Trigonometric Values for Special Angles

In the TMUA and ESAT, you are expected to know the exact values of trigonometric functions for several key angles. While these can be memorised, it is far more reliable to understand how to derive them using two specific 'standard triangles'. You should also be able to identify these values on the graphs of $\sin$, $\cos$, and $\tan$.

Deriving Values for $45^\circ$

To find the trigonometric values for $45^\circ$ (or $\frac{\pi}{4}$ radians), we consider an isosceles right-angled triangle. If we set the two shorter sides to have a length of 1, we can use Pythagoras' theorem to find the hypotenuse: $\sqrt{1^2 + 1^2} = \sqrt{2}$.

From this triangle, we can immediately read off the ratios for an angle of $45^\circ$:

1. $\sin 45^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}}$
2. $\cos 45^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}}$
3. $\tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1}{1} = 1$

Note that $\frac{1}{\sqrt{2}}$ is often written in its rationalised form as $\frac{\sqrt{2}}{2}$. Both forms are mathematically equivalent and acceptable in the exam.

Deriving Values for $30^\circ$ and $60^\circ$

For the angles $30^\circ$ ($\frac{\pi}{6}$ radians) and $60^\circ$ ($\frac{\pi}{3}$ radians), we use half of an equilateral triangle. We start with an equilateral triangle where every side has a length of 2. By dropping a perpendicular line from one vertex to the opposite side, we split the triangle into two congruent right-angled triangles.

This process creates a triangle with angles of $30^\circ$ and $60^\circ$. The base is halved to 1, the hypotenuse remains 2, and the vertical height is calculated via Pythagoras as $\sqrt{2^2 - 1^2} = \sqrt{3}$. Using this triangle, we find:

For $60^\circ$:

1. $\sin 60^\circ = \frac{\sqrt{3}}{2}$
2. $\cos 60^\circ = \frac{1}{2}$
3. $\tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3}$

For $30^\circ$:

1. $\sin 30^\circ = \frac{1}{2}$
2. $\cos 30^\circ = \frac{\sqrt{3}}{2}$
3. $\tan 30^\circ = \frac{1}{\sqrt{3}}$ (which is also $\frac{\sqrt{3}}{3}$)

Values for $0^\circ$ and $90^\circ$

The boundary values are best understood by looking at the unit circle or the trigonometric graphs. As the angle $\theta$ approaches $0^\circ$, the vertical component (sine) vanishes, while the horizontal component (cosine) reaches its maximum.

• At $0^\circ$: $\sin 0^\circ = 0$, $\cos 0^\circ = 1$, and $\tan 0^\circ = \frac{0}{1} = 0$.
• At $90^\circ$: $\sin 90^\circ = 1$, $\cos 90^\circ = 0$. For the tangent function, we observe that $\tan 90^\circ = \frac{1}{0}$, which is undefined. This corresponds to the vertical asymptote on the graph of $y = \tan x$.

Graph Identification and Extension

You should be able to locate these exact values on the standard sketches of $\sin x$, $\cos x$, and $\tan x$. For example, the point $(60^\circ, 0.5)$ sits on the cosine curve, and $(45^\circ, 1)$ sits on the tangent curve. Knowing these values allows you to find related angles in other quadrants, such as $120^\circ$ or $225^\circ$, by applying the symmetries and periodicities of the functions.