Once again, Matheteus and Geometros reunite under the leafy canopy of the Academy’s grove, their minds eager to explore the intricate tapestry of trigonometry and its connection to the eternal Forms.

### Matheteus:

Ah, Geometros, shall we unravel more threads from the spool of trigonometric wisdom?

### Geometros:

Indeed, Matheteus, my mind thirsts for further elucidation. What shall it be today?

### Matheteus:

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

### Geometros:

Ah, these foundations are becoming like old friends. What new revelations can we expect?

### Matheteus:

We shall extend our understanding by examining an angle in the act of doubling—when $$A = B$$.

### Geometros:

A splendid idea! Proceed.

### Matheteus:

We start with the formula for $$\cos(A + B)$$:

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

If $$A = B$$, it transforms into:

$\cos(2A) = \cos^2(A) – \sin^2(A)$

### Geometros:

A marvelous simplification! And how might we further this discovery?

### Matheteus:

Ah, we call upon Pythagoras and his transcendent identity—$$\sin^2(A) + \cos^2(A) = 1$$:

By rearranging, we find two additional expressions for $$\cos(2A)$$:

$\cos(2A) = 2\cos^2(A) – 1$
$\cos(2A) = 1 – 2\sin^2(A)$

### Geometros:

Ah, three forms for $$\cos(2A)$$! A divine triptych!

### Matheteus:

And we must not neglect the sine. Recall the formula for $$\sin(A + B)$$:

$\sin(2A) = 2\sin(A)\cos(A)$

### Geometros:

The symmetry is overwhelming. It’s as if the heavens themselves have revealed their calculations to us.

### Matheteus:

With these double-angle formulas, we glimpse the inner workings of the eternal Forms, illustrating the unfathomable unity and consistency of the cosmos.

### Geometros:

Verily, Matheteus. The elegance of these equations guides us ever closer to the celestial harmony that governs all.

### Conclusion:

As the dialogue concludes, the scholars find themselves in awe, realizing that each equation is not just a mathematical truth but a clue, a stepping stone on their ceaseless journey toward understanding the eternal Forms.

### Exercise Section

Here are 10 exercises to further test your understanding of these formulas:

Exercise 1: Compute $$\cos(\frac{\pi}{3} + \frac{\pi}{3})$$ using the double angle formula.

Using $$\cos(2A) = \cos^2(A) – \sin^2(A)$$, we find $$\cos(\frac{2\pi}{3}) = \cos^2(\frac{\pi}{3}) – \sin^2(\frac{\pi}{3}) = \frac{1}{4} – \frac{3}{4} = -\frac{1}{2}$$.

Exercise 2: Compute $$\sin(\frac{\pi}{4} + \frac{\pi}{4})$$ using $$\sin(2A) = 2\sin(A)\cos(A)$$.

We find $$\sin(\frac{\pi}{2}) = 2\sin(\frac{\pi}{4})\cos(\frac{\pi}{4}) = 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = 1$$.

Exercise 3: Compute $$\cos(\frac{\pi}{3} + \frac{\pi}{3})$$ using the double angle formula.

Using $$\cos(2A) = \cos^2(A) – \sin^2(A)$$, we find $$\cos(\frac{2\pi}{3}) = \cos^2(\frac{\pi}{3}) – \sin^2(\frac{\pi}{3}) = \frac{1}{4} – \frac{3}{4} = -\frac{1}{2}$$.

Exercise 4: Compute $$\cos(\pi)$$ using any double angle formula.

Using $$\cos(2A) = 1 – 2\sin^2(A)$$, we find $$\cos(\pi) = 1 – 2\sin^2(\frac{\pi}{2}) = 1 – 2 \times 1 = -1$$.

Exercise 5: Compute $$\sin(\frac{5\pi}{6})$$ using $$\sin(2A) = 2\sin(A)\cos(A)$$.

We find $$\sin(\frac{5\pi}{6}) = 2\sin(\frac{5\pi}{12})\cos(\frac{5\pi}{12})$$ (Use trigonometric values of $$\frac{5\pi}{12}$$ for an exact solution).

Exercise 6: Simplify $$\cos(2\theta) + \sin^2(\theta)$$.

Using $$\cos(2\theta) = \cos^2(\theta) – \sin^2(\theta)$$ and $$\sin^2(\theta) + \cos^2(\theta) = 1$$, the expression simplifies to 1.

Exercise 7: Compute $$\cos(\frac{\pi}{6} + \frac{\pi}{6})$$ using any double angle formula.

Using $$\cos(2A) = 2\cos^2(A) – 1$$, we find $$\cos(\frac{\pi}{3}) = 2 \times \left(\frac{\sqrt{3}}{2}\right)^2 – 1 = \frac{1}{2}$$.

Exercise 8: Compute $$\cos(\frac{\pi}{3} – \frac{\pi}{3})$$.

Using $$\cos(A – B) = \cos(A)\cos(B) + \sin(A)\sin(B)$$, we find $$\cos(0) = \cos(\frac{\pi}{3})\cos(\frac{\pi}{3}) + \sin(\frac{\pi}{3})\sin(\frac{\pi}{3}) = 1$$.

Exercise 9: Compute $$\cos(\frac{\pi}{4} + \frac{\pi}{4})$$ using $$\cos(2A) = 1 – 2\sin^2(A)$$.

We find $$\cos(\frac{\pi}{2}) = 1 – 2\sin^2(\frac{\pi}{4}) = 1 – 2 \times \frac{1}{2} = 0$$.

Exercise 10: Find $$\sin(3\pi)$$ using $$\sin(2A) = 2\sin(A)\cos(A)$$.

We know that $$\sin(3\pi) = \sin(2\pi + \pi)$$.

Using $$\sin(2A + B) = \sin(2A)\cos(B) + \cos(2A)\sin(B)$$, we get:

$$\sin(3\pi) = 2\sin(\pi)\cos(\pi) + \cos(2\pi)\sin(\pi)$$

$$= 0 + 1 \times 0$$

$$= 0$$