Question

Hey ^^"

Prove the following identity :

$\sin( \alpha \pm \beta ) = \sin( \alpha ) \cos( \beta ) \pm \sin( \beta ) \cos( \alpha )$

$\cos( \alpha \pm \beta ) = \cos( \alpha ) \cos( \beta ) \mp \sin( \alpha ) \sin( \beta )$

Answer 1

Hey

$\triangle abd \sim \triangle ced$

$= > \frac{ab}{ad} = \frac{ce}{cd}$

$= > ce = \frac{ab.cd}{ad}$

Now,

$\sin(x - y) = \frac{ce}{ac}$

$= \frac{ab.cd}{ad.ac}$

$= \frac{ab.bd}{ac.cd} - \frac{ab.bc}{ac.cd}$

$= \sin(x) \cos(y ) - \sin(y) \cos(x) -> (I)$

$put \: x = 0$

$\sin( - y) = - \sin(y)$

$put \: y = -y$

$\sin(x + y) =\sin(x) \cos(y ) + \sin(y) \cos(x)$

$put \: x = ( \frac{ \pi }{2}-x) \: in \: I \: to \: get \: the \: second \: result$

Answer 2

Answer:

△abd∼△ced

= > \frac{ab}{ad} = \frac{ce}{cd}=>

ad

ab

=

cd

ce

= > ce = \frac{ab.cd}{ad}=>ce=

ad

ab.cd

Now,

\sin(x - y) = \frac{ce}{ac}sin(x−y)=

ac

ce

= \frac{ab.cd}{ad.ac}=

ad.ac

ab.cd

= \frac{ab.bd}{ac.cd} - \frac{ab.bc}{ac.cd}=

ac.cd

ab.bd

−

ac.cd

ab.bc

= \sin(x) \cos(y ) - \sin(y) \cos(x) -> (I)=sin(x)cos(y)−sin(y)cos(x)−>(I)

put \: x = 0putx=0

\sin( - y) = - \sin(y)sin(−y)=−sin(y)

put \: y = -yputy=−y

\sin(x + y) =\sin(x) \cos(y ) + \sin(y) \cos(x)sin(x+y)=sin(x)cos(y)+sin(y)cos(x)

put \: x = ( \frac{ \pi }{2}-x) \: in \: I \: to \: get \: the \: second \: resultputx=(

2

π

−x)inI