Google
Câu hỏi: Nếu \(z = - \sin \varphi - i\cos \varphi \) thì acgumen của z bằng:
(A) \(- {\pi \over 2} + \varphi + k2\pi \left( {k \in\mathbb Z} \right)\);
(B) \(- {\pi \over 2} - \varphi + k2\pi \left( {k \in\mathbb Z} \right)\);
(C) \({\pi \over 2} + \varphi + k2\pi \left( {k \in\mathbb Z} \right)\);
(D) \(\pi - \varphi + k2\pi \left( {k \in\mathbb Z} \right)\).
(A) \(- {\pi \over 2} + \varphi + k2\pi \left( {k \in\mathbb Z} \right)\);
(B) \(- {\pi \over 2} - \varphi + k2\pi \left( {k \in\mathbb Z} \right)\);
(C) \({\pi \over 2} + \varphi + k2\pi \left( {k \in\mathbb Z} \right)\);
(D) \(\pi - \varphi + k2\pi \left( {k \in\mathbb Z} \right)\).
Lời giải chi tiết
Ta có
\(\eqalign{ & z = - \sin \varphi - i\cos \varphi \cr &= - \cos \left( {{\pi \over 2} - \varphi } \right) - i\sin \left({{\pi \over 2} - \varphi } \right)\cr & = \cos \left({\pi + {\pi \over 2} - \varphi } \right) + i\sin \left({\pi + {\pi \over 2} - \varphi } \right) \cr & = \cos \left({{{3\pi } \over 2} - \varphi } \right) + i\sin \left({{{3\pi } \over 2} - \varphi } \right) \cr} \)
Argumen của z bằng \({{3\pi } \over 2} - \varphi + k2\pi \) \(= - {\pi \over 2} - \varphi + \left( {k + 1} \right)2\pi \)\(= - {\pi \over 2} - \varphi + 2l\pi, l\in Z \)
Chọn (B).
Ta có
\(\eqalign{ & z = - \sin \varphi - i\cos \varphi \cr &= - \cos \left( {{\pi \over 2} - \varphi } \right) - i\sin \left({{\pi \over 2} - \varphi } \right)\cr & = \cos \left({\pi + {\pi \over 2} - \varphi } \right) + i\sin \left({\pi + {\pi \over 2} - \varphi } \right) \cr & = \cos \left({{{3\pi } \over 2} - \varphi } \right) + i\sin \left({{{3\pi } \over 2} - \varphi } \right) \cr} \)
Argumen của z bằng \({{3\pi } \over 2} - \varphi + k2\pi \) \(= - {\pi \over 2} - \varphi + \left( {k + 1} \right)2\pi \)\(= - {\pi \over 2} - \varphi + 2l\pi, l\in Z \)
Chọn (B).