Question

Cho \(\left\{{}\begin{matrix}\dfrac{\pi}{2}< \alpha< 2\pi\\tan\left(\alpha+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

Tính \(P=cos\left(\alpha-\dfrac{\pi}{6}\right)+sin\alpha\)

Answer 1

\(tan\left(\alpha+\dfrac{\Pi}{4}\right)=1\)

\(\Leftrightarrow a+\dfrac{\Pi}{4}=\dfrac{\Pi}{4}+k\Pi\)

hay \(a=k\Pi\)

mà \(\dfrac{\Pi}{2}< a< 2\Pi\)

nên \(a=\Pi\)

\(P=\cos\left(\dfrac{5\Pi}{6}\right)+sin\Pi=\dfrac{-\sqrt{3}}{2}\)