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:
\[ \sin(\frac{\pi}{2} – \frac{\pi}{6}) = \sin(\frac{\pi}{2})\cos(\frac{\pi}{6}) – \cos(\frac{\pi}{2})\sin(\frac{\pi}{6}) \]
\[= 1 \times \frac{\sqrt{3}}{2} – 0 \]
\[= \frac{\sqrt{3}}{2} \]
