(3) $$ \frac{\sin 4 heta-\sin 2 heta}{\cos 4 heta+\cos 2 heta}= an heta $$

Question

(3) $$ \frac{\sin 4 	heta-\sin 2 	heta}{\cos 4 	heta+\cos 2 	heta}=	an 	heta $$

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We start with the equation

$$
\frac{\sin 4\theta - \sin 2\theta}{\cos 4\theta + \cos 2\theta} = \tan \theta.
$$

**Step 1. Use sum-to-product identities**

Recall the identities

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

For our numerator, let $$A = 4\theta$$ and $$B = 2\theta$$. Then

$$
\sin 4\theta - \sin 2\theta = 2 \cos\left(\frac{4\theta+2\theta}{2}\right) \sin\left(\frac{4\theta-2\theta}{2}\right)
= 2 \cos(3\theta) \sin\theta.
$$

For the denominator, with the same choices we have

$$
\cos 4\theta + \cos 2\theta = 2 \cos\left(\frac{4\theta+2\theta}{2}\right) \cos\left(\frac{4\theta-2\theta}{2}\right)
= 2 \cos(3\theta) \cos\theta.
$$

**Step 2. Substitute back into the equation**

Substituting the results into the original equation gives

$$
\frac{2 \cos(3\theta) \sin\theta}{2 \cos(3\theta) \cos\theta} = \tan \theta.
$$

Assuming that the factors being cancelled are nonzero (i.e. $$\cos(3\theta) \neq 0$$ and $$\cos\theta \neq 0$$), we cancel to obtain

$$
\frac{\sin\theta}{\cos\theta} = \tan \theta.
$$

Since $$\frac{\sin\theta}{\cos\theta}$$ is exactly $$\tan \theta,$$ the equation becomes

$$
\tan \theta = \tan \theta,
$$

which is an identity.

**Step 3. Determine the restrictions**

While the simplified equation holds for all $$\theta$$, we must note the restrictions from the cancellation:

1. We canceled a factor $$2 \cos(3\theta)$$, so we must have
   $$
   \cos(3\theta) \neq 0.
   $$
   That is, 
   $$
   3\theta \neq \frac{\pi}{2} + k\pi, \quad k\in\mathbb{Z}.
   $$
   Equivalently,
   $$
   \theta \neq \frac{\pi}{6} + \frac{k\pi}{3}.
   $$

2. We also cancelled $$\cos\theta$$ (implicitly by writing $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$), so we require
   $$
   \cos\theta \neq 0.
   $$
   That is,
   $$
   \theta \neq \frac{\pi}{2} + k\pi, \quad k\in\mathbb{Z}.
   $$

Also, originally the denominator $$\cos 4\theta + \cos 2\theta$$ must not be zero. Notice that

$$
\cos 4\theta + \cos 2\theta = 2\cos(3\theta)\cos\theta,
$$

so the condition $$\cos 4\theta + \cos 2\theta \neq 0$$ is equivalent to

$$
\cos(3\theta) \neq 0 \quad \text{and} \quad \cos\theta \neq 0.
$$

**Final Answer**

The equation

$$
\frac{\sin 4\theta - \sin 2\theta}{\cos 4\theta + \cos 2\theta} = \tan \theta
$$

is identically true for all values of $$\theta$$ except when

$$
\theta = \frac{\pi}{2} + k\pi \quad \text{or} \quad \theta = \frac{\pi}{6} + \frac{k\pi}{3}, \quad k\in\mathbb{Z}.
$$
The equation $$ \frac{\sin 4\theta - \sin 2\theta}{\cos 4\theta + \cos 2\theta} = \tan \theta $$ is true for all $$\theta$$ except when $$\theta = \frac{\pi}{2} + k\pi$$ or $$\theta = \frac{\pi}{6} + \frac{k\pi}{3}$$, where $$k$$ is any integer.

(3) \( \frac{\sin 4 \theta-\sin 2 \theta}{\cos 4 \theta+\cos 2 \theta}=\tan \theta \)

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