To verify the trigonometric identity:
[tex]\[
\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta
\][/tex]
we can simplify both the numerator and the denominator to see if they equal \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).
Starting with the numerator:
[tex]\[
\sin \theta - 2 \sin^3 \theta
\][/tex]
We can factor out \(\sin \theta\):
[tex]\[
\sin \theta (1 - 2 \sin^2 \theta)
\][/tex]
Using the Pythagorean identity \(\sin^2 \theta = 1 - \cos^2 \theta\), substitute for \(\sin^2 \theta\):
[tex]\[
\sin \theta \left(1 - 2(1 - \cos^2 \theta) \right) = \sin \theta (1 - 2 + 2 \cos^2 \theta) = \sin \theta (-1 + 2 \cos^2 \theta)
\][/tex]
The numerator is now:
[tex]\[
\sin \theta (2 \cos^2 \theta - 1)
\][/tex]
Next, let's simplify the denominator:
[tex]\[
2 \cos^3 \theta - \cos \theta
\][/tex]
We can factor out \(\cos \theta\):
[tex]\[
\cos \theta (2 \cos^2 \theta - 1)
\][/tex]
So, the denominator is:
[tex]\[
\cos \theta (2 \cos^2 \theta - 1)
\][/tex]
Putting both the simplified numerator and denominator together, we get:
[tex]\[
\frac{\sin \theta (2 \cos^2 \theta - 1)}{\cos \theta (2 \cos^2 \theta - 1)}
\][/tex]
Since \(2 \cos^2 \theta - 1\) is common in both the numerator and denominator, it cancels out:
[tex]\[
\frac{\sin \theta}{\cos \theta}
\][/tex]
We recognize that:
[tex]\[
\frac{\sin \theta}{\cos \theta} = \tan \theta
\][/tex]
Therefore, we have shown that:
[tex]\[
\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta
\][/tex]
Thus, the given trigonometric identity is correct.
