Solving Inverse Trigonometric Equations

Let's take all this highfalutin knowledge and use it in some real, live problems. Otherwise, it's just hanging out at the mall all day spending money on handbags.

Sample Problem

Find .

First, we need to ask ourselves: where is cosine positive? (Is it near Waldo?)

Cosine is positive in Quadrants I and IV, but inverse cosine is limited to the range [0, π], so we're in Quadrant I.

Now we ask ourselves: for what angle is ?

Knowing your reference triangles will be really helpful here (along with an extra-large pizza, hold the anchovies).

Sample Problem

Find .

Where is sine positive?

Sine is positive in Quadrants I and II, but inverse sine is limited by . Therefore, we must be partying in Quadrant I. We ask ourselves: for what angle is ?

Think: reference triangles.

Sample Problem

Find tan-1(1).

Where is tangent positive?

Tangent is positive in Quadrants I and III, but inverse tangent is limited to the range . That's just how he rolls. We must be in Quadrant I.

Again, we ask ourselves: for what angle is tan θ = 1?

Reference triangles, anyone?

tan-1(1) = 45°

Sample Problem

Find .

Ask yourself: where is cosine negative? (That's so rad...)

Cosine is negative in Quadrants II and III, but inverse cosine is limited to [0, π]. So our angle's gotta be in Quadrant II.

We ask ourselves: for what angle is ? Man, we're sure asking ourselves a lot of questions.

Think: reference triangles (if you're tired of thinking, chill with "Angry Birds").

If you're stuck on the 150°, here's what you can do. The reference angle is 30°, but we're in Quadrant II right now. That means we need to subtract our reference angle from 180° to get the actual angle.

180° – 30° = 150°

Sample Problem

Find

Whoa. What we're really looking for is the tangent of the angle whose cosine is negative one-half. Let's work from the inside out. (This approach works in math, and maybe psychology.)

Solve the inside first. For what angle is ?

Once again, it's back to those reference triangles, but now ask yourself that weighty question: where is cosine negative?

But inverse cosine is bounded by [0, π], which means our angle is in Quadrant II.

Now we find tan(120°).

Tangent is negative in Quadrant II, so we'll have a negative answer.

So .