Question

Me ayudan porfaaa aaaas ​

Answer 1

Respuesta:

Solucion:

pecado ( θ - α )pecado ( θ - β)=unsegundo

⇒asen(θ−alfa)=bsen(θ−β) Letasen(θ−α)=bsin(θ−β)=k1 ⇒a=k1sin(θ−α),b=k1sin(θ−β) Similarmente, ⇒c=k2cos(θ−α),d=k2cos(θ−β)Ahora, ` (ac + bd) / (ad + bc) = (k_1sin (theta-alpha) k_2cos (theta-alpha) + k_1sin (theta-beta) k_2cos (theta -beta)) / (k_1sin (theta-alfa) k_2cos (theta-beta) + k_1sin (theta-beta) k_2cos (theta-alfa)) ` =k1k2(sin(θ−alfa)cos(θ−alfa)+sin(θ−β)cos(θ−β))k1k2(sin(θ−alfa)cos(θ−β)+sin(θ−β)cos(θ−alfa)) =12(sin(2(θ−alfa))+sin(2(θ−β)))sin(θ−alfa+θ−β)⇒unpecado ( θ - α )=segundopecado ( θ - β)

unpecado ( θ - α )=segundopecado ( θ - β)=k1

⇒ a =k1pecado ( θ - α ) , b =k1pecado ( θ - β)

⇒ c =k2cos ( θ - α ) , d=k2cos ( θ - β)

a c + b duna d+ b c=k1pecado ( θ - α )k2cos ( θ - α ) +k1pecado ( θ - β)k2cos ( θ - β)k1pecado ( θ - α )k2cos ( θ - β) +k1pecado ( θ - β)k2cos ( θ - α )

=k1k2( sin ( θ - α ) cos ( θ - α ) + sin ( θ - β) cos ( θ - β) )k1k2( sin ( θ - α ) cos ( θ - β) + pecado ( θ - β) cos ( θ - α ) )

=12( sin ( 2 ( θ - α ) ) + sin ( 2 ( θ - β)) ) )pecado ( θ - α + θ - β)

=sin(2(θ−alfa))+sin(2(θ−β))2sin(2θ−(α+β))

=2sin(2(θ−alfa+θ−β)2)cos(2(θ−alfa−θ+β)2)2sin(2θ−(α+β))

=2sin((2θ−(alfa+β))cos(β−alfa))2sin(2θ−(alfa+β))

=cos(β−alfa)

=cos(−(β−alfa))...[Comocos(−θ)=cosθ]

=cos(α−β)Entonces,