Trigonometry, Part 4

I will now do some examples of using the trig identities covered in the previous posts. But before I do, I want to show you a table that gives the values of the main trig functions for common angles:

Angle, πœƒsinπœƒcosπœƒtanπœƒ
0010
πœ‹/6, 30Β°1/2√3Μ…/21/√3Μ…
πœ‹/4, 45Β°1/√2Μ…1/√2Μ…1
πœ‹/3, 60°√3Μ…/21/2√3Μ…
πœ‹/2, 90Β°10Undefined
πœ‹, 180Β°0-10
3πœ‹/2, 270Β°-10Undefined

Now let’s do some examples:

  1. If cos(πœƒ) = 0.8829 and πœƒ is in the first quadrant, find cos(3πœ‹/2 – πœƒ).

According to the identity developed before, cos(3πœ‹/2 – πœƒ) = -sinπœƒ. But what is sin(πœƒ)? To find this, we need the Pythagorean identity, sinΒ²(πœƒ) + cosΒ²(πœƒ) = 1:

\[\begin{equation*}
\text{sin}^{2} \theta +\text{cos}^{2} \theta \ =1\ \ \ \Longrightarrow \ \ \ \text{sin}( \theta ) =\sqrt{1-\text{cos}^{2} \theta }
\end{equation*}\] \[\begin{equation*}
\text{sin}( \theta ) =\sqrt{1-0.8829^{2}} =0.4696
\end{equation*}\]

Therefore, cos(3πœ‹/2 – πœƒ) = -sinπœƒ = -0.4696

2. If sin(πœƒ) = 0.1736, and πœƒ is in the first quadrant, find tan(3πœ‹/2 + πœƒ).

According to the identity developed before, tan(3πœ‹/2 + πœƒ) = -1/tanπœƒ = -cosπœƒ/sinπœƒ. Again, we need the Pythagorean identity to find cosπœƒ:

\[\begin{equation*}
\text{cos}( \theta ) =\sqrt{1-0.1736^{2}} =0.9848
\end{equation*}\]

Therefore, tan(3πœ‹/2 + πœƒ) = -0.9848/0.1736 = -5.6729.

3. Find the exact value of cos(5πœ‹/6) (without a calculator).

The clues here are that I have been developing trig identities and I just gave you a table of exact values of trig functions for common angles. Perhaps we can equate 5πœ‹/6 in terms of a common angle. Notice that 6πœ‹/6 – πœ‹/6 = 5πœ‹/6. That is 5πœ‹/6 = πœ‹ – πœ‹/6. We have a trig identity for this: cos(πœ‹ – πœƒ) = -cosπœƒ. In our case πœƒ = πœ‹/6 and cos(πœ‹/6) = √3Μ…/2 so cos(5πœ‹/6) = -√3Μ…/2.

By the way, you can find a decimal approximation to √3Μ…/2, but this will just be an approximation no matter how many decimal places you include since √3Μ…/2 is an irrational number and has non-repeating decimals. So the exact value is √3Μ…/2 since we have agreed that √3Μ… is the notation for the exact square root of 3.

4. If sinπœƒ = 4/5 and πœ‹/2 < πœƒ < πœ‹, find the exact value of cosπœƒ

The part πœ‹/2 < πœƒ < πœ‹ means that the angle is in the second quadrant which means the cosine is negative. As before, we can find the cosine via the Pythagorean identity:

\[\begin{equation*}
\text{cos}( \theta ) =\sqrt{1-(4/5)^{2}} =3/5
\end{equation*}\]

But as the angle is in the second quadrant, cosπœƒ = – 3/5.