Bonsoir,
1. On sait que :
cos(a)*cos(b)+sin(a)*sin(b) = cos(a-b)
cos(a)*cos(b)-sin(a)*sin(b) = cos(a+b)
Donc :
A = cos(π/12)*cos(5π/12)+sin(π/12)*sin(5π/12) = cos((5π/12)-(π/12)) = cos(4π/12) = cos(π/3) = 1/2
B = cos(π/12)*cos(5π/12)-sin(π/12)*sin(5π/12) = cos((5π/12)+(π/12)) = cos(6π/12) = cos(π/2) = 0
2. On a donc le système suivant :
[tex] \left \{ {{(E1):cos( \frac{ \pi}{12})*cos( \frac{5 \pi}{12})+sin( \frac{ \pi}{12})*sin( \frac{5 \pi}{12}) = \frac{1}{2}} \atop {(E2):cos( \frac{ \pi}{12})*cos( \frac{5 \pi}{12})-sin( \frac{ \pi}{12})*sin( \frac{5 \pi}{12}) =0}} \right. [/tex]
Donc (E1)+(E2) ⇒ 2(cos(π/12)*cos(5π/12)) = 1/2 ⇒ cos(π/12)*cos(5π/12) = 1/4
Et (E1)-(E2) ⇒ 2(sin(π/12)*sin(5π/12)) = 1/2 ⇒ sin(π/12)*sin(5π/12) = 1/4
Donc sin(π/12)*sin(5π/12) = cos(π/12)*cos(5π/12) = 1/4
