Mostre que R_theta_1 R_theta_2 = R_theta_1 + theta 2

Passo 1

Fala aeee, tudo na paz? 😊

Uma matriz R_theta é do tipo:

R_theta = [[[c,o,s,,(,theta,),-,s,e,n,(,theta,)],[s,e,n,(,theta,),c,o,s,,(,theta,)]]]

Portanto:

R_theta_1 = [[[cos (theta)_1,-,s,e,n,(theta)_1],[s,e,n,(theta)_1,cos (theta)_1]]]

R_theta_2 = [[[cos (theta)_2,-,s,e,n,(theta)_2],[s,e,n,(theta)_2,cos (theta)_2]]]

Passo 2

R_theta_1 . R_theta_2 = [[[cos (theta)_1,-,s,e,n,(theta)_1],[s,e,n,(theta)_1,cos (theta)_1]]] x [[[cos (theta)_2,-,s,e,n,(theta)_2],[s,e,n,(theta)_2,cos (theta)_2]]]

R_theta_1 . R_theta_2 = [[[cos (theta)_1,cos (theta)_2,-,s,e,n,(theta)_1,s,e,n,(theta)_2,- s e n (theta)_2 cos (theta)_1,-,s,e,n,(theta)_1,cos (theta)_2],[s,e,n,(theta)_1,cos (theta)_2,+,,s,e,n,(theta)_2,cos (theta)_1,- s e n (theta)_1 s e n (theta)_2 + cos (theta)_1,cos (theta)_2]]]

Passo 3

Agora, devemos lembrar da soma de arcos:

cos (theta_1 + theta_2) = cos (theta)_1 cos (theta)_2 - s e n (theta)_1 s e n (theta)_2

s e n (theta_1 + theta_2) = s e n (theta)_1 cos (theta)_2 + s e n (theta)_2 cos (theta)_1

Logo:

R_theta_1 + theta_2 = [[[cos (theta_1 + theta_2),-,s,e,n,(theta_1 + theta_2)],[s,e,n,(theta_1 + theta_2),cos (theta_1 + theta_2)]]] =

R_theta_1 + theta_2 = [[[cos (theta)_1,cos (theta)_2,-,s,e,n,(theta)_1,s,e,n,(theta)_2,- s e n (theta)_2 cos (theta)_1,-,s,e,n,(theta)_1,cos (theta)_2],[s,e,n,(theta)_1,cos (theta)_2,+,,s,e,n,(theta)_2,cos (theta)_1,- s e n (theta)_1 s e n (theta)_2 + cos (theta)_1,cos (theta)_2]]]

R_theta_1 R_theta_2 = R_theta_1 + theta 2

Resposta

R_theta_1 R_theta_2 = R_theta_1 + theta 2 Demonstração