Nội dung bài giảng
Tính các giá trị lượng giác của các góc sau
2250; -2250; 7500; -5100
\({{5\pi } \over 3};\,\,{{11\pi } \over 6};\,\,{{ - 10\pi } \over 3};\,\,\, - {{17\pi } \over 3}\)
Đáp án
+
\(\eqalign{
& \sin {225^0} = \sin ( - {135^0} + {360^0})\cr& = \sin ( - {135^0}) = - {{\sqrt 2 } \over 2} \cr
& \cos {225^0} = \cos ( - {135^0} + {360^0}) \cr&= \cos ( - {135^0}) = - {{\sqrt 2 } \over 2} \cr
& \tan ( - {225^0}) = \cot {225^0} = 1 \cr} \)
+
\(\eqalign{
& \sin ( - {225^0}) = \sin ({135^0} - {360^0}) = \sin {135^0} = {{\sqrt 2 } \over 2} \cr
& cos( - {225^0}) = \cos ({135^0} - {360^0}) = \cos {135^0} = -{{\sqrt 2 } \over 2} \cr
& \tan ( - {225^0}) = - 1 = \cot ( - 225) \cr} \)
+
\(\eqalign{
& \sin {750^0} = \sin ({30^0} + {720^0}) = \sin {30^0} = {1 \over 2} \cr
& \cos {750^0} = \cos {30^0} = {{\sqrt 3 } \over 2} \cr
& \tan {750^0} = \tan {30^0} = {{\sqrt 3 } \over 2} \cr
& \cot {750^0} = \cot {30^0} = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin ( - {510^0}) = \sin ( - {150^0} - {360^0})\cr& = \sin ( - {150^0}) = - {1 \over 2} \cr
& \cos ( - {510^0}) = \cos ( - {150^0}) = - {{\sqrt 3 } \over 2} \cr
& \tan ( - {510^0}) = {1 \over {\sqrt 3 }} \cr
& \cot ( - {510^0}) = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin {{5\pi } \over 3} = \sin ( - {\pi \over 3} + 2\pi ) = \sin ( - {\pi \over 3}) = - {{\sqrt 3 } \over 2} \cr
& \cos {{5\pi } \over 3} = \cos ( - {\pi \over 3}) = {1 \over 2} \cr
& \tan ({{5\pi } \over 3}) = - \sqrt 3 \cr
& \cot {{5\pi } \over 3} = - {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin {{11\pi } \over 6} = \sin ( - {\pi \over 6} + 2\pi ) = \sin ( - {\pi \over 6}) = - {1 \over 2} \cr
& \cos {{11\pi } \over 6} = {{\sqrt 3 } \over 2} \cr
& \tan {{11\pi } \over 6} = - {1 \over {\sqrt 3 }} \cr
& \cot {{11\pi } \over 6} = - \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin ( - {{10\pi } \over 3}) = \sin ({{2\pi } \over 3} - 4\pi ) = \sin {{2\pi } \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos ( - {{10\pi } \over 3}) = \cos {{2\pi } \over 3} = - {1 \over 2} \cr
& \tan ( - {{10\pi } \over 3}) = - \sqrt 3 \cr
& \cot ( - {{10\pi } \over 3}) = - {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin ( - {{17\pi } \over 3}) = \sin ({\pi \over 3} - 6\pi ) = \sin {\pi \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos ( - {{17\pi } \over 3}) = \cos {\pi \over 3} = {1 \over 2} \cr
& \tan ( - {{17\pi } \over 3}) = \sqrt 3 \cr
& \cot ( - {{17\pi } \over 3}) = {1 \over {\sqrt 3 }} \cr} \)