Home / We now seek the minimum value of $ f(x) = 1 - \sin 3x \sin x $. Since $ -1 \leq \sin 3x \sin x \leq 1 $, the maximum of $ \sin 3x \sin x $ is 1, so the minimum of $ f(x) $ is:

We now seek the minimum value of $ f(x) = 1 - \sin 3x \sin x $. Since $ -1 \leq \sin 3x \sin x \leq 1 $, the maximum of $ \sin 3x \sin x $ is 1, so the minimum of $ f(x) $ is:

February 22, 2026

$We now seek the minimum value of $ f(x) = 1 - \sin 3x \sin x $. Since $ -1 \leq \sin 3x \sin x \leq 1 $, the maximum of $ \sin 3x \sin x $ is 1, so the minimum of $ f(x) $ is:$

Let $ A = 4x $, $ B = 2x $:
\cos 4x - \cos 2x = -2 \sin(3x)\sin(x)
f(x) = 1 + rac{1}{2}(-2 \sin 3x \sin x) = 1 - \sin 3x \sin x
- 1 = 0
We verify this value is attainable. Suppose $ \sin 3x = 1 $ and $ \sin x = 1 $. Then $ x = rac{\pi}{2} + 2\pi n $. Check $ \sin 3x = \sin\left(rac{3\pi}{2}
ight) = -1