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 | sinπ | cosπ | tanπ |

0 | 0 | 1 | 0 |

π/6, 30Β° | 1/2 | β3Μ /2 | 1/β3Μ |

π/4, 45Β° | 1/β2Μ | 1/β2Μ | 1 |

π/3, 60Β° | β3Μ /2 | 1/2 | β3Μ |

π/2, 90Β° | 1 | 0 | Undefined |

π, 180Β° | 0 | -1 | 0 |

3π/2, 270Β° | -1 | 0 | Undefined |

Now let’s do some examples:

- If cos(π) = 0.8829 and π is in the first quadrant, find cos(3π/2 – π).

According to the identity developed before, cos(3π/2 β π) = -sinπ. But what is sin(π)? To find this, we need the Pythagorean identity, sinΒ²(π) + cosΒ²(π) = 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/2 β π) = -sinπ = -0.4696

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

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

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

\end{equation*}\]

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

3. Find the exact value of cos(5π/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/6 in terms of a common angle. Notice that 6π/6 – π/6 = 5π/6. That is 5π/6 = π – π/6. We have a trig identity for this: cos(π β π) = -cosπ. In our case π = π/6 and cos(π/6) = β3Μ /2 so cos(5π/6) = -β3Μ /2.

By the way, you can find a decimal approximation to β3Μ /2, but this will just be an approximation no matter how many decimal places you include since β3Μ /2 is an irrational number and has non-repeating decimals. So the exact value is β3Μ /2 since we have agreed that β3Μ is the notation for the exact square root of 3.

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

The part π/2 < π < π 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 = – 3/5.