If $\alpha + \beta + \gamma = 2 heta$, then $\cos{ heta + \cos{( heta - \alpha)}} + \cos{( heta - \beta)} + \cos{( heta - \gamma)}$ is equal to

Question

If $\alpha + \beta + \gamma = 2	heta$, then $\cos{	heta + \cos{(	heta - \alpha)}} + \cos{(	heta - \beta)} + \cos{(	heta - \gamma)}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$\ {\cos{(	heta - \alpha)}} + \cos{(	heta - \beta)} + \cos{	heta + cos(	heta - \gamma)}$
$= 2\cos{\left( 	heta - \left( \frac{\alpha + \beta}{2} ight) ight)\cos{\left( \frac{\beta - \alpha}{2} ight) + 2\cos{\left( \frac{\gamma}{2} ight)\cos\left( 	heta - \frac{\gamma}{2} ight)}}}$
$= 2\cos{\left( \frac{\gamma}{2} ight)\cos{\left( \frac{\beta - \alpha}{2} ight) + 2\cos\left( \frac{\gamma}{2} ight)\cos\left( \frac{\alpha + \beta}{2} ight)}}$
$= \lbrack\because 2	heta = \alpha + \beta + \gammabrack$
$= 2\cos{\left( \frac{\gamma}{2} ight)\left\lbrack 2\cos{\frac{\alpha}{2}\cos\frac{\beta}{2}} ightbrack}$
$= 4\cos{\frac{\alpha}{2}\cos{\frac{\beta}{2}\cos\frac{\gamma}{2}\ }}$

Q. If $\alpha + \beta + \gamma = 2\theta$, then $\cos{\theta + \cos{(\theta - \alpha)}} + \cos{(\theta - \beta)} + \cos{(\theta - \gamma)}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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