Question

If $f(x) = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+\tan x}$, then the value of $f'(\frac{\pi}{6})$ is equal to ______.

• A. 0
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Hint:

The factorization $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$ is frequently paired with $\sin^2 x + \cos^2 x = 1$ to vastly simplify massive trigonometric fractions!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must simplify the given trigonometric function $f(x)$, find its first derivative $f'(x)$, and evaluate it at $x = \pi/6$.

Step 2: Detailed Explanation:
Let's simplify $f(x)$ by converting $\tan x$ and $\cot x$ into sine and cosine:
$f(x) = \frac{\sin^2 x}{1 + \frac{\cos x}{\sin x}} + \frac{\cos^2 x}{1 + \frac{\sin x}{\cos x}}$
Find a common denominator in the lower fractions:
$f(x) = \frac{\sin^2 x}{\frac{\sin x + \cos x}{\sin x}} + \frac{\cos^2 x}{\frac{\cos x + \sin x}{\cos x}}$
Bring the denominators up:
$f(x) = \frac{\sin^3 x}{\sin x + \cos x} + \frac{\cos^3 x}{\cos x + \sin x}$
Since they have the same denominator:
$f(x) = \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x}$
Use the sum of cubes factorization: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.
$f(x) = \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x}$
Cancel out the common term $(\sin x + \cos x)$:
$f(x) = \sin^2 x - \sin x \cos x + \cos^2 x$
Since $\sin^2 x + \cos^2 x = 1$:
$f(x) = 1 - \sin x \cos x$
Multiply and divide the second term by 2 to use the double angle formula ($\sin 2x = 2\sin x \cos x$):
$f(x) = 1 - \frac{1}{2} \sin(2x)$
Now differentiate with respect to $x$:
$f'(x) = 0 - \frac{1}{2} (\cos(2x) \cdot 2)$
$f'(x) = -\cos(2x)$
Evaluate the derivative at $x = \pi/6$:
$f'\left(\frac{\pi}{6}\right) = -\cos\left(2 \cdot \frac{\pi}{6}\right)$
$f'\left(\frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{3}\right)$
Since $\cos(\pi/3) = \frac{1}{2}$:
$f'\left(\frac{\pi}{6}\right) = -\frac{1}{2}$

Step 3: Final Answer:
The value is $-\frac{1}{2}$, matching option (c).