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

The exact value of $\cos^2 73^\circ + \cos^2 47^\circ + (\cos 73^\circ \cdot \cos 47^\circ)$ is

Minimum value of $8 \cos^2 x + 18 \sec^2 x$ for all $x \in R$ wherever it is defined,is:

If $a \cos^3 \alpha + 3a \cos \alpha \sin^2 \alpha = m$ and $a \sin^3 \alpha + 3a \cos^2 \alpha \sin \alpha = n$,then $(m + n)^{2/3} + (m - n)^{2/3}$ is equal to

If $m$ and $M$ are the minimum and the maximum values of $4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$ for $x \in R$,then $M - m$ is equal to

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module