Let A and B denote the statements:A: cos alph | vedclass

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

Question

(B) Given $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$.Multiplying by $2$,we get $2[\cos (\alpha - \beta) +

Vedclass · Accepted Answer

both are true

Vedclass · Answer

(B) Given $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$.Multiplying by $2$,we get $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] = -3$.Adding $3$ to both sides: $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] + 3 = 0$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we can write $3 = (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + (\sin^2 \gamma + \cos^2 \gamma)$.Substituting this into the equation:$(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + 2\sin \alpha \sin \beta + 2\sin \beta \sin \gamma + 2\sin \gamma \sin \alpha) + (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2\cos \alpha \cos \beta + 2\cos \beta \cos \gamma + 2\cos \gamma \cos \alpha) = 0$.This simplifies to $(\sin \alpha + \sin \beta + \sin \gamma)^2 + (\cos \alpha + \cos \beta + \cos \gamma)^2 = 0$.Since the sum of squares is zero,each term must be zero:$\sin \alpha + \sin \beta + \sin \gamma = 0$ and $\cos \alpha + \cos \beta + \cos \gamma = 0$.Therefore,both statements $A$ and $B$ are true.

Let $A$ and $B$ denote the statements:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then:

Similar Questions

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-px+q=0$,then $\sin^2(A+B)$ is equal to

$ABC$ is a triangle such that $\sin(2A + B) = \sin(C - A) = -\sin(B + 2C) = \frac{1}{2}$. If $A, B,$ and $C$ are in $A.P.$,then $A, B,$ and $C$ are:

If $m \cdot \tan (\theta-30^{\circ})=n \cdot \tan (\theta+120^{\circ})$,then $\frac{m+n}{m-n}=$

$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = $

If $2 \sin \theta + 3 \cos \theta = 2$ and $\theta \neq (2n + 1) \frac{\pi}{2}$,then find the value of $3 \sin \theta - 2 \cos \theta$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $A$ and $B$ denote the statements:$A: \cos \alpha + \cos \beta + \cos \gamma = 0$$B: \sin \alpha + \sin \beta + \sin \gamma = 0$If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then: