If sin theta = -frac{4}{5} and theta lies in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\sin 	heta = -\frac{4}{5}$ and $	heta$ lies in the $III$ quadrant $(180^\circ < 	heta < 270^\circ)$.Since $	heta$ is in the $III$ quadr

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{5}}$

Vedclass · Answer

(B) Given $\sin 	heta = -\frac{4}{5}$ and $	heta$ lies in the $III$ quadrant $(180^\circ Since $	heta$ is in the $III$ quadrant,$\cos 	heta$ is negative.$\cos 	heta = -\sqrt{1 - \sin^2 	heta} = -\sqrt{1 - (-\frac{4}{5})^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.We need to find $\cos \frac{	heta}{2}$. Using the half-angle formula: $\cos \frac{	heta}{2} = \pm \sqrt{\frac{1 + \cos 	heta}{2}}$.Since $180^\circ This means $\frac{	heta}{2}$ lies in the $II$ quadrant,where cosine is negative.Therefore,$\cos \frac{	heta}{2} = -\sqrt{\frac{1 + (-3/5)}{2}} = -\sqrt{\frac{2/5}{2}} = -\sqrt{\frac{1}{5}} = -\frac{1}{\sqrt{5}}$.

If $\sin \theta = -\frac{4}{5}$ and $\theta$ lies in the third quadrant,then $\cos \frac{\theta}{2} = $

Similar Questions

Which of the following relations is correct for $0^{\circ} < \theta < 90^{\circ}$?

Let $f(x) = \sin^2 x + \cos^4 x + 2$ and $g(x) = \cos(\cos x) + \cos(\sin x)$. If the periods of $f(x)$ and $g(x)$ are $T_1$ and $T_2$ respectively,then:

If $2 \sec 2\alpha = \tan \beta + \cot \beta$,then one of the values of $\alpha + \beta$ is

If $A$ lies in the third quadrant and $3 \tan A - 4 = 0$,then find the value of $5 \sin 2A + 3 \sin A + 4 \cos A$.

If $\sin 600^{\circ} \cos 30^{\circ} + \cos 120^{\circ} \sin 150^{\circ} = k$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module