Si quelqu’un pouvais m’aider.. svp
Mathématiques
Nijika
Question
Si quelqu’un pouvais m’aider.. svp
1 Réponse
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1. Réponse ProfdeMaths1
cos(2001π/4)=cos(π/4)=√2/2
sin(2001π/4)=sin(π/4)=√2/2
cos(199π/6)=cos(-π/6)=√3/2
sin(199π/4)=sin(-π/6)=-1/2
cos(-77π/3)=cos(π/3)=1/2
sin(-77π/3)=sin(π/3)=√3/2
cos(-41π/2)=cos(-π/2)=0
sin(-41π/2)=sin(-π/2)=-1
cos(106π)=cos(0)=1
sin(106π)=sin(0)=0