$cosx=\dfrac{5}{13}$
Vì $(-\dfrac{\pi}{2}<x<0)$ ⇒ $x∈(IV)$ ⇒ $\begin{cases} cosx>0 \\ sinx<0 \end{cases}$
Ta có: $sin^2x+cos^2x=1$
→ $sin^2x+(\dfrac{5}{13})^2=1$
⇔ \(\left[ \begin{array}{l}sinx=\dfrac{12}{13}(loại)\\sinx=-\dfrac{12}{13}(nhận)\end{array} \right.\)
$tanx=-\dfrac{12}{13}÷\dfrac{5}{13}=-\dfrac{12}{5}$
Tính theo yêu câu:
$sin2x=2sinx×cosx=2×(-\dfrac{12}{13})×\dfrac{5}{13}=-\dfrac{120}{169}$
$cos2x=cos^2x-sin^2x=(\dfrac{5}{13})^2-(-\dfrac{12}{13})^2=-\dfrac{119}{169}$
$tan2x=\dfrac{2tanx}{1-tan^2x}=\dfrac{2×(-\dfrac{12}{5})}{1-(-\dfrac{12}{5})^2}=\dfrac{120}{119}$
