The angle x lies in the interval π/2 ≤ x ≤ π, and sin² x = 8/9. Without using a calculator, determine the value of sin 2x, cos 2x, x/2, and sin 3x.

Steps to Solve

1. Find sin x and cos x

Given that $\sin^2 x = \frac{8}{9}$, we take the square root.

Thus,

$$ \sin x = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} $$

Since $x$ lies in the interval $\left[\frac{\pi}{2}, \pi\right]$, $\sin x$ is positive.

To find $\cos x$, we use the identity $\sin^2 x + \cos^2 x = 1$.

So,

$$ \cos^2 x = 1 - \sin^2 x = 1 - \frac{8}{9} = \frac{1}{9} $$

Therefore,

$$ \cos x = -\sqrt{\frac{1}{9}} = -\frac{1}{3} $$

2. Calculate sin 2x

Using the double angle identity for sine,

$$ \sin 2x = 2 \sin x \cos x $$

Substituting the values:

$$ \sin 2x = 2 \left(\frac{2\sqrt{2}}{3}\right)\left(-\frac{1}{3}\right) = -\frac{4\sqrt{2}}{9} $$

3. Calculate cos 2x

Using the double angle identity for cosine,

$$ \cos 2x = \cos^2 x - \sin^2 x $$

Substituting the values:

$$ \cos 2x = \left(-\frac{1}{3}\right)^2 - \left(\frac{2\sqrt{2}}{3}\right)^2 = \frac{1}{9} - \frac{8}{9} = -\frac{7}{9} $$

4. Calculate sin 3x

Using the identity for sine of a triple angle:

$$ \sin 3x = 3\sin x - 4\sin^3 x $$

We first find $\sin^3 x$:

$$ \sin^3 x = \left(\frac{2\sqrt{2}}{3}\right)^3 = \frac{8\sqrt{2}}{27} $$

Now substituting:

$$ \sin 3x = 3\left(\frac{2\sqrt{2}}{3}\right) - 4\left(\frac{8\sqrt{2}}{27}\right) = 2\sqrt{2} - \frac{32\sqrt{2}}{27} $$

To combine the terms:

$$ \sin 3x = \frac{54\sqrt{2}}{27} - \frac{32\sqrt{2}}{27} = \frac{22\sqrt{2}}{27} $$

5. Calculate x/2

Since $\sin^2 x = \frac{8}{9}$, we can find:

$$ x = \arcsin\left(\frac{2\sqrt{2}}{3}\right) \text{ within the specified interval.} $$

Thus,

$$ \frac{x}{2} \text{ is half of the angle. } $$

Since we do not require the exact angle, we denote it as $\frac{x}{2}$.