Question

If Sin α + Sin β = a and Cos α + Cos β = b show that Cos (α + β) = ( b^2- a^2 ) / ( b^2 + a^2 )​

Answer 1

Step-by-step explanation:

Given,

$\sin( \alpha ) + \sin( \beta ) = a \: \: and \: \: \cos( \alpha ) + \cos( \beta ) = b$

Now, squaring both sides, we have,

${a}^{2} = \sin ^{2} ( \alpha ) + \sin ^{2} ( \beta ) + 2 \sin( \alpha ) \sin( \beta ) \: \: and \: \: {b}^{2} = \cos^{2} ( \alpha ) + \cos^{2} ( \beta ) + 2 \cos( \alpha ) \cos( \beta )$

Now, RHS=

$\frac{ {b}^{2} - {a}^{2} }{ {b}^{2} + {a}^{2} } = \frac{ \cos^{2} ( \alpha ) + \cos^{2} ( \beta ) +2 \cos( \alpha ) \cos( \beta ) - \sin^{2} ( \alpha ) - \sin ^{2} ( \beta ) - 2 \sin( \alpha ) \sin( \beta ) }{ \: \cos^{2} ( \alpha ) + \cos^{2} ( \beta ) +2 \cos( \alpha ) \cos( \beta ) + \sin^{2} ( \alpha ) + \sin ^{2} ( \beta ) + 2 \sin( \alpha ) \sin( \beta )\: }$

$= > \frac{ {b}^{2} - {a}^{2} }{ {b}^{2} + {a}^{2} } = \frac{ \cos( 2\alpha ) + \cos( 2\beta ) + 2 \cos( \alpha + \beta ) }{2 + 2 \cos( \alpha - \beta ) }$

$\frac{ {b}^{2} - {a}^{2} }{ {b}^{2} + {a}^{2} } = \frac{2 \cos( \alpha + \beta ) \cos( \alpha - \beta ) + 2 \cos( \alpha + \beta ) }{2 + 2 \cos( \alpha - \beta ) }$

$\frac{ {b}^{2} - {a}^{2} }{ {b}^{2} + {a}^{2} } = \frac{2 \cos( \alpha + \beta )( \cos( \alpha - \beta ) + 1) }{2(1 + \cos( \alpha - \beta )) }$

$= > \frac{ {b}^{2} - {a}^{2} }{ {b}^{2} + {a}^{2} } = \cos( \alpha + \beta )$

=LHS

Since, LHS=RHS,

Hence proved