cos2(\(\frac{\pi}{4}\)+ \(\alpha\)) - cos2(\(\frac{\pi}{4}\)- \(\alpha\))
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HQ
Hà Quang Minh
Giáo viên
25 tháng 8 2023
\(a,cos2\alpha=2cos^2\alpha-1=\dfrac{2}{5}\\ \Leftrightarrow cos^2\alpha=\dfrac{7}{10}\Rightarrow cos\alpha=\pm\dfrac{\sqrt{70}}{10}\)
Vì \(-\dfrac{\pi}{2}< \alpha< 0\Rightarrow cos\alpha=\dfrac{\sqrt{70}}{10}\)
Ta có:
\(sin^2\alpha+cos^2\alpha=1\\ \Rightarrow sin^2\alpha=1-\dfrac{7}{10}=\dfrac{3}{10}\\ \Rightarrow sin\alpha=\pm\sqrt{30}10\)
Vì \(-\dfrac{\pi}{2}< \alpha< 0\Rightarrow sin\alpha=-\dfrac{\sqrt{30}}{10}\)
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\dfrac{\sqrt{30}}{10}}{\dfrac{-\sqrt{70}}{10}}=-\dfrac{\sqrt{21}}{7}\\ cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{-\dfrac{\sqrt{21}}{7}}=-\dfrac{\sqrt{21}}{3}\)
HQ
Hà Quang Minh
Giáo viên
25 tháng 8 2023
\(b,sin^22\alpha+cos^22\alpha=1\\ \Rightarrow cos2\alpha=\sqrt{1-\left(-\dfrac{4}{9}\right)^2}=\pm\dfrac{\sqrt{65}}{9}\)
Vì \(\dfrac{\pi}{2}< \alpha< \dfrac{3\pi}{4}\Rightarrow\pi< 2\alpha< \dfrac{3\pi}{2}\Rightarrow cos2\alpha=-\dfrac{\sqrt{65}}{9}\)
\(cos2\alpha=2cos^2\alpha-1=-\dfrac{\sqrt{65}}{9}\\ \Rightarrow cos\alpha=\pm\sqrt{\dfrac{9-\sqrt{65}}{18}}\)
Vì \(\dfrac{\pi}{2}< \alpha< \dfrac{3\pi}{4}\Rightarrow cos\alpha=-\sqrt{\dfrac{9-\sqrt{65}}{18}}\)
\(sin^2\alpha+cos^2\alpha=1\\ \Rightarrow sin^2\alpha=\dfrac{9+\sqrt{65}}{18}\\ \Rightarrow sin\alpha=\pm\sqrt{\dfrac{9+\sqrt{65}}{18}}\)
Vì \(\dfrac{\pi}{2}< \alpha< \dfrac{3\pi}{4}\Rightarrow sin\alpha=\sqrt{\dfrac{9+\sqrt{65}}{18}}\)
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{\dfrac{9+\sqrt{65}}{18}}}{-\sqrt{\dfrac{9-\sqrt{65}}{18}}}\approx-4,266\\ cot\alpha=\dfrac{1}{tan\alpha}\approx-0,234\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 6 2020
\(\frac{\pi}{2}< a< \frac{3\pi}{2}\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{3}}{2}\)
\(A=cosa.cos\frac{4\pi}{3}+sina.sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}.\left(-\frac{1}{2}\right)+\frac{1}{2}.\left(-\frac{\sqrt{3}}{2}\right)=0\)
\(B=cos\left(2a+2019.2\pi\right)=cos2a=1-2sin^2a=1-2\left(\frac{1}{2}\right)^2=\frac{1}{2}\)
QT
Quoc Tran Anh Le
Giáo viên
22 tháng 9 2023
\(a)\;sin(\alpha + \beta ).sin(\alpha - \beta ) = \;\frac{1}{2}.\left[ {cos\left( {\alpha + \beta - \alpha + \beta } \right) - cos\left( {\alpha + \beta + \alpha - \beta } \right)} \right]\)
\(\begin{array}{l} = \;\frac{1}{2}.(cos2\beta - cos2\alpha ) = \;\frac{1}{2}.(1 - 2si{n^2}\beta - 1 + 2si{n^2}\alpha )\\ = si{n^2}\alpha - si{n^2}\beta \end{array}\)
\(\begin{array}{l}b)\;co{s^4}\alpha - co{s^4}\left( {\alpha - \frac{\pi }{2}} \right) = \;co{s^4}\alpha - si{n^4}\alpha \\ = \;(co{s^2}\alpha + si{n^2}\alpha )(co{s^2}\alpha - si{n^2}\alpha )\\ = \;co{s^2}\alpha -si{n^2}\alpha = cos2\alpha .\end{array}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2019
\(\dfrac{1+cos2a-sin2a}{1+cos2a+sin2a}=\dfrac{2cos^2a-2sina.cosa}{2cos^2a+2sinacosa}\)
\(=\dfrac{2cosa\left(cosa-sina\right)}{2cosa\left(cosa+sina\right)}=\dfrac{cosa-sina}{cosa+sina}=\dfrac{\sqrt{2}sin\left(\dfrac{\pi}{4}-a\right)}{\sqrt{2}cos\left(\dfrac{\pi}{4}-a\right)}=tan\left(\dfrac{\pi}{4}-a\right)\)
\(\dfrac{1+cos2a-cosa}{sin2a-sina}=\dfrac{2cos^2a-cosa}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)
AS
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 4 2019
\(0< a< \frac{\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina>0\\cosa>0\end{matrix}\right.\)
\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cos^2a=\frac{1}{1+tan^2a}\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}\)
\(\Rightarrow cosa=\frac{1}{2}\Rightarrow sina=cosa.tana=\frac{\sqrt{3}}{2}\)
\(cos2a=2cos^2a-1=-\frac{1}{2}\)
\(sin2a=2sina.cosa=\frac{\sqrt{3}}{2}\)
\(\Rightarrow sin\left(2a-\frac{\pi}{3}\right)=sin2a.cos\frac{\pi}{3}-cos2a.sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)
\(tan\left(a+\frac{\pi}{4}\right)=\frac{tana+tan\frac{\pi}{4}}{1-tana.tan\frac{\pi}{4}}=-2-\sqrt{3}\)