Trigonometry Solution – Problem 16: Find the range of for

Question

Range of Trigonometric Function

Find the range of $f(x) = \sin^2 x - \cos^2 x$ for $x \in \left[\frac{\pi}{12}, \frac{2\pi}{3}\right]$ .

Simplify the function:

$f(x) = \sin^2 x - \cos^2 x = -\cos 2x$

For $x \in \left[\frac{\pi}{12}, \frac{2\pi}{3}\right]$ :

$2x \in \left[\frac{\pi}{6}, \frac{4\pi}{3}\right]$

In this interval, $\cos 2x \in \left[-1, \frac{\sqrt{3}}{2}\right]$ .

Therefore:

$f(x) = -\cos 2x \in \left[-\frac{\sqrt{3}}{2}, 1\right]$

$\left[-\frac{\sqrt{3}}{2}, 1\right]$

Translate the question into conditions (2 pts): domain/range/period/symmetry/monotonicity constraints are written correctly.
Key transformation or rewrite (2 pts): rewrite the function into a standard form (e.g. amplitude/phase/period) or reduce using identities.
Correct interval/parameter reasoning (2 pts): derive the correct inequalities or argument interval.
Final answer (1 pt): state the final set / interval / period clearly.

Listing properties ("periodic", "even") without applying them to the given function.