Problem 16, Chapter 7: Algebra and Trigonometry
Uploaded: 2019-06-24T18:48:31-08:00
Duration: 4 min 58 s
Channel: Mihir Nayar
Description: 5-16 Solving Basic Trigonometric Equations Solve the given equation. tanθ=-(1)/(3)

5-16 Solving Basic Trigonometric Equations Solve the given...

Key Concepts

Periodicity of Tangent

The tangent function has a period of ?, meaning that if an angle satisfies the equation, then adding or subtracting any multiple of ? will yield another valid solution. This periodic nature is applied to express the general solution of equations involving the tangent function.

Reference Angle

The reference angle is the acute angle formed by the terminal side of an angle and the horizontal axis. When solving equations like these, the reference angle is calculated first using the absolute value of the trigonometric function, and then adjusted to the appropriate quadrants based on the given sign of the function.

Quadrant Analysis

Quadrant analysis involves determining which quadrants an angle falls into based on the sign of the trigonometric function. For the tangent function, which is negative when the sine and cosine have opposite signs, it is important to identify the correct quadrants to adjust the reference angle accordingly, ensuring all solutions are accounted for.

Trigonometric Equations

Trigonometric equations involve functions such as sine, cosine, and tangent, and are solved by finding all angle measures that satisfy the given condition. This process often involves using algebraic manipulation, inverse trigonometric functions, and identities to derive the general solution over the domain of the function.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as arctan, are used to determine the principal value of an angle that produces a given trigonometric value. They provide the initial or reference angle from which all possible solutions can be generated, while keeping in mind the range restrictions of the inverse functions.

Transcript

00:01 Hey guys, today we're solving problem number 16 on page 605 of the textbook, which gives us this simple trigonometric equation.

00:11 The tangent of theta equals negative one -third.

00:15 So the first thing you're going to want to notice is that you have a tangent trigonometric function and the inverse tangent should equal a negative 1 over 3.

00:26 Now i'm noticing this.

00:28 If we look at our unit circle, clearly we're not going to, like get like a 5 pi over 6 as our answer or like a 7 pi over 4.

00:35 However, we are going to be dealing with those specific quadrants.

00:39 We'll be dealing with quadrants 2 and 4.

00:41 And that's because tangent is the same as sign over cosine as we're in up here.

00:47 And in order for it to be negative, you need to have opposite sign and cosine signs.

00:52 So we're in difference 1 and 3.

00:56 So automatically, you know you're only going to have one, so like one solution.

01:02 In the period of tangent.

01:04 And remember that the period of tangent, this is an important distinction between sine, cosine, and tangent.

01:11 For tangent, the period is equivalent to not 2 pi.

01:17 It's actually equivalent to pi radiance.

01:21 Again, not 2 pi radiance, pi radiance for tangent.

01:25 That's going to be an important distinction later on.

01:27 So in order to solve this problem, you simply write that theta is the inverse tangent.

01:35 So on your calculator, make sure in radiance mode when you do this, put in tan negative 1 of negative 1 over 3, and your answer ends up being negative 0 .32.

02:02 So on our unit circle, that would be somewhere around here, let's say.

02:07 So there's our negative 0 .322.

02:15 But the problem with using that as kind of our base value to add something to is that it's not between our range of tangent.

02:23 And remember the range for tangent, so the range, since the period is pi, the range can only be as big as pie itself.

02:34 And that range is going to be from 0 to pi.

02:38 And unfortunately, when dealing with negative numbers, tangents just don't work.

02:43 So what we're going to have to do with this negative point 322 is just add pi to it, and you'd end up somewhere around here...

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