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2250; -2250; 7500; -5100
\({{5\pi } \over 3};\,\,{{11\pi } \over 6};\,\,{{ – 10\pi } \over 3};\,\,\, – {{17\pi } \over 3}\)
Đáp án
+
\(\eqalign{
& \sin {225^0} = \sin ( – {135^0} + {360^0})\cr& = \sin ( – {135^0}) = – {{\sqrt 2 } \over 2} \cr
& \cos {225^0} = \cos ( – {135^0} + {360^0}) \cr&= \cos ( – {135^0}) = – {{\sqrt 2 } \over 2} \cr
& \tan ( – {225^0}) = \cot {225^0} = 1 \cr} \)
+
\(\eqalign{
& \sin ( – {225^0}) = \sin ({135^0} – {360^0}) = \sin {135^0} = {{\sqrt 2 } \over 2} \cr
& cos( – {225^0}) = \cos ({135^0} – {360^0}) = \cos {135^0} = -{{\sqrt 2 } \over 2} \cr
& \tan ( – {225^0}) = – 1 = \cot ( – 225) \cr} \)
+
\(\eqalign{
& \sin {750^0} = \sin ({30^0} + {720^0}) = \sin {30^0} = {1 \over 2} \cr
& \cos {750^0} = \cos {30^0} = {{\sqrt 3 } \over 2} \cr
& \tan {750^0} = \tan {30^0} = {{\sqrt 3 } \over 2} \cr
& \cot {750^0} = \cot {30^0} = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin ( – {510^0}) = \sin ( – {150^0} – {360^0})\cr& = \sin ( – {150^0}) = – {1 \over 2} \cr
& \cos ( – {510^0}) = \cos ( – {150^0}) = – {{\sqrt 3 } \over 2} \cr
& \tan ( – {510^0}) = {1 \over {\sqrt 3 }} \cr
& \cot ( – {510^0}) = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin {{5\pi } \over 3} = \sin ( – {\pi \over 3} + 2\pi ) = \sin ( – {\pi \over 3}) = – {{\sqrt 3 } \over 2} \cr
& \cos {{5\pi } \over 3} = \cos ( – {\pi \over 3}) = {1 \over 2} \cr
& \tan ({{5\pi } \over 3}) = – \sqrt 3 \cr
& \cot {{5\pi } \over 3} = – {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin {{11\pi } \over 6} = \sin ( – {\pi \over 6} + 2\pi ) = \sin ( – {\pi \over 6}) = – {1 \over 2} \cr
& \cos {{11\pi } \over 6} = {{\sqrt 3 } \over 2} \cr
& \tan {{11\pi } \over 6} = – {1 \over {\sqrt 3 }} \cr
& \cot {{11\pi } \over 6} = – \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin ( – {{10\pi } \over 3}) = \sin ({{2\pi } \over 3} – 4\pi ) = \sin {{2\pi } \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos ( – {{10\pi } \over 3}) = \cos {{2\pi } \over 3} = – {1 \over 2} \cr
& \tan ( – {{10\pi } \over 3}) = – \sqrt 3 \cr
& \cot ( – {{10\pi } \over 3}) = – {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin ( – {{17\pi } \over 3}) = \sin ({\pi \over 3} – 6\pi ) = \sin {\pi \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos ( – {{17\pi } \over 3}) = \cos {\pi \over 3} = {1 \over 2} \cr
& \tan ( – {{17\pi } \over 3}) = \sqrt 3 \cr
& \cot ( – {{17\pi } \over 3}) = {1 \over {\sqrt 3 }} \cr} \)