Đề bài - bài 20 trang 200 sgk đại số 10 nâng cao

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

+

\(\eqalign{
& \sin {225^0} = \sin ( - {135^0} + {360^0})\cr& = \sin ( - {135^0}) =-\sin 135^0= - {{\sqrt 2 } \over 2} \cr
& \cos {225^0} = \cos ( - {135^0} + {360^0}) \cr&= \cos ( - {135^0}) = \cos 135^0=- {{\sqrt 2 } \over 2} \cr} \)

\(\tan {225^0} = \frac{{\sin {{225}^0}}}{{\cos {{225}^0}}} \)\(= \left( { - \frac{{\sqrt 2 }}{2}} \right):\left( { - \frac{{\sqrt 2 }}{2}} \right) = 1\)

\(\cot {225^0} = \frac{1}{{\tan {{225}^0}}} = 1\)

+

\(\eqalign{
& \sin ( - {225^0}) = \sin ({135^0} - {360^0})\cr & = \sin {135^0} = {{\sqrt 2 } \over 2} \cr
& cos( - {225^0}) = \cos ({135^0} - {360^0}) \cr &= \cos {135^0} = -{{\sqrt 2 } \over 2} \cr} \)

\(\begin{array}{l}
\tan {\left( { - 225} \right)^0} = \frac{{\sin \left( { - {{225}^0}} \right)}}{{\cos \left( { - {{225}^0}} \right)}}\\
= \frac{{\sqrt 2 }}{2}:\left( { - \frac{{\sqrt 2 }}{2}} \right) = - 1\\
\cot \left( { - {{225}^0}} \right) = \frac{1}{{\tan \left( { - {{225}^0}} \right)}} = - 1
\end{array}\)

+

\(\eqalign{
& \sin {750^0} = \sin ({30^0} + {720^0})\cr & = \sin {30^0} = {1 \over 2} \cr
& \cos {750^0} = \cos {30^0} = {{\sqrt 3 } \over 2} \cr
& \tan {750^0} = \tan {30^0} = {{\sqrt 3 } \over 2} \cr
& \cot {750^0} = \cot {30^0} = \sqrt 3 \cr} \)

+

\(\eqalign{
& \sin ( - {510^0}) = \sin ( - {150^0} - {360^0})\cr& = \sin ( - {150^0}) = - {1 \over 2} \cr
& \cos ( - {510^0}) = \cos ( - {150^0}) = - {{\sqrt 3 } \over 2} \cr
& \tan ( - {510^0}) = {1 \over {\sqrt 3 }} \cr
& \cot ( - {510^0}) = \sqrt 3 \cr} \)

+

\(\eqalign{
& \sin {{5\pi } \over 3} = \sin ( - {\pi \over 3} + 2\pi ) \cr &= \sin ( - {\pi \over 3}) = - {{\sqrt 3 } \over 2} \cr
& \cos {{5\pi } \over 3} = \cos ( - {\pi \over 3}) = {1 \over 2} \cr
& \tan ({{5\pi } \over 3}) = - \sqrt 3 \cr
& \cot {{5\pi } \over 3} = - {1 \over {\sqrt 3 }} \cr} \)

+

\(\eqalign{
& \sin {{11\pi } \over 6} = \sin ( - {\pi \over 6} + 2\pi ) \cr &= \sin ( - {\pi \over 6}) = - {1 \over 2} \cr
& \cos {{11\pi } \over 6}= \cos \left( { - \frac{\pi }{6}} \right)= {{\sqrt 3 } \over 2} \cr
& \tan {{11\pi } \over 6} = - {1 \over {\sqrt 3 }} \cr
& \cot {{11\pi } \over 6} = - \sqrt 3 \cr} \)

+

\(\eqalign{
& \sin ( - {{10\pi } \over 3}) = \sin ({{2\pi } \over 3} - 4\pi )\cr &= \sin {{2\pi } \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos ( - {{10\pi } \over 3}) = \cos {{2\pi } \over 3} = - {1 \over 2} \cr
& \tan ( - {{10\pi } \over 3}) = - \sqrt 3 \cr
& \cot ( - {{10\pi } \over 3}) = - {1 \over {\sqrt 3 }} \cr} \)

+

\(\eqalign{
& \sin ( - {{17\pi } \over 3}) = \sin ({\pi \over 3} - 6\pi )\cr & = \sin {\pi \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos ( - {{17\pi } \over 3}) = \cos {\pi \over 3} = {1 \over 2} \cr
& \tan ( - {{17\pi } \over 3}) = \sqrt 3 \cr
& \cot ( - {{17\pi } \over 3}) = {1 \over {\sqrt 3 }} \cr} \)