## Trigonometry Tutorial

#### Intro

This tutorial is Part 1 of 2 on Solving Trigonometric Equations. In this tutorial I will solve several different trigonometric equations using different techniques such as factoring, substitution and converting the equation to all one trigonometric function using identities. Solving trigonometric equations involves knowledge of several mathematical concepts such as factoring techniques we learn in an algebra class, trigonometric identities and functions and common angle values. Furthermore knowledge of inverse trigonometric functions, recognizing equation forms such as a quadratic equation and creativity are also helpful.

A free online Trigonometry book which I used as a reference can be obtained by going to www.stitz-zeager.com.

#### Sample Problem

Find all of the exact solutions of the equations and then list those solutions which lie in the interval .

#### Solution

Since we have a single trig function our objective is to isolate the trig function using algebra then solve for x.

Step 1) Isolate the trig function using algebra.

Subtract 1 from each side;

Divide each side by

Step 2) Replace the argument of 2x with a general letter such as u. Then solve for u in the interval adding , k an integer, to each value. Remember that the trig functions sin, cos, csc and sec are periodic in .

Solve .

The angles at which cos is is or . These values can be obtained from a figure of the unit circle with common angle values or from memory. Therefore

Step 3) Replace u with the original argument of 2x and solve for x.

All exact solutions is

You may be wondering why I put the solution for all values over one denominator. I did this to make the calculations easier to find all values of x in the interval .

Substituting 0, 1, 2, 3, … in for k until x is greater than gives

The solutions in is
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To solve this equation we must recognize that the left hand side of (1) resembles a sum to product identity. Our objective is to write the left hand side as the product of two sine functions using the sum to product formulas then solve for x.

The sum to product identity that we need is

Comparing the left sides of (1) and (2) we see that and . Substituting these values into (2) and using the identity we have

The angles at which sine is 0 is 0 or . Since these two values are apart we only need to add to the smaller value.

Note that is not needed because generates the same values when k is a multiple of 4. So my solution for all values is

All exact solutions is

Substituting 0, 1, 2, 3, … in for k until x is greater than gives

The solutions in is
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The objective here is to write the expression on the left side of the equation as a sine or cosine function of the form or . Note that this method can only be applied if the arguments are the same. The left side resembles part of the sum or difference formulas for sine or cosine. I pick cosine.

Set the left side of (1) equal to the right side of (2) we have

By comparing each side and equating coefficients we see that

To solve for A use the identity

Multiply each side by .

Substituting into (3) gives

I will pick A = 2. You could use A = -2. Now solve for .

The value of that satisfies both equations is . Either angle will work. I pick . Substituting A and into we have

Now solve for x.

The angles for which cosine is is .

All exact solutions is

Substituting 0, 1, 2, 3, … in for k in (4) and -1, 0, 1, 2, 3, … in for k in (5) until x is greater than gives

The solutions in is

Can you see why I started off with k = -1 in (5)? Remember we want to list those solutions which are in the interval . Do not always assume that k starts at zero.
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To solve this equation we need to move all terms to one side of the equal sign and get the same trig function using a Pythagorea identity. Then realize that the new equation has the form of a quadratic equation which then can be solved for using substitution, factoring directly or the quadratic formula.

The Pythagorean identity that we need is .

Note that we can factor (1) directly by considering tan(x) as the variable so

If you cannot see how I factored you can use the substitution method that I will demonstrate next. Remember that tan and cot are periodic every .

Solving using the substitution method

Replacing w with

Note that there is no common angle for so we use the arctan function, aka inverse trig functions. Remember that the inverse functions are defined over specific intervals and only gives one angle. We know that tangent is positive in quadrants 1 and 3; however, the arctan function gives only the first quadrant angle. Remember that if the two angles are apart we only need to add to the smaller angle.

All exact solutions is

The solutions in is
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To solve this equation we move all terms to the left side of the equal sign and realize that the new equation has the form of a quadratic equation which can be solved by factoring if possible or by using the quadratic formula.

Since (1) does not factor use the quadratic formula treating as the variable

All exact solutions is

The solutions in is