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Câu hỏi: Tính các giá trị lượng giác của các góc sau
2250; -2250; 7500; -5100
\({{5\pi } \over 3}; {{11\pi } \over 6}; {{ - 10\pi } \over 3}; - {{17\pi } \over 3}\)
2250; -2250; 7500; -5100
\({{5\pi } \over 3}; {{11\pi } \over 6}; {{ - 10\pi } \over 3}; - {{17\pi } \over 3}\)
Lời giải chi tiết
+
\(\eqalign{
& \sin {225^0} = \sin (- {135^0} + {360^0})\cr& = \sin (- {135^0}) =-\sin 135^0= - {{\sqrt 2 } \over 2} \cr
& \cos {225^0} = \cos (- {135^0} + {360^0}) \cr&= \cos (- {135^0}) = \cos 135^0=- {{\sqrt 2 } \over 2} \cr } \)
\(\tan {225^0} = \frac{{\sin {{225}^0}}}{{\cos {{225}^0}}} \)\(= \left( { - \frac{{\sqrt 2 }}{2}} \right):\left({ - \frac{{\sqrt 2 }}{2}} \right) = 1\)
\(\cot {225^0} = \frac{1}{{\tan {{225}^0}}} = 1\)
+
\(\eqalign{
& \sin (- {225^0}) = \sin ({135^0} - {360^0})\cr & = \sin {135^0} = {{\sqrt 2 } \over 2} \cr
& cos(- {225^0}) = \cos ({135^0} - {360^0}) \cr &= \cos {135^0} = -{{\sqrt 2 } \over 2} \cr } \)
\(\begin{array}{l}
\tan {\left({ - 225} \right)^0} = \frac{{\sin \left({ - {{225}^0}} \right)}}{{\cos \left({ - {{225}^0}} \right)}}\\
= \frac{{\sqrt 2 }}{2}:\left({ - \frac{{\sqrt 2 }}{2}} \right) = - 1\\
\cot \left({ - {{225}^0}} \right) = \frac{1}{{\tan \left({ - {{225}^0}} \right)}} = - 1
\end{array}\)
+
\(\eqalign{
& \sin {750^0} = \sin ({30^0} + {720^0})\cr & = \sin {30^0} = {1 \over 2} \cr
& \cos {750^0} = \cos {30^0} = {{\sqrt 3 } \over 2} \cr
& \tan {750^0} = \tan {30^0} = {{\sqrt 3 } \over 2} \cr
& \cot {750^0} = \cot {30^0} = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin (- {510^0}) = \sin (- {150^0} - {360^0})\cr& = \sin (- {150^0}) = - {1 \over 2} \cr
& \cos (- {510^0}) = \cos (- {150^0}) = - {{\sqrt 3 } \over 2} \cr
& \tan (- {510^0}) = {1 \over {\sqrt 3 }} \cr
& \cot (- {510^0}) = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin {{5\pi } \over 3} = \sin (- {\pi \over 3} + 2\pi) \cr &= \sin (- {\pi \over 3}) = - {{\sqrt 3 } \over 2} \cr
& \cos {{5\pi } \over 3} = \cos (- {\pi \over 3}) = {1 \over 2} \cr
& \tan ({{5\pi } \over 3}) = - \sqrt 3 \cr
& \cot {{5\pi } \over 3} = - {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin {{11\pi } \over 6} = \sin (- {\pi \over 6} + 2\pi) \cr &= \sin (- {\pi \over 6}) = - {1 \over 2} \cr
& \cos {{11\pi } \over 6} = \cos \left({ - \frac{\pi }{6}} \right) = {{\sqrt 3 } \over 2} \cr
& \tan {{11\pi } \over 6} = - {1 \over {\sqrt 3 }} \cr
& \cot {{11\pi } \over 6} = - \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin (- {{10\pi } \over 3}) = \sin ({{2\pi } \over 3} - 4\pi)\cr &= \sin {{2\pi } \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos (- {{10\pi } \over 3}) = \cos {{2\pi } \over 3} = - {1 \over 2} \cr
& \tan (- {{10\pi } \over 3}) = - \sqrt 3 \cr
& \cot (- {{10\pi } \over 3}) = - {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin (- {{17\pi } \over 3}) = \sin ({\pi \over 3} - 6\pi)\cr & = \sin {\pi \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos (- {{17\pi } \over 3}) = \cos {\pi \over 3} = {1 \over 2} \cr
& \tan (- {{17\pi } \over 3}) = \sqrt 3 \cr
& \cot (- {{17\pi } \over 3}) = {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin {225^0} = \sin (- {135^0} + {360^0})\cr& = \sin (- {135^0}) =-\sin 135^0= - {{\sqrt 2 } \over 2} \cr
& \cos {225^0} = \cos (- {135^0} + {360^0}) \cr&= \cos (- {135^0}) = \cos 135^0=- {{\sqrt 2 } \over 2} \cr } \)
\(\tan {225^0} = \frac{{\sin {{225}^0}}}{{\cos {{225}^0}}} \)\(= \left( { - \frac{{\sqrt 2 }}{2}} \right):\left({ - \frac{{\sqrt 2 }}{2}} \right) = 1\)
\(\cot {225^0} = \frac{1}{{\tan {{225}^0}}} = 1\)
+
\(\eqalign{
& \sin (- {225^0}) = \sin ({135^0} - {360^0})\cr & = \sin {135^0} = {{\sqrt 2 } \over 2} \cr
& cos(- {225^0}) = \cos ({135^0} - {360^0}) \cr &= \cos {135^0} = -{{\sqrt 2 } \over 2} \cr } \)
\(\begin{array}{l}
\tan {\left({ - 225} \right)^0} = \frac{{\sin \left({ - {{225}^0}} \right)}}{{\cos \left({ - {{225}^0}} \right)}}\\
= \frac{{\sqrt 2 }}{2}:\left({ - \frac{{\sqrt 2 }}{2}} \right) = - 1\\
\cot \left({ - {{225}^0}} \right) = \frac{1}{{\tan \left({ - {{225}^0}} \right)}} = - 1
\end{array}\)
+
\(\eqalign{
& \sin {750^0} = \sin ({30^0} + {720^0})\cr & = \sin {30^0} = {1 \over 2} \cr
& \cos {750^0} = \cos {30^0} = {{\sqrt 3 } \over 2} \cr
& \tan {750^0} = \tan {30^0} = {{\sqrt 3 } \over 2} \cr
& \cot {750^0} = \cot {30^0} = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin (- {510^0}) = \sin (- {150^0} - {360^0})\cr& = \sin (- {150^0}) = - {1 \over 2} \cr
& \cos (- {510^0}) = \cos (- {150^0}) = - {{\sqrt 3 } \over 2} \cr
& \tan (- {510^0}) = {1 \over {\sqrt 3 }} \cr
& \cot (- {510^0}) = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin {{5\pi } \over 3} = \sin (- {\pi \over 3} + 2\pi) \cr &= \sin (- {\pi \over 3}) = - {{\sqrt 3 } \over 2} \cr
& \cos {{5\pi } \over 3} = \cos (- {\pi \over 3}) = {1 \over 2} \cr
& \tan ({{5\pi } \over 3}) = - \sqrt 3 \cr
& \cot {{5\pi } \over 3} = - {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin {{11\pi } \over 6} = \sin (- {\pi \over 6} + 2\pi) \cr &= \sin (- {\pi \over 6}) = - {1 \over 2} \cr
& \cos {{11\pi } \over 6} = \cos \left({ - \frac{\pi }{6}} \right) = {{\sqrt 3 } \over 2} \cr
& \tan {{11\pi } \over 6} = - {1 \over {\sqrt 3 }} \cr
& \cot {{11\pi } \over 6} = - \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin (- {{10\pi } \over 3}) = \sin ({{2\pi } \over 3} - 4\pi)\cr &= \sin {{2\pi } \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos (- {{10\pi } \over 3}) = \cos {{2\pi } \over 3} = - {1 \over 2} \cr
& \tan (- {{10\pi } \over 3}) = - \sqrt 3 \cr
& \cot (- {{10\pi } \over 3}) = - {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin (- {{17\pi } \over 3}) = \sin ({\pi \over 3} - 6\pi)\cr & = \sin {\pi \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos (- {{17\pi } \over 3}) = \cos {\pi \over 3} = {1 \over 2} \cr
& \tan (- {{17\pi } \over 3}) = \sqrt 3 \cr
& \cot (- {{17\pi } \over 3}) = {1 \over {\sqrt 3 }} \cr} \)