Question

$y = sin^2(\frac{15\pi}{8}-4x) - sin^2(\frac{17\pi}{8}-4x)$. Find range of $y$.

Answer 1

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

Answer 2

Let the given function be $y = \sin^2(\frac{15\pi}{8}-4x) - \sin^2(\frac{17\pi}{8}-4x)$.

We use the trigonometric identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$. Let $A = \frac{15\pi}{8}-4x$ and $B = \frac{17\pi}{8}-4x$.

Calculate $A+B$: $A+B = (\frac{15\pi}{8}-4x) + (\frac{17\pi}{8}-4x) = \frac{15\pi+17\pi}{8} - 8x = \frac{32\pi}{8} - 8x = 4\pi - 8x$.

Calculate $A-B$: $A-B = (\frac{15\pi}{8}-4x) - (\frac{17\pi}{8}-4x) = \frac{15\pi}{8} - 4x - \frac{17\pi}{8} + 4x = \frac{15\pi-17\pi}{8} = \frac{-2\pi}{8} = -\frac{\pi}{4}$.

Substitute these into the identity: $y = \sin(4\pi - 8x)\sin(-\frac{\pi}{4})$.

Simplify the sine terms: $\sin(4\pi - 8x) = \sin(-(8x - 4\pi)) = -\sin(8x - 4\pi)$. Since $\sin(\theta - 2n\pi) = \sin(\theta)$ for any integer $n$, $\sin(8x - 4\pi) = \sin(8x - 2 \times 2\pi) = \sin(8x)$. So, $\sin(4\pi - 8x) = -\sin(8x)$.

$\sin(-\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.

Substitute the simplified terms back into the expression for $y$: $y = (-\sin(8x)) \times (-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}}\sin(8x)$.

To find the range of $y$, we consider the range of $\sin(8x)$. For any real number $x$, $8x$ can take any real value. The range of the sine function for real input is $[-1, 1]$. So, $-1 \le \sin(8x) \le 1$.

Multiply the inequality by $\frac{1}{\sqrt{2}}$: $-\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{2}}\sin(8x) \le \frac{1}{\sqrt{2}}$.

Thus, the range of $y$ is $[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$. This can also be written as $[-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}]$.