Giải câu 5 bài 3: Công thức lượng giác sgk Đại số 10 trang 154

Ta có:

\(sin\,2a=2\,sin\,a\,cos\,a\)

\(cos\,2a=cos^2\,a-sin^2\,a=2cos^2\,a-1=1-2sin^2a\)

\(tan\,2a=\frac{sin\,2a}{cos\,2a}\)

a) \(\sin a = -0,6; \pi  < a < {{3\pi } \over 2}\)

\(\pi  < a < {{3\pi } \over 2} \Rightarrow \cos\, a < 0\)

và \(\sin a = -0,6=-\frac{3}{5} \Rightarrow \cos a =  - {4 \over 5}\)

\(\Rightarrow \sin 2{\rm{a}} = 2.( - 0,6).\left( { - {4 \over 5}} \right) = {{24} \over {25}}\)

\(\cos 2a = 1 - 2\sin^2a = 1 - 2{\left( { - {3 \over 5}} \right)^2} = 1 - {{18} \over {25}}= {7 \over {25}}\)

\(\tan 2a = {{\sin 2a} \over {\cos 2a}} = {{24} \over {25}}.{{25} \over 7} = {{24} \over 7}\)

b) \(\cos a =  - {5 \over {13}}; {\pi  \over 2} < a < \pi\)

\({\pi  \over 2} < a < \pi \Rightarrow \sin a > 0; \tan a < 0\)

\(\cos a =  - {5 \over {13}}\Rightarrow \sin {\rm{a}} = {{12} \over {13}}\)

\(\Rightarrow \sin 2{\rm{a}} = 2.{{12} \over {13}}.\left( { - {5 \over {13}}} \right) =  - {{120} \over {169}}\)

\(\cos 2a = 2.{\cos ^2}a - 1 = 2.{{25} \over {169}} - 1 =  - {{119} \over {169}}\)

\(\tan 2a = {{\sin 2a} \over {\cos 2a}} = \left( { - {{120} \over {169}}} \right).\left( { - {{169} \over {119}}} \right) = {{120} \over {119}}\)

c) \(\sin {\rm{a}} + {\mathop{\rm cosa}\nolimits}  = {1 \over 2}\) và  \({{3\pi } \over 4} < a < \pi\)

\({{3\pi } \over 4} < a < \pi \Rightarrow \sin a > 0; \cos a < 0\)

\(\left\{\begin{matrix}cos^2\,a+sin^2\,a=1 & \\ sin\,a+cos\,a=\frac{1}{2} & \end{matrix}\right. \Rightarrow \left\{ \matrix{\cos a = {{1 - \sqrt 7 } \over 4} \hfill \cr \sin a = {{1 + \sqrt 7 } \over 4} \hfill \cr} \right.\)

\(\Rightarrow \sin 2a = 2.{{1 + \sqrt 7 } \over 4}.{{1 - \sqrt 7 } \over 4} = {{ - 3} \over 4}\)

\(\cos 2a = 1 - 2{\sin ^2}a = 1 - 2{\left( {{{1 + \sqrt 7 } \over 4}} \right)^2} = {{  \sqrt 7 } \over 4}\)

\(\tan 2a =  - {{3\sqrt 7 } \over 7}\)