Question

Sin a = - 1/3; p < a < 3p/2; cos^2 a/2 = ?

Answer 1

$\pi$

α - угол третьей четверти , ⇒  Cosα < 0

$Cos\alpha=-\sqrt{1-Sin^{2}\alpha}=-\sqrt{1-(-\frac{1}{3})^{2}}=-\sqrt{1-\frac{1}{9}}=-\sqrt{\frac{8}{9}}=-\frac{2\sqrt{2}}{3}\\\\Cos^{2}\frac{\alpha }{2}=\frac{1-\frac{2\sqrt{2}}{3} }{2}=\frac{3-2\sqrt{2}}{6}$

Answer 2

To find the value of $\cos^2\left(\frac{a}{2}\right)$, we can use the double angle formula for cosine:

$\cos(2\theta) = 2\cos^2(\theta) - 1$

In this case, $\theta = \frac{a}{2}$, so we can rewrite the formula as:

$\cos(a) = 2\cos^2\left(\frac{a}{2}\right) - 1$

Now, we know that $\sin(a) = -\frac{1}{3}$. Since $a$ is in the second quadrant ($p < a < \frac{3}{2}p$), we can use the Pythagorean identity to find $\cos(a)$:

$\sin^2(a) + \cos^2(a) = 1$

Plugging in the value of $\sin(a)$:

$\left(-\frac{1}{3}\right)^2 + \cos^2(a) = 1$

Solving for $\cos(a)$:

$\frac{1}{9} + \cos^2(a) = 1$

Subtracting $\frac{1}{9}$ from both sides:

$\cos^2(a) = 1 - \frac{1}{9}$

$\cos^2(a) = \frac{8}{9}$

Now, we have $\cos^2(a)$, but we want $\cos^2\left(\frac{a}{2}\right)$. We can use the double angle formula we derived earlier to find $\cos^2\left(\frac{a}{2}\right)$:

$\cos(a) = 2\cos^2\left(\frac{a}{2}\right) - 1$

Plugging in the value of $\cos(a)$ we found:

$\frac{8}{9} = 2\cos^2\left(\frac{a}{2}\right) - 1$

Now, add 1 to both sides:

$\frac{8}{9} + 1 = 2\cos^2\left(\frac{a}{2}\right)$

$\frac{17}{9} = 2\cos^2\left(\frac{a}{2}\right)$

Finally, divide by 2:

$\cos^2\left(\frac{a}{2}\right) = \frac{17}{18}$

So, $\cos^2\left(\frac{a}{2}\right) = \frac{17}{18}$ when $\sin(a) = -\frac{1}{3}$ and $p < a < \frac{3}{2}p$.