Can u prove this equation using only LHS #[cos(alpha)+-isin(alpha)]/[cos(beta)+-isin(beta)]#= #cos(alpha-beta)+-sin(alpha-beta)#?

#[cos(alpha)+-isin(alpha)]/[cos(beta)+-isin(beta)]#= #cos(alpha-beta)+-sin(alpha-beta)#?

1 Answer
Jan 7, 2018

See the proof below

Explanation:

Apply Euler's Identity

#e^(itheta)=costheta+isintheta#

#e^(-itheta)=costheta-isintheta#

Therefore,

#LHS=(cosalpha+-isinalpha)/(cosbeta+-isinbeta)#

#=e^(+-ialpha)/e^(+-ibeta)#

#=e^(+-i(alpha-beta))#

#=cos(alpha-beta)+-isin(alpha-beta)#

#=RHS#

#QED#