Problem and Solution
We use the trigonometric formulae for $ \cos(A - B) $ and $ \cos(A + B) $ to solve the given problems.
Part (a): Show that $ \cos(A - B) - \cos(A + B) = 2 \sin A \sin B $
Using the sum-to-product identities:
$ \cos(A - B) = \cos A \cos B + \sin A \sin B, $
$ \cos(A + B) = \cos A \cos B - \sin A \sin B. $
Subtracting these two:
$ \cos(A - B) - \cos(A + B) = (\cos A \cos B + \sin A \sin B) - (\cos A \cos B - \sin A \sin B). $
Simplify:
$ \cos(A - B) - \cos(A + B) = 2 \sin A \sin B. $
Part (b): Show that $ \cos 5\theta - \cos 9\theta = 2 \sin 7\theta \sin 2\theta $
Using the result from Part (a) with $ A = 7\theta $ and $ B = 2\theta $:
$ \cos(A - B) - \cos(A + B) = 2 \sin A \sin B. $
Substitute $ A = 7\theta $ and $ B = 2\theta $:
$ \cos(7\theta - 2\theta) - \cos(7\theta + 2\theta) = 2 \sin(7\theta) \sin(2\theta). $
Simplify:
$ \cos 5\theta - \cos 9\theta = 2 \sin 7\theta \sin 2\theta. $
Part (c): Solve $ \cos 5\theta - \cos 9\theta = \sqrt{3} \sin 7\theta $ for $ 0 < \theta \leq \frac{\pi}{3} $
Using the result from Part (b):
$ \cos 5\theta - \cos 9\theta = 2 \sin 7\theta \sin 2\theta. $
Substitute into the equation:
$ 2 \sin 7\theta \sin 2\theta = \sqrt{3} \sin 7\theta. $
Solve $ \sin 7\theta = 0 $
We know that $ \sin x = 0 $ when:
$ x = n\pi, \quad n \in \mathbb{Z}. $
For $ \sin 7\theta = 0 $, we have:
$ 7\theta = n\pi, \quad n \in \mathbb{Z}. $
Solve for $ \theta $:
$ \theta = \frac{n\pi}{7}. $
Restrict the solution to $ 0 < \theta \leq \frac{\pi}{3} $
From the given condition, $ 0 < \theta \leq \frac{\pi}{3} $. Substituting $ \theta = \frac{n\pi}{7} $, we require:
$ 0 < \frac{n\pi}{7} \leq \frac{\pi}{3}. $
Divide through by $ \pi $:
$ 0 < \frac{n}{7} \leq \frac{1}{3}. $
Multiply through by $ 7 $:
$ 0 < n \leq \frac{7}{3}. $
Since $ n $ must be an integer, the only possible values are $ n = 1, \text{ or } n=2 $.
Substitute $ n = 1 $ back into $ \theta = \frac{n\pi}{7} $:
$ \theta = \frac{\pi}{7}. $
Substitute $ n = 2 $ back into $ \theta = \frac{n\pi}{7} $:
$ \theta = \frac{2\pi}{7}. $
Divide through by $ \sin 7\theta $ (assuming $ \sin 7\theta \neq 0 $):
$ 2 \sin 2\theta = \sqrt{3}. $
Solve for $ \sin 2\theta $:
$ \sin 2\theta = \frac{\sqrt{3}}{2}. $
The angle $ 2\theta $ corresponding to $ \sin 2\theta = \frac{\sqrt{3}}{2} $ is:
$ 2\theta = \frac{\pi}{3} \text{ (or) } \frac{2\pi}{3}. $
Solve for $ \theta $:
$ \theta = \frac{\pi}{6} \text{ (or) } \frac{\pi}{3}. $
Final Answer
$ \boxed{\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{7}, \frac{2\pi}{7}.} $
Part (d): Evaluate $ \int_0^{\frac{\pi}{7}} 8 \sin 7x \cos 2x \tan 2x \, dx $
We are tasked with evaluating:
$ I = \int_0^{\frac{\pi}{7}} 8 \sin 7x \cos 2x \tan 2x \, dx. $
Step 1: Simplify the integrand
Recall that $ \tan 2x = \frac{\sin 2x}{\cos 2x} $. Substituting this into the integral:
$ I = \int_0^{\frac{\pi}{7}} 8 \sin 7x \cos 2x \cdot \frac{\sin 2x}{\cos 2x} \, dx. $
Simplify:
$ I = \int_0^{\frac{\pi}{7}} 8 \sin 7x \sin 2x \, dx. $
Step 2: Use the product-to-sum identities
Using the product-to-sum identity:
$ \sin A \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right], $
we rewrite the integrand:
$ \sin 7x \sin 2x = \frac{1}{2} \left[ \cos(7x - 2x) - \cos(7x + 2x) \right]. $
Substitute this into the integral:
$ I = \int_0^{\frac{\pi}{7}} 8 \cdot \frac{1}{2} \left[ \cos(5x) - \cos(9x) \right] \, dx. $
Simplify:
$ I = 4 \int_0^{\frac{\pi}{7}} \cos(5x) \, dx - 4 \int_0^{\frac{\pi}{7}} \cos(9x) \, dx. $
Step 3: Integrate each term
The integral of $ \cos(kx) $ is:
$ \int \cos(kx) \, dx = \frac{\sin(kx)}{k}. $
Using this, we compute each term:
1. For $ \int_0^{\frac{\pi}{7}} \cos(5x) \, dx $:
$ \int_0^{\frac{\pi}{7}} \cos(5x) \, dx = \left[ \frac{\sin(5x)}{5} \right]_0^{\frac{\pi}{7}} = \frac{\sin\left(\frac{5\pi}{7}\right)}{5} - \frac{\sin(0)}{5}. $
Simplify:
$ \int_0^{\frac{\pi}{7}} \cos(5x) \, dx = \frac{\sin\left(\frac{5\pi}{7}\right)}{5}. $
2. For $ \int_0^{\frac{\pi}{7}} \cos(9x) \, dx $:
$ \int_0^{\frac{\pi}{7}} \cos(9x) \, dx = \left[ \frac{\sin(9x)}{9} \right]_0^{\frac{\pi}{7}} = \frac{\sin\left(\frac{9\pi}{7}\right)}{9} - \frac{\sin(0)}{9}. $
Simplify:
$ \int_0^{\frac{\pi}{7}} \cos(9x) \, dx = \frac{\sin\left(\frac{9\pi}{7}\right)}{9}. $
Step 4: Combine the results
Substitute back into $ I $:
$ I = 4 \left( \frac{\sin\left(\frac{5\pi}{7}\right)}{5} \right) - 4 \left( \frac{\sin\left(\frac{9\pi}{7}\right)}{9} \right). $
Step 5: Evaluate numerically
Using numerical approximation (to 3 decimal places):
$ \sin\left(\frac{5\pi}{7}\right) \approx 0.7818, \quad \sin\left(\frac{9\pi}{7}\right) \approx -0.7818. $
Substitute these values:
$ I = 4 \left( \frac{0.7818}{5} \right) - 4 \left( \frac{-0.7818}{9} \right). $
Simplify:
$ I = 4 \cdot 0.1563 + 4 \cdot 0.0868. $
$ I = 0.6254 + 0.3474 = 0.9728. $
Final Answer
$ \boxed{I \approx 0.973} $
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