To solve the given trigonometric expression [tex]\(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x\)[/tex], we can use the sum-to-product identities for synthesizing and simplifying the trigonometric functions.
First, recall the sum-to-product identities for sine and cosine:
1. [tex]\(\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)\)[/tex]
2. [tex]\(\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)\)[/tex]
Let's apply these identities to the numerator and the denominator of the given expression:
### Numerator:
[tex]\[
\sin 5x + \sin 3x = 2 \sin \left(\frac{5x + 3x}{2}\right) \cos \left(\frac{5x - 3x}{2}\right)
\][/tex]
[tex]\[
= 2 \sin \left(\frac{8x}{2}\right) \cos \left(\frac{2x}{2}\right)
\][/tex]
[tex]\[
= 2 \sin 4x \cos x
\][/tex]
### Denominator:
[tex]\[
\cos 5x + \cos 3x = 2 \cos \left(\frac{5x + 3x}{2}\right) \cos \left(\frac{5x - 3x}{2}\right)
\][/tex]
[tex]\[
= 2 \cos \left(\frac{8x}{2}\right) \cos \left(\frac{2x}{2}\right)
\][/tex]
[tex]\[
= 2 \cos 4x \cos x
\][/tex]
Now, substituting these into the original expression, we have:
[tex]\[
\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \frac{2 \sin 4x \cos x}{2 \cos 4x \cos x}
\][/tex]
Since [tex]\(2\)[/tex] and [tex]\(\cos x\)[/tex] are common factors in both the numerator and the denominator, they cancel each other out:
[tex]\[
\frac{2 \sin 4x \cos x}{2 \cos 4x \cos x} = \frac{\sin 4x}{\cos 4x}
\][/tex]
By the definition of the tangent function [tex]\(\tan x = \frac{\sin x}{\cos x}\)[/tex], we get:
[tex]\[
\frac{\sin 4x}{\cos 4x} = \tan 4x
\][/tex]
Hence, we have shown that:
[tex]\[
\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x
\][/tex]
