Question

sendo sen a = 1/2 e cos b= 1/3 calcule cos(a+b).
Por favor me ajudem , nao consigo desenvolver.

Answer 1

cos(a+b)=cosa.cosb-sena.senb
sen²a+cos²a=1
(1/2)²+cos²a=1
1/4+cos²a=1
cos²a=1-1/4
cos²a=3/4
cosa=√3/4
cosa=√3/2

sen²b+cos²b=1
sen²b+(1/3)²=1
sen²b+1/9=1
sen²b=1-1/9
sen²b=8/9
senb=√8/9
senb=√2²x2/√9
senb=2√2/3

Fica:
cos(a+b)=√3/2.1/3-1/2.2√2/3
cos(a+b)=√3/6-√2/3
cos(a+b)=[(√3-2√2):6]
cos(a+b)=[(1,73-2,82):6]
cos(a+b)=[-1,09/6]
cos(a+b)=-0,18