Question

If 180° <theta<270°, cot theta = 4/3, then sin theta/ 2 =​

Answer 1

sorry for the late

the answer is minus 3 by 5

hope it's helpful

Answer 2

Answer:

Required value of $\frac{1}{ \sqrt{10} } \: \: or \: \: \frac{3}{ \sqrt{10} }$

Step-by-step explanation:

Here given,$\cot(\theta) = \frac{4}{3}$

We know,

$\cot(\theta) = \frac{ \cos(\theta) }{ \sin(\theta) } = \frac{ \sqrt{1 - {sin}^{2}\theta } }{ \sin(\theta) }$

Let,$\sin(\theta) = x$

So,$\frac{ \sqrt{1 - {x}^{2} } }{x} = \frac{4}{3}$

We are taking square in both side,

$\frac{1 - {x}^{2} }{ {x}^{2} } = \frac{16}{9}$

By cross multiplication,

$9 - 9 {x}^{2} = 16 {x}^{2}$

$16 {x}^{2} + 9 {x}^{2} = 9$

$25 {x}^{2} = 9$

${x}^{2} = \frac{9}{25}$

$x = \frac{3}{5}$

So,$\sin(\theta) = \frac{3}{5}$

Again we know,

$\sin(\theta) = 2 \sin( \frac{\theta}{2} ) \cos( \frac{\theta}{2} ) = 2 \sin( \frac{\theta}{2} ) \sqrt{1 - {sin}^{2} \frac{\theta}{2} }$

Let,$\sin( \frac{\theta}{2} ) = y$

So,$2y \sqrt{1 - {y}^{2} } = \frac{3}{5}$

We are taking square in both side,

$4 {y}^{2} (1 - {y}^{2} ) = \frac{9}{25}$

By cross multiplication,

$100 {y}^{4} - 100 {y}^{2} + 9 = 0$

Let,${y}^{2} = p$

Now,

$100 {p}^{2} - 100p + 9 = 0$

By Sridhar Acharya rule,

$p = \frac{9}{10}$

or $p = \frac{1}{10}$

Therefore,

${y}^{2} = p = \frac{1}{10 } \\ y = \frac{1}{ \sqrt{10} }$

and

${y}^{2} = p = \frac{9}{10} \\ y = \frac{3}{ \sqrt{10} }$

$\sin( \frac{\theta}{2} ) = \frac{1}{ \sqrt{10} } \: \: or \: \: \frac{3}{ \sqrt{10} }$