cos A + cos (240^circ + A) + cos (240^circ - | vedclass

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

Question

(B) Using the sum-to-product formula $\cos(X + Y) + \cos(X - Y) = 2 \cos X \cos Y$,we have:$\cos(240^\circ + A) + \cos(240^\circ - A) = 2 \cos 240^\ci

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Using the sum-to-product formula $\cos(X + Y) + \cos(X - Y) = 2 \cos X \cos Y$,we have:$\cos(240^\circ + A) + \cos(240^\circ - A) = 2 \cos 240^\circ \cos A$Since $\cos 240^\circ = \cos(180^\circ + 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$,Substituting this value:$2 \cos 240^\circ \cos A = 2 \left( -\frac{1}{2} \right) \cos A = -\cos A$Now,substituting back into the original expression:$\cos A + (-\cos A) = 0$.

$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $

Similar Questions

$\cos^2 \alpha + \cos^2(\alpha + 120^\circ) + \cos^2(\alpha - 120^\circ)$ is equal to

The value of $\sin 10^\circ + \sin 20^\circ + \sin 30^\circ + ... + \sin 360^\circ$ is

If $x + \frac{1}{x} = 2\cos \theta ,$ then ${x^3} + \frac{1}{{{x^3}}} = $

If $90^\circ < A < 180^\circ$ and $\sin A = \frac{4}{5}$,then $\tan \frac{A}{2}$ is equal to

If $A + B + C = 180^o,$ then the value of $(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module