\(\cos^2a=1-\dfrac{25}{169}=\dfrac{144}{169}\)
mà \(\cos a< 0\)
nên \(\cos a=-\dfrac{12}{13}\)
\(\sin\left(\text{Δ}+\dfrac{\Pi}{3}\right)=\sin\text{Δ}\cdot\dfrac{1}{2}+\cos\text{Δ}\cdot\dfrac{\sqrt{3}}{2}\)
\(=\dfrac{-5}{13}\cdot\dfrac{1}{2}+\dfrac{-12}{13}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{-5-12\sqrt{3}}{26}\)
Có :
\(cos^2a=1-\dfrac{25}{169}=\dfrac{144}{169}\)
-> Mà cos a < 0 , nên cos a = \(-\dfrac{12}{13}\)
Ta có :
\(sin\left(\Delta+\dfrac{11}{3}\right)=sin\Delta\times\dfrac{1}{2}+cos\Delta\times\dfrac{\sqrt{3}}{2}\)
Vậy :
\(\dfrac{-5}{13}\times\dfrac{1}{2}+\dfrac{-12}{13}\times\dfrac{\sqrt{3}}{2}=\dfrac{-5-12\sqrt{3}}{26}\)
