cho \(\pi< \alpha< 2\pi\) và \(cos\alpha=\frac{2}{3}\). Tính GTBT \(A=tan\left(\alpha-3\pi\right)-tan2\alpha\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 4 2021
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 5 2020
\(\pi< a< \frac{3\pi}{2}\Rightarrow2\pi< 2a< 3\pi\Rightarrow sin2a>0\)
\(cot2a=\frac{1}{2}\Rightarrow sin2a=\frac{1}{\sqrt{1+cot^22a}}=\frac{2\sqrt{5}}{5}\)
\(cos\left(a+\frac{\pi}{3}\right)+cos\left(a-\frac{\pi}{3}\right)=2cosa.cos\frac{\pi}{3}=cosa\)
\(tan\left(\frac{\pi}{2}-a\right)+tan\left(\frac{\pi}{2}+\frac{a}{2}\right)=\frac{-sin\frac{a}{2}}{cos\left(\frac{\pi}{2}-a\right).cos\left(\frac{\pi}{2}+\frac{a}{2}\right)}=\frac{sin\frac{a}{2}}{sina.sin\frac{a}{2}}=\frac{1}{sina}\)
\(\Rightarrow M=sina.cosa=\frac{1}{2}sin2a=\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow2a+b=7\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 6 2020
\(\frac{\pi}{2}< a< \frac{3\pi}{2}\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{3}}{2}\)
\(A=cosa.cos\frac{4\pi}{3}+sina.sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}.\left(-\frac{1}{2}\right)+\frac{1}{2}.\left(-\frac{\sqrt{3}}{2}\right)=0\)
\(B=cos\left(2a+2019.2\pi\right)=cos2a=1-2sin^2a=1-2\left(\frac{1}{2}\right)^2=\frac{1}{2}\)
LH
16 tháng 5 2020
--.-- \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ
LH
16 tháng 5 2020
\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)
\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)
\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)
\(\cos2a=2\cos^2a-1=\frac{7}{25}\)
\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)
\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)
\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)
\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)
\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)
Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)
\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)
QT
Quoc Tran Anh Le
Giáo viên
22 tháng 9 2023
a, Ta có: \({\sin ^2}x + co{s^2}x = 1\)
\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha + {\left( {\frac{1}{3}} \right)^2} = 1\\ \Leftrightarrow \sin \alpha = \pm \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} = \pm \frac{{2\sqrt 2 }}{3}\end{array}\)
Vì \( - \frac{\pi }{2} < \alpha < 0\) nên \(sin\alpha < 0 \Rightarrow \sin \alpha = - \frac{{2\sqrt 2 }}{3}\).
\(b)\;\,sin2\alpha = 2sin\alpha .cos\alpha = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} = - \frac{{4\sqrt 2 }}{9}\)
\(c)\;cos(\alpha + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6 + 1}}{6}\).
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
Ta có:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3 + 3\sqrt 2 }}{6}\)
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} = - \frac{{3 + \sqrt 6 }}{6}\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
BT
11 tháng 5 2017
a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).
BT
11 tháng 5 2017
b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).
25 tháng 7 2018
bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)
\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\)
\(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)
QT
Quoc Tran Anh Le
Giáo viên
21 tháng 9 2023
\(\cos \alpha = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = - \frac{{12}}{{13}}\) (vì \(\pi < \alpha < \frac{{3\pi }}{2}\))
\(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)
\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha + \sin \frac{\pi }{4}sin\alpha = \frac{{ - 17\sqrt 2 }}{{26}}\)
\(\left\{{}\begin{matrix}\pi< a< 2\pi\\cosa>0\end{matrix}\right.\) \(\Rightarrow\frac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\Rightarrow sina=-\sqrt{1-cos^2a}=-\frac{\sqrt{5}}{3}\)
\(A=tan\left(a-3\pi\right)-tan2a=tana-tan2a\)
\(=\frac{sina}{cosa}-\frac{sin2a}{cos2a}=\frac{sina}{cosa}-\frac{2sina.cosa}{2cos^2a-1}\) (thay số và bấm máy)