If cos(theta_1) + cos(theta_2) + cos(theta_3) | vedclass

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

Question

(D) Given,$\cos(	heta_1) + \cos(	heta_2) + \cos(	heta_3) + \cos(	heta_4) = -4$. Since the range of the cosine function is $[-1, 1]$,the only way t

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given,$\cos(	heta_1) + \cos(	heta_2) + \cos(	heta_3) + \cos(	heta_4) = -4$. Since the range of the cosine function is $[-1, 1]$,the only way their sum can be $-4$ is if each term is equal to $-1$. Thus,$\cos(	heta_1) = \cos(	heta_2) = \cos(	heta_3) = \cos(	heta_4) = -1$. This implies $	heta_1 = 	heta_2 = 	heta_3 = 	heta_4 = (2n+1)\pi$ for some integer $n$. Taking $	heta_i = \pi$,we have $\frac{	heta_i}{2} = \frac{\pi}{2}$. Then,$\cot(\frac{	heta_i}{2}) = \cot(\frac{\pi}{2}) = 0$. Therefore,$\cot(\frac{	heta_1}{2}) + \cot(\frac{	heta_2}{2}) + \cot(\frac{	heta_3}{2}) + \cot(\frac{	heta_4}{2}) = 0 + 0 + 0 + 0 = 0$.

If $\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_3) + \cos(\theta_4) = -4$,then the value of $\cot(\frac{\theta_1}{2}) + \cot(\frac{\theta_2}{2}) + \cot(\frac{\theta_3}{2}) + \cot(\frac{\theta_4}{2}) = $

Similar Questions

The value of $\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$ is

$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=$

If $\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1$,then the value of $\left[ \frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} \right]$ is (where $[.]$ denotes the greatest integer function).

The value of $\text{cosec}10^{\circ} - \sqrt{3} \text{sec}10^{\circ}$ is equal to:

The value of $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module