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

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

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Answer 1 · 2018-01-07T07:36:39

See the proof below

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#