Giá trị biểu thức P= \(\left(sin2a+sin2b\right)^2+\left(cos2a+cos2b\right)^2\) BIẾT a-b=\(\frac{\pi}{6}\) là
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NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 6 2020
\(\pi< a< \frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\) \(\Rightarrow sin2a=2sina.cosa>0\)
\(\Rightarrow sin2a=\sqrt{1-cos^22a}=\frac{3\sqrt{7}}{8}\)
\(cos2a=1-2sin^2a=\frac{1}{8}\)
\(\Leftrightarrow sin^2a=\frac{7}{16}\Rightarrow sina=-\frac{\sqrt{7}}{4}\)
\(\Rightarrow M=\frac{-\frac{\sqrt{7}}{4}-\frac{3\sqrt{7}}{8}}{-\frac{\sqrt{7}}{4}+\frac{3\sqrt{7}}{8}}=...\)
\(sinx\left(1-tan^2\frac{x}{2}\right)=sinx\left(1-\frac{sin^2\frac{x}{2}}{cos^2\frac{x}{2}}\right)=sinx\left(1-\frac{1-cosx}{1+cosx}\right)\)
\(=sinx\left(\frac{1+cosx-\left(1-cosx\right)}{1+cosx}\right)=\frac{2sinx.cosx}{1+cosx}\)
\(1-sin2x.sin3x-cos2x.cos3x=1-\left(cos3x.cos2x+sin3x.sin2x\right)=1-cos\left(3x-2x\right)=1-cosx\)
\(\Rightarrow\frac{1-sin2x.sin3x-cos2x.cos3x}{sinx\left(1-tan^2\frac{x}{2}\right)}=\frac{1-cosx}{\frac{2sinx.cosx}{1+cosx}}=\frac{\left(1-cosx\right)\left(1+cosx\right)}{2sinx.cosx}\)
\(=\frac{1-cos^2x}{2sinx.cosx}=\frac{sin^2x}{2sinx.cosx}=\frac{sinx}{2cosx}=\frac{1}{2}tanx\)
RH
18 tháng 2 2022
b)\(P=cos2a-cos(\dfrac{\pi}{3}-a) \\=2cos^2a-1-cos\dfrac{\pi}{3}cosa-sin\dfrac{\pi}{3}sina \\=2.(\dfrac{-2}{5})^2-1-\dfrac{1}{2}.\dfrac{-2}{5}-\dfrac{\sqrt3}{2}.\dfrac{-\sqrt{21}}{5} \\=\dfrac{-24+15\sqrt7}{50}\)
DT
18 tháng 2 2022
a, Vì : \(\pi< a< \dfrac{3\pi}{2}\) nên \(cos\alpha< 0\) mà \(cos^2\alpha=1-sin^2\alpha=1-\dfrac{4}{25}=\dfrac{21}{25},\)
do đó : \(cos\alpha=-\dfrac{\sqrt{21}}{5}\)
từ đó suy ra : \(tan\alpha=\dfrac{2}{\sqrt{21}},cot\alpha=\dfrac{\sqrt{21}}{2}\)
DT
0
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 5 2019
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 5 2019
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 6 2020
\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(cos^2\left(a-\frac{\pi}{4}\right)-sin^2\left(a-\frac{\pi}{4}\right)=cos\left(2a-\frac{\pi}{2}\right)\)
\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 5 2020
\(sin\left(\frac{\pi}{7}\right)H=sin\left(\frac{\pi}{7}\right)cos\left(\frac{2\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{4\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)\)
\(=\frac{1}{2}\left[sin\left(\frac{3\pi}{7}\right)-sin\left(\frac{\pi}{7}\right)+sin\left(\frac{5\pi}{7}\right)-sin\left(\frac{3\pi}{7}\right)+sin\pi-sin\left(\frac{5\pi}{7}\right)\right]\)
\(=-\frac{1}{2}sin\left(\frac{\pi}{7}\right)\)
\(\Rightarrow H=-\frac{1}{2}\)
\(sinA+sinB+sinC=2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)+2sin\left(\frac{C}{2}\right)cos\left(\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}cos\left(\frac{A-B}{2}\right)+2cos\left(\frac{A+B}{2}\right)cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left[cos\left(\frac{A-B}{2}\right)+cos\left(\frac{A+B}{2}\right)\right]\)
\(=4cos\frac{C}{2}cos\frac{A}{2}cos\frac{B}{2}\)
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) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
\(P=sin^22a+cos^22a+sin^22b+cos^22b+2sin2a.sin2b+2cos2a.cos2b\)
\(P=2+2\left(sin2a.sin2b+cos2a.cos2b\right)=2+2cos\left(2a-2b\right)\)
\(P=2+2cos\frac{\pi}{3}=3\)