Question

Differentiate \( \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) \) with respect to x.

Hint:

Before differentiating complex trigonometric or inverse trigonometric functions, always check if the expression can be simplified using identities. This can often reduce a complicated chain rule problem to a very simple derivative.

Answer 1

Step 1: Understanding the Concept:
The problem asks for the derivative of a function involving an inverse trigonometric function. A direct application of the chain rule is possible but complicated. A more efficient method is to first simplify the expression inside the \( \tan^{-1} \) using trigonometric identities.
Step 2: Key Formula or Approach:
We will use the following half-angle identities:
1. \( \sin x = 2\sin(x/2)\cos(x/2) \)
2. \( 1 + \cos x = 2\cos^2(x/2) \)
After simplification, we will use the derivative formula \( \frac{d}{dx}(\tan^{-1}(\tan u)) = \frac{du}{dx} \), provided u is in the principal domain.
Step 3: Detailed Explanation or Calculation:
Let \( y = \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) \).
First, simplify the argument of \( \tan^{-1} \):
\[ \frac{\sin x}{1 + \cos x} = \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} \] Assuming \( \cos(x/2) \neq 0 \), we can cancel \( 2\cos(x/2) \) from the numerator and denominator:
\[ \frac{\sin(x/2)}{\cos(x/2)} = \tan(x/2) \] Now, substitute this simplified expression back into the function y:
\[ y = \tan^{-1}(\tan(x/2)) \] For \( x \in (-\pi, \pi) \), \( x/2 \in (-\pi/2, \pi/2) \), which is the principal value range for the tangent function. In this range, \( \tan^{-1}(\tan u) = u \).
So, we can simplify the function to:
\[ y = \frac{x}{2} \] Now, differentiate y with respect to x:
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2}\right) = \frac{1}{2} \] Step 4: Final Answer:
The derivative of \( \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) \) with respect to x is \( \frac{1}{2} \).