Given: $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$ 6.1.1 Use the given identity to derive a formula for $\cos(\alpha + \beta)$ - NSC Mathematics - Question 6 - 2021 - Paper 2

To simplify the expression $2\cos 6x \cos 4x - \cos 10x + 2\sin^2 x$, we start by applying the product-to-sum identities:

1. For $2\cos A \cos B$, the identity is:
   $2\cos A \cos B = \cos(A+B) + \cos(A-B)$
   Therefore,
   $2\cos 6x \cos 4x = \cos(10x) + \cos(2x)$

2. Substitute that into the original expression:
   $\cos(10x) + \cos(2x) - \cos(10x) + 2\sin^2 x$

3. The $\cos(10x)$ terms cancel:
   $\cos(2x) + 2\sin^2 x$

4. Recall that $\sin^2 x = 1 - \cos^2 x$, we can use the Pythagorean identity:
   Thus,
   $\cos(2x) + 2(1 - \cos^2 x) = \cos(2x) + 2 - 2\cos^2 x$

Therefore, the final simplified expression is:

$\cos(2x) + 2 - 2\cos^2 x$

To solve the equation $\tan x = 2\sin 2x$, we start by expressing $\sin 2x$ in terms of $\tan x$:

1. Use the identity $\sin 2x = 2\sin x \cos x$:
   Therefore,
   $\tan x = 4\sin x \cos x$

2. Replace $\tan x$ with $\frac{\sin x}{\cos x}$:
   $\frac{\sin x}{\cos x} = 4 \sin x \cos x$
   Cross-multiplying gives:
   $\sin x = 4\sin x \cos^2 x$

3. Factor out $\sin x$:
   $\sin x (1 - 4\cos^2 x) = 0$
   This leads to two equations:

   • $\sin x = 0$
   • $1 - 4\cos^2 x = 0$, thus $\cos^2 x = \frac{1}{4}$, hence $\cos x = \pm \frac{1}{2}$.

4. For $\sin x = 0$, we have: $x = k\pi, \quad k \in \mathbb{Z}$
   For $\cos x = -\frac{1}{2}$, we have: $x = 120^{\circ} + k \cdot 360^{\circ} \quad \text{or} \quad x = 240^{\circ} + k \cdot 360^{\circ}, \quad k \in \mathbb{Z}$
   But we need to ensure $\cos x < 0$ which applies here.

Thus, the general solutions are: $x = k\pi ext{ and } x = 120^{\circ} + k\cdot 360^{\circ} ext{ or } x = 240^{\circ} + k\cdot 360^{\circ}, \quad k \in \mathbb{Z}$