Question

The number of solutions that the equation $\sin 5\theta \cos 3\theta = \sin 9\theta \cos 7\theta$ has in $\left[0, \frac{\pi}{4}\right]$

Answer 1

5

Answer 2

Solution:

Given the equation

$$\sin5\theta \cos3\theta = \sin9\theta \cos7\theta,$$

apply the identity

$$\sin A \cos B = \frac{1}{2}\left[\sin(A+B) + \sin(A-B)\right].$$

Thus,

$$\sin5\theta \cos3\theta = \frac{1}{2}\left[\sin(8\theta) + \sin(2\theta)\right]$$

and

$$\sin9\theta \cos7\theta = \frac{1}{2}\left[\sin(16\theta) + \sin(2\theta)\right].$$

Cancelling $\frac{1}{2}$ and $\sin(2\theta)$ from both sides, we have:

$$\sin(8\theta)=\sin(16\theta).$$

Recall that $\sin a=\sin b$ if either:

1. $a = b + 2\pi k$, or
2. $a = \pi - b + 2\pi k$, for any integer $k$.

Case 1: $8\theta = 16\theta + 2\pi k$

$$8\theta - 16\theta = 2\pi k \quad \Rightarrow \quad -8\theta = 2\pi k \quad \Rightarrow \quad \theta = -\frac{k\pi}{4}.$$

We need $\theta \in \left[0, \frac{\pi}{4}\right]$. Thus:

• For $k=0$: $\theta=0$,
• For $k=-1$: $\theta=\frac{\pi}{4}$.

Case 2: $8\theta = \pi - 16\theta + 2\pi k$

$$8\theta + 16\theta = \pi + 2\pi k \quad \Rightarrow \quad 24\theta = \pi(1+2k) \quad \Rightarrow \quad \theta = \frac{\pi(1+2k)}{24}.$$

Again, for $\theta \in \left[0, \frac{\pi}{4}\right]$ (i.e. $\theta \le \frac{\pi}{4}=\frac{6\pi}{24}$):

$$0 \le \frac{\pi(1+2k)}{24} \le \frac{\pi}{4} \quad \Rightarrow \quad 0 \le 1+2k \le 6.$$

Solve for integer $k$:

• For $k=0$: $1+0=1$ ⇒ $\theta=\frac{\pi}{24}$,
• For $k=1$: $1+2=3$ ⇒ $\theta=\frac{3\pi}{24}=\frac{\pi}{8}$,
• For $k=2$: $1+4=5$ ⇒ $\theta=\frac{5\pi}{24}$.

These three solutions do not duplicate the ones from Case 1.

Total number of solutions: $2$ (from Case 1) $+ 3$ (from Case 2) $= 5$.

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Explanation (minimal):

• Express both products using $\sin A \cos B$ identity.
• Cancel common terms to get $\sin 8\theta = \sin 16\theta$.
• Solve using $\sin a = \sin b$ conditions.
• Count all solutions in $\left[0, \frac{\pi}{4}\right]$.