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