cos alpha sin (beta-gamma) + cos beta sin (ga | vedclass

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

Question

(A) Expand the expression using the formula $\sin (A-B) = \sin A \cos B - \cos A \sin B$: $\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamm

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Expand the expression using the formula $\sin (A-B) = \sin A \cos B - \cos A \sin B$: $\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta)$ Distribute the terms: $\cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta$ Grouping the terms,we see that each term cancels out: $(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$

$\cos \alpha \sin (\beta-\gamma) + \cos \beta \sin (\gamma-\alpha) + \cos \gamma \sin (\alpha-\beta)$ is equal to

Similar Questions

Evaluate: $\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma )$

$\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = $

Prove that $\frac{\sin (x+y)}{\sin (x-y)} = \frac{\tan x + \tan y}{\tan x - \tan y}$.

If $\cos (A-B)=3/5$ and $\tan A \tan B=2$,then which one of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module