In the Academy’s grove, two scholars, Matheteus and Geometros, find themselves engaged in spirited conversation.

### Matheteus:

Ah, Geometros, have you not marveled at the wondrous interplay of angles and circles? I speak, of course, of the harmonic functions of sine and cosine.

### Geometros:

Indeed, Matheteus, for their laws unfurl the very fabric of the cosmos. Pray, speak more on this subject.

### Matheteus:

Gladly. You see, when two angles join in celestial union, or when they part ways, their sines and cosines respond thusly:

$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$
$\cos(A + B) = \cos(A)\cos(B) – \sin(A)\sin(B)$
$\sin(A – B) = \sin(A)\cos(B) – \cos(A)\sin(B)$
$\cos(A – B) = \cos(A)\cos(B) + \sin(A)\sin(B)$

### Geometros:

Ah, a divine symmetry! But what, dear Matheteus, are the higher implications?

### Matheteus:

Consider this: the angles, when they join or separate, hold true to these laws, as if obeying a higher order. Is this not evidence of the Forms—the perfect, unchanging ideals that exist in the realm beyond?

### Geometros:

Indeed, it points to a cosmic harmony, a celestial choreography directed by the Architect of Forms.

### Matheteus:

Before we depart this enlightening conversation, shall we example concrete examples to solidify our understanding?

### Geometros:

Ah, examples-the shadows on the wall of the cave that point us toward the higher Forms. Proceed, Matheteus.

### Matheteus:

For our first example, let’s consider angles $$A = \frac{\pi}{6}$$ and $$B = \frac{\pi}{4}$$.

### Geometros:

A more natural choice, Matheteus. The language of radians, after all, resonates with the Forms.

### Matheteus:

First, we shall find $$\sin(A + B)$$ using the sacred formula:

$\sin\left(\frac{\pi}{6} + \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{4}\right)$

$= \frac{1}{2} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}$

$= \frac{\sqrt{2} + \sqrt{6}}{4}$

### Geometros:

Ah, there it is—a result that resonates in harmony with the Forms!

### Matheteus:

Now, let’s explore $$\cos(A – B)$$ using our second holy formula:
$\cos\left(\frac{\pi}{6} – \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{4}\right)$

$= \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} + \frac{1}{2} \times \frac{\sqrt{2}}{2}$

$= \frac{\sqrt{6} + \sqrt{2}}{4}$

### Geometros:

Sublime! The numbers, they sing in the language of the spheres, guided by the Divine Forms!

So the scholars part, their dialogue serving as a terrestrial mirror to the celestial harmonies, resonant with the eternal Forms and the Divine Choreography of the cosmos.

## Exercises

1. Compute $$\sin(\frac{\pi}{6} + \frac{\pi}{4})$$.

Click to reveal solution

Using the angle addition formula for sine:

$\sin(\frac{\pi}{6} + \frac{\pi}{4}) = \sin(\frac{\pi}{6})\cos(\frac{\pi}{4}) + \cos(\frac{\pi}{6})\sin(\frac{\pi}{4})$

$= \frac{1}{2} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}$

$= \frac{\sqrt{2} + \sqrt{6}}{4}$

2. Compute $$\cos(\frac{\pi}{3} – \frac{\pi}{6})$$.

Click to reveal solution

Using the angle subtraction formula for cosine:

$\cos(\frac{\pi}{3} – \frac{\pi}{6}) = \cos(\frac{\pi}{3})\cos(\frac{\pi}{6}) + \sin(\frac{\pi}{3})\sin(\frac{\pi}{6})$

$= \frac{1}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \times \frac{1}{2}$

$= \frac{1}{2}$

3. Compute $$\sin(2\pi + \frac{\pi}{4})$$.

Click to reveal solution

Since $$\sin(2\pi) = 0$$ and $$\cos(2\pi) = 1$$, using the angle addition formula for sine:

$\sin(2\pi + \frac{\pi}{4}) = \sin(2\pi)\cos(\frac{\pi}{4}) + \cos(2\pi)\sin(\frac{\pi}{4})$

$= 0 + 1 \times \frac{\sqrt{2}}{2}$

$= \frac{\sqrt{2}}{2}$

4. Compute $$\cos(-\frac{\pi}{2} + \frac{\pi}{3})$$.

Click to reveal solution

Using the angle addition formula for cosine:

$\cos(-\frac{\pi}{2} + \frac{\pi}{3}) = \cos(-\frac{\pi}{2})\cos(\frac{\pi}{3}) – \sin(-\frac{\pi}{2})\sin(\frac{\pi}{3})$

$= 0 – (-1) \times \frac{\sqrt{3}}{2}$

$= \frac{\sqrt{3}}{2}$

5. Compute $$\sin(\frac{\pi}{2} – \frac{\pi}{6})$$.

Click to reveal solution

Using the angle subtraction formula for sine:

$\sin(\frac{\pi}{2} – \frac{\pi}{6}) = \sin(\frac{\pi}{2})\cos(\frac{\pi}{6}) – \cos(\frac{\pi}{2})\sin(\frac{\pi}{6})$

$= 1 \times \frac{\sqrt{3}}{2} – 0$

$= \frac{\sqrt{3}}{2}$