Question

Simplify : \( \tan^{-1} \left( \frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x} \right) \), \( 0 < x < \frac{\pi}{4} \).

Hint:

When you see \( \cos \theta \pm \sin \theta \), dividing by \( \cos \theta \) is almost always the first step to convert it to a tangent form.
Always verify the range of the argument for inverse functions to ensure the result is correct.

Answer 1

Concept:
• Conversion to tangent form: To simplify an expression involving \( \tan^{-1} \), we try to express the inner term as \( \tan \theta \).
• Trigonometric Identity: \( \tan \left( \frac{\pi}{4} - \theta \right) = \frac{1 - \tan \theta}{1 + \tan \theta} \).

Step 1: Divide numerator and denominator by \( \cos 2x \)
Divide both the numerator and the denominator inside the brackets by \( \cos 2x \): \[ \frac{\frac{\cos 2x}{\cos 2x} - \frac{\sin 2x}{\cos 2x}}{\frac{\cos 2x}{\cos 2x} + \frac{\sin 2x}{\cos 2x}} = \frac{1 - \tan 2x}{1 + \tan 2x} \]

Step 2: Rewrite the expression using tangent identity
We know that \( \tan \frac{\pi}{4} = 1 \). The expression can be written as: \[ \frac{\tan \frac{\pi}{4} - \tan 2x}{1 + \tan \frac{\pi}{4} \tan 2x} \] Using the formula \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \), we get: \[ = \tan \left( \frac{\pi}{4} - 2x \right) \]

Step 3: Simplify the inverse tangent function
Substitute the result back into the original expression: \[ \tan^{-1} \left[ \tan \left( \frac{\pi}{4} - 2x \right) \right] \] Check the range: Given \( 0 < x < \frac{\pi}{4} \), then \( 0 < 2x < \frac{\pi}{2} \). Subtracting from \( \frac{\pi}{4} \): \[ \frac{\pi}{4} - \frac{\pi}{2} < \frac{\pi}{4} - 2x < \frac{\pi}{4} - 0 \] \[ -\frac{\pi}{4} < \frac{\pi}{4} - 2x < \frac{\pi}{4} \] Since this is within the principal value branch \( (-\frac{\pi}{2}, \frac{\pi}{2}) \): \[ = \frac{\pi}{4} - 2x \]