Question

Find $\sin \left( {\dfrac{\theta }{2}} \right)$ , if$\cos \theta = - \dfrac{4}{5}$ and $- 270 < \theta < \- 180$.

Answer 1

To solve this problem, here we are using trigonometric identities and then we will substitute the value of $\cos \theta$ i.e,$- \dfrac{4}{5}$ to find the value of $\sin \left( {\dfrac{\theta }{2}} \right)$ and we clearly know that there are six trigonometric ratios i.e, sine$\left( {\sin } \right)$ , cosine$\left( {\cos } \right)$, tangent$\left( {\tan } \right)$, cosec$\left( {\csc } \right)$ , secant$\left( {\sec } \right)$ , cotangent $\left( {\cot } \right)$ .

Complete step by step solution:
In this problem, we have given$\cos \theta = - \dfrac{4}{5}$and the value of $\theta$ lies between$- 270$ and$- 180$ and to solve the value of$\sin \left( {\dfrac{\theta }{2}} \right)$, we are using the identity,
$\cos 2x = 1 - 2{\sin ^2}x$ . Here the value of $x$ is$\dfrac{\theta }{2}$ , now, we will substitute it in place of $x$and on substituting we get,
$\cos 2 \times \dfrac{\theta }{2} = 1 - 2{\sin ^2} \times \dfrac{\theta }{2}$
Now, the equation becomes,
$\Rightarrow \cos \theta = 1 - 2{\sin ^2}\left( {\dfrac{\theta }{2}} \right)$
Firstly, we will rearrange the equation as,
$\Rightarrow 2{\sin ^2}\left( {\dfrac{\theta }{2}} \right) = 1 - \cos \theta$
Now, we will substitute the value of $\cos \theta$ in the above equation,
$\Rightarrow 2{\sin ^2}\left( {\dfrac{\theta }{2}} \right) = 1 - \left( { - \dfrac{4}{5}} \right)$
On further solving, we get,

$$\Rightarrow 2{\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{9}{5} \\\ \Rightarrow {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{9}{5} \times \dfrac{1}{2} \\\ \Rightarrow {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{9}{{10}} \\\$$

Now, on taking square root on both sides, we get,
$\Rightarrow \sin \dfrac{\theta }{2} = \sqrt {\dfrac{9}{{10}}} \\\ \Rightarrow \sin \dfrac{\theta }{2} = \pm \dfrac{3}{{\sqrt {10} }} \\\$
But, we have also given that$- 270 < \theta < \- 180$, if we divide it by $2$,$\dfrac{{ - 270}}{2} < \dfrac{\theta }{2} < \dfrac{{ - 180}}{2}$ we get,$- 135 < \dfrac{\theta }{2} < \- 90$ , which means that $\dfrac{\theta }{2}$ lies in the third quadrant, hence the value of $\sin \dfrac{\theta }{2}$ is$- \dfrac{3}{{\sqrt {10} }}$.

Note: There are some rules for the trigonometric ratios in the quadrants, in the first quadrants all the trigonometric ratios are positive, in the second quadrant only the values of sin are positive, in the third quadrant only the values of tan are positive and in the fourth quadrant only the values of cos are positive and when the angle formed moving anticlockwise then it is positive and if it is formed by moving clockwise then it is negative.