Question

prove that sin 2 alpha + sin 2 Beta + sin 2 Gama equal to 2 into sin alpha + sin beta + sin Gamma into one + cos alpha + cos beta + cos gamma. if alpha + beta + Gama equal to zero

Answer 1

p prove that sin theta + alpha + sin theta 2 + theta + sin theta 2 + gana equal to 2 into sin alpha + sin theta + sin come up EP 1 + cos theta + cos theta + cos alpha beta gamma is equal to zero

Answer 2

Given :

$\alpha + \gamma + \beta = 0$

To Prove :

$sin2\alpha + sin2\beta + sin 2 \gamma =2 ( sin\alpha +sin\beta + sin\gamma )(1+ cos \alpha + cos \beta + cos \gamma )$

Solution :

RHS:

$=2 ( sin\alpha +sin\beta + sin\gamma )(1+ cos \alpha + cos \beta + cos \gamma )$

$=2(sin\alpha+ sin\alpha cos\alpha +sin\alpha cos \beta + sin\alpha cos \gamma + sin\beta + sin\beta cos \alpha + sin\beta cos \beta + sin\alpha cos \gamma + sin\gamma + sin\gamma cos \alpha + sin\gamma cos \beta + sin\gamma cos \gamma )$$= 2(sin\alpha cos\alpha + sin\beta cos \beta + sin\gamma cos \gamma )+ 2[sin\alpha ( sn\alpha cos\beta + cos\alpha sin\beta ) +sin\beta+ ( sin\alpha cos \gamma + cos \alpha sin\gamma ) + sin\gamma +( sin\beta cos \gamma + cos\beta sin\gamma)$                                                                                             -(1)

∴$\alpha + \beta + \gamma = 0$

So we can write :

$\alpha + \beta = -\gamma$

We can also write :

$sin(\alpha + \beta ) = - sin \gamma$

Similarly :

$sin(\beta + \gamma ) = -sin\alpha$

and also :

$sin(\alpha + \gamma ) = - sin \gamma$

Now ;

$2[sin\alpha + sin \beta + sin \gamma + sin ( \alpha + \beta ) + sin ( \beta + \gamma ) + sin (\alpha + \gamma )]$

$=2(sin\alpha + sin \beta + sin \gamma -sin\gamma - sin\beta - sin\alpha )\\=o$

On putting this  obtained value in eq 1 we get :

RHS $=2(sin\alpha cos\alpha + sin\beta cos\beta + sin\gamma cos \gamma )$

      $= sin2\alpha + sin2\beta + sin2\gamma$

      = LHS

Since , RHS comes equal to the LHS .

Hence , proved