Double-Angle Formula Given and , find .

Trigonometry Solution – Problem 6: find

Double-Angle Formula

Given $\sin\alpha - \cos\alpha = \frac{1}{3}$ and $\alpha \in (0,\pi)$ , find $\cos 2\alpha$ .

Squaring $\sin\alpha - \cos\alpha = \frac{1}{3}$ :

$1 - 2\sin\alpha\cos\alpha = \frac{1}{9}$

$\sin 2\alpha = 2\sin\alpha\cos\alpha = \frac{8}{9} > 0$

Since $\alpha \in (0,\pi)$ and $\sin\alpha, \cos\alpha$ have different signs, $\alpha \in (\frac{\pi}{2},\pi)$ .

Therefore $\cos\alpha - \sin\alpha < 0$ , so:

$\cos 2\alpha = (\cos\alpha - \sin\alpha)(\cos\alpha + \sin\alpha) < 0$

$\cos 2\alpha = -\sqrt{1 - \sin^2 2\alpha} = -\sqrt{1 - \frac{64}{81}} = -\frac{\sqrt{17}}{9}$

$-\frac{\sqrt{17}}{9}$

Correct identity setup (2 pts): choose an appropriate sum/difference, double-angle, or auxiliary-angle idea and set up the key equation(s).
Correct algebra / trig simplification (2 pts): transform expressions without sign mistakes.
Solve for target quantity (2 pts): isolate the requested value and handle any constraints if needed.
Final answer (1 pt): clearly state the result in the required form.