Question

Sin a/2=8/17; п<а<2п; найти cos a и tg a​

Answer 1

$sin\frac{a}{2} = \frac{8}{17} => cos\frac{a}{2} = -\sqrt{1 - (\frac{8}{17})^2} = -\frac{15}{17} \\cosa = cos^2\frac{a}{2} - sin^2\frac{a}{2} = \frac{161}{289}\\ tan(a) = \sqrt{\frac{1}{cos^2(a)} - 1 } = \sqrt{\frac{289^2}{161^2} - 1} = \sqrt{\frac{450*128}{161^2} } = \frac{240}{161}$

Answer 2

To find the values of $\cos(a)$ and $\tan(a)$ given $\sin(a/2) = 8/17$ and $\pi < a < 2\pi$, you can use the half-angle formulas for sine, cosine, and tangent.

The half-angle formulas state:

$\sin\left(\frac{a}{2}\right) = \pm \sqrt{\frac{1 - \cos(a)}{2}}$

From the given information, we have:

$\sin\left(\frac{a}{2}\right) = \frac{8}{17}$

Using the half-angle formula for sine, we can solve for $\cos(a)$:

$\frac{8}{17} = \pm \sqrt{\frac{1 - \cos(a)}{2}}$

Squaring both sides, we get:

$\left(\frac{8}{17}\right)^2 = \frac{1 - \cos(a)}{2}$

$\frac{64}{289} = \frac{1 - \cos(a)}{2}$

Now, solve for $\cos(a)$:

$1 - \cos(a) = \frac{128}{289}$

$\cos(a) = 1 - \frac{128}{289}$

$\cos(a) = \frac{161}{289}$

Now that you know $\cos(a)$, you can find $\tan(a)$ using the identity:

$\tan(a) = \frac{\sin(a)}{\cos(a)}$

Given $\sin(a/2) = \frac{8}{17}$, you can use the double-angle formula for sine:

$\sin(a) = 2\sin\left(\frac{a}{2}\right)\cos\left(\frac{a}{2}\right)$

Substitute the given value of $\sin(a/2) = \frac{8}{17}$ and $\cos(a) = \frac{161}{289}$ into the formula:

$\sin(a) = 2 \times \frac{8}{17} \times \sqrt{\frac{1 - \frac{161}{289}}{2}}$

Solving this will give you the value of $\sin(a)$. Once you have both $\sin(a)$ and $\cos(a)$, you can find $\tan(a)$ using the formula mentioned above.