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: $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$Multiply by $2$: $2[\cos (\alpha - \beta) + \cos (\b

Vedclass · Accepted Answer

Both are true

Vedclass · Answer

(B) Given: $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$Multiply by $2$: $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] = -3$$\Rightarrow 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 $\sin A = \sin B$ and $\cos A = \cos B,$ then

The equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$ is only possible when

The value of $\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 179^{\circ} = $

If $5 \cos \theta + 12 \sin \theta = 13,$ then $\tan \theta = ?$

Given that $\pi < \alpha < \frac{3\pi}{2},$ then the expression $\sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\cos^2 \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)$ is equal to

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: