Trigonometry, Part 5

Now let’s turn to solving equations with trig functions. Before we get into it though, let me introduce the inverse trig functions.

Just like square and square root, logs and exponentials, add and subtract, etc are inverse operations, there are corresponding inverse functions for the trig functions. These functions are identified usually by a -1 exponent. For example, sin-1(0.25) is asking the question: what angle has a sine of 0.25? Since sin-1x is an inverse function, then sin-1(sin x) = sin(sin-1 x) = x. That is, they undo each other. The same notation is used for the inverse cosine and tangent: cos-1x, tan-1x. Some older calculators may use the notation arcsin(arcsine), arccos(arccosine), and arctan(arc tangent), or they may further abbreviate these as asin, acos, atan. However, these are the same things as the corresponding inverse functions.

Solving trig equations, for the most part, use the same skills you already know for other types of algebraic equations. However, you need to be aware of the cyclic nature of trig functions and the varying signs of these functions in different quadrants, in order to get the correct and complete answers. For example, solve

√2̅cos(x) + 1 = 0, 0 ≤ x ≤ 2𝜋

First notice that frequently, a domain of the equation is specified. This is the 0 ≤ x ≤ 2𝜋 part. In most trig equations, where the unknown is in the trig expression, a domain needs to be specified or there will be an infinite number of solutions. By the way, the expression that is being acted upon by the trig function is called the argument of that function. For example the argument of sin(x² + 7) is x² + 7. In our example, the argument of the cosine is simply x.

So to solve this, you use your algebra skills to get,

\[\begin{equation*}
\text{cos}( x) =-\frac{1}{\sqrt{2}}
\end{equation*}\]

So we can take the inverse cosine of both side and use our calculator to find x, and you would find that x =3𝜋/4. However, let’s look at the equation before we take the inverse cosine. You may be asked to solve this without a calculator. You should notice that the cos(x) is equal to an entry in the table of common values I presented in my last post (except for the minus sign). It appears that our answers are associated with the angle 𝜋/4.

Now we have two trig identities that will solve this for us, namely

cos(𝜋 – 𝜃) = -cos𝜃
cos(𝜋 + 𝜃) = -cos𝜃

So, as we want the negative of cos(𝜋/4), then the two angles that will solve this equation within the domain 0 ≤ x ≤ 2𝜋 are 𝜋 – 𝜋/4 = 3𝜋/4 and 𝜋 + 𝜋/4 = 5𝜋/4. Notice that your calculator only gave one of these answers.

Instead of using the trig identities directly, I prefer to use the unit circle to guide me to all correct answers. I highly recommend drawing a unit circle for a particular problem to help guide you to the correct solutions. For this problem:

For practice, draw the unit circle for the same problem, with the provided domain as -𝜋 ≤ x ≤ 𝜋. You should be able to see that the two answers would be -3𝜋/4 and 3𝜋/4.

I will do more examples in my next post.