Bài 18: Tính giá trị lượng giác của góc α trong mỗi trường hợp sau:
a) \(\cos \alpha = {1 \over 4};\,\,\sin \alpha < 0\)
b) \(\sin = – {1 \over 3};\,{\pi \over 2} < \alpha < {{3\pi } \over 2}\)
c) \(\tan \alpha = {1 \over 2};\, – \pi < \alpha < 0\)
Đáp án
a) Ta có:
\(\eqalign{
& \sin \alpha = – \sqrt {1 – {{\cos }^2}\alpha } = – \sqrt {1 – {1 \over {16}}} = – {{\sqrt {15} } \over 4} \cr
& \tan \alpha = {{\sin \alpha } \over {\cos \alpha }} = – \sqrt {15} \cr
& \cot \alpha = {1 \over {\tan \alpha }} = – {{\sqrt {15} } \over 5} \cr} \)
b) Ta có:
\(\eqalign{
& \,{\pi \over 2} < \alpha < {{3\pi } \over 2} \Rightarrow \cos \alpha = – \sqrt {1 – {{\sin }^2}\alpha } = – {{2\sqrt 2 } \over 3} \cr
& \tan \alpha = {{\sin \alpha } \over {\cos \alpha }} = {1 \over {2\sqrt 2 }} = {{\sqrt 2 } \over 4} \cr
& \cot \alpha = 2\sqrt 2 \cr} \)
c) Ta có:
\(\eqalign{
& \left\{ \matrix{
– \pi < \alpha < 0 \hfill \cr
\tan \alpha = {1 \over 2} \hfill \cr} \right. \Rightarrow \cos \alpha < 0\cr& \Rightarrow \cos \alpha = – {1 \over {\sqrt {1 + {{\tan }^2}\alpha } }} = – {{2\sqrt 5 } \over 5} \cr
& \sin \alpha = \tan \alpha .\cot \alpha = – {{\sqrt 5 } \over 5} \cr
& \cot \alpha = {1 \over {\tan \alpha }} = 2 \cr} \)
Bài 19: Đơn giản các biểu thức
a) \(\sqrt {{{\sin }^4}\alpha + {{\sin }^2}\alpha {{\cos }^2}\alpha } \)
b) \({{1 – \cos \alpha } \over {{{\sin }^2}\alpha }} – {1 \over {1 + \cos \alpha }}\,\,(\sin \alpha \ne 0)\)
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c) \({{1 – {{\sin }^2}\alpha {{\cos }^2}\alpha } \over {{{\cos }^2}\alpha }} – {\cos ^2}\alpha \,\,\,(cos\alpha \ne 0)\)
Đáp án
a) Ta có:
\(\eqalign{
& \sqrt {{{\sin }^4}\alpha + {{\sin }^2}\alpha {{\cos }^2}\alpha } = \sqrt {{{\sin }^2}\alpha ({{\sin }^2}\alpha + {{\cos }^2}\alpha )} \cr
& = \sqrt {{{\sin }^2}\alpha } = |\sin \alpha | \cr} \)
b) Ta có:
\(\eqalign{
& {{1 – \cos \alpha } \over {{{\sin }^2}\alpha }} – {1 \over {1 + \cos \alpha }}= {{1 – \cos \alpha } \over {1 – {{\cos }^2}\alpha }} – {1 \over {1 + \cos \alpha }} \cr
& = {1 \over {1 + \cos \alpha }} – {1 \over {1 + \cos \alpha }} = 0 \cr} \)
c) Ta có:
\(\eqalign{
& {{1 – {{\sin }^2}\alpha{{\cos }^2}\alpha} \over {{{\cos }^2}\alpha}} – {\cos ^2}\alpha\cr&= {1 \over {{{\cos }^2}\alpha }} – {\sin ^2}\alpha – {\cos ^2}\alpha \cr
& = {1 \over {{{\cos }^2}\alpha }} – 1 = {\tan ^\alpha } \cr} \)
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Bài 20: 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
+
\(\eqalign{
& \sin {225^0} = \sin ( – {135^0} + {360^0})\cr& = \sin ( – {135^0}) = – {{\sqrt 2 } \over 2} \cr
& \cos {225^0} = \cos ( – {135^0} + {360^0}) \cr&= \cos ( – {135^0}) = – {{\sqrt 2 } \over 2} \cr
& \tan ( – {225^0}) = \cot {225^0} = 1 \cr} \)
+
\(\eqalign{
& \sin ( – {225^0}) = \sin ({135^0} – {360^0}) = \sin {135^0} = {{\sqrt 2 } \over 2} \cr
& cos( – {225^0}) = \cos ({135^0} – {360^0}) = \cos {135^0} = -{{\sqrt 2 } \over 2} \cr
& \tan ( – {225^0}) = – 1 = \cot ( – 225) \cr} \)
+
\(\eqalign{
& \sin {750^0} = \sin ({30^0} + {720^0}) = \sin {30^0} = {1 \over 2} \cr
& \cos {750^0} = \cos {30^0} = {{\sqrt 3 } \over 2} \cr
& \tan {750^0} = \tan {30^0} = {{\sqrt 3 } \over 2} \cr
& \cot {750^0} = \cot {30^0} = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin ( – {510^0}) = \sin ( – {150^0} – {360^0})\cr& = \sin ( – {150^0}) = – {1 \over 2} \cr
& \cos ( – {510^0}) = \cos ( – {150^0}) = – {{\sqrt 3 } \over 2} \cr
& \tan ( – {510^0}) = {1 \over {\sqrt 3 }} \cr
& \cot ( – {510^0}) = \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin {{5\pi } \over 3} = \sin ( – {\pi \over 3} + 2\pi ) = \sin ( – {\pi \over 3}) = – {{\sqrt 3 } \over 2} \cr
& \cos {{5\pi } \over 3} = \cos ( – {\pi \over 3}) = {1 \over 2} \cr
& \tan ({{5\pi } \over 3}) = – \sqrt 3 \cr
& \cot {{5\pi } \over 3} = – {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin {{11\pi } \over 6} = \sin ( – {\pi \over 6} + 2\pi ) = \sin ( – {\pi \over 6}) = – {1 \over 2} \cr
& \cos {{11\pi } \over 6} = {{\sqrt 3 } \over 2} \cr
& \tan {{11\pi } \over 6} = – {1 \over {\sqrt 3 }} \cr
& \cot {{11\pi } \over 6} = – \sqrt 3 \cr} \)
+
\(\eqalign{
& \sin ( – {{10\pi } \over 3}) = \sin ({{2\pi } \over 3} – 4\pi ) = \sin {{2\pi } \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos ( – {{10\pi } \over 3}) = \cos {{2\pi } \over 3} = – {1 \over 2} \cr
& \tan ( – {{10\pi } \over 3}) = – \sqrt 3 \cr
& \cot ( – {{10\pi } \over 3}) = – {1 \over {\sqrt 3 }} \cr} \)
+
\(\eqalign{
& \sin ( – {{17\pi } \over 3}) = \sin ({\pi \over 3} – 6\pi ) = \sin {\pi \over 3} = {{\sqrt 3 } \over 2} \cr
& \cos ( – {{17\pi } \over 3}) = \cos {\pi \over 3} = {1 \over 2} \cr
& \tan ( – {{17\pi } \over 3}) = \sqrt 3 \cr
& \cot ( – {{17\pi } \over 3}) = {1 \over {\sqrt 3 }} \cr} \)
