a) \(\cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } \)
b) \(\sqrt {{{\sin }^2}\alpha } = \sin \alpha \)
c) \(\tan \alpha = {{\sqrt {1 - {{\cos }^2}\alpha } } \over {\cos \alpha }}\)
a) Ta có:
\(\eqalign{
& \cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } \Leftrightarrow \cos \alpha = \sqrt {{{\cos }^2}\alpha } \cr
& \Leftrightarrow \cos \alpha \ge 0 \cr} \)
⇔ M(x, y) thỏa mãn x2 + y2 = 1; x ≥ 0
b) Ta có:
\(\sqrt {{{\sin }^2}\alpha } = \sin \alpha \Leftrightarrow \sin \alpha \ge 0\)
⇔ M(x, y) thỏa mãn x2 + y2 = 1; y ≥ 0
c) Ta có:
\(\tan \alpha = {{\sqrt {1 - {{\cos }^2}\alpha } } \over {\cos \alpha }} \Leftrightarrow \left\{ \matrix{
\sin \alpha \ge 0 \hfill \cr
\cos \alpha \ne 0 \hfill \cr} \right.\)
⇔ M(x, y) thỏa mãn x2 + y2 = 1, y ≥ 0; y ≠ 1
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