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BT
18 tháng 5 2017
\(A=\dfrac{3sin\alpha-cos\alpha}{sin\alpha+cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-1}{\dfrac{sin\alpha}{cos\alpha}-1}=\dfrac{3tan\alpha-1}{tan\alpha-1}\)\(=\dfrac{3\sqrt{2}-1}{\sqrt{2}-1}=5+2\sqrt{2}\).
HT
18 tháng 5 2016
cotα = \(\frac{1}{3}\) \(\Leftrightarrow\frac{cos\alpha}{\sin\alpha}=\frac{1}{3}\Leftrightarrow\sin\alpha=3\cos\alpha\)
cotα =\(\frac{1}{\tan\alpha}=\frac{1}{3}\Rightarrow\tan\alpha=3\)
T = \(\frac{2016}{\sin^2\alpha-\sin\alpha\cos\alpha-\cos^2\alpha}=\frac{2016}{9\cos^2\alpha-3\cos^2\alpha-\cos^2\alpha}\) \(=\frac{2016}{5\cos^2\alpha}=\frac{2016}{5}\times\frac{1}{\cos^2\alpha}=\frac{2016}{5}\times\left(1+\tan^2\alpha\right)\) \(=\frac{2016}{5}\left(1+9\right)=4032\)
NK
30 tháng 3 2017
Chia tử và mẫu cho cosa ta có:
B=\(\dfrac{4\tan a+5}{2\tan a-3}\). Vì \(\cot a=\dfrac{1}{2}\) nên \(\tan a=2\)
=> B=13
30 tháng 3 2017
chia tử và mẫu của B cho sina khác 0\(B=\dfrac{4\dfrac{sina}{sina}+5\dfrac{cosa}{sina}}{2\dfrac{sina}{sina}-3\dfrac{cosa}{sina}}=\dfrac{4+5cota}{2-3cota}=\dfrac{4+5\dfrac{1}{2}}{2-3\dfrac{1}{2}}=13\)
vay B = 13
CM
29 tháng 4 2018
Chọn D.
Ta có : sin2a = 2.sina. cosa và sin2a = 1 - cos2a.
Do đó;
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 4 2021
\(P.sin\left(\dfrac{\pi}{7}\right)=sin\dfrac{\pi}{7}.cos\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}\)
\(\Leftrightarrow P.sin\dfrac{\pi}{7}=\dfrac{1}{2}sin\dfrac{2\pi}{7}cos\dfrac{2\pi}{7}cos\dfrac{4\pi}{7}\)
\(\Leftrightarrow P.sin\dfrac{\pi}{7}=\dfrac{1}{4}sin\dfrac{4\pi}{7}cos\dfrac{4\pi}{7}\)
\(\Leftrightarrow P.sin\dfrac{\pi}{7}=\dfrac{1}{8}sin\dfrac{8\pi}{7}=\dfrac{1}{8}sin\left(\pi+\dfrac{\pi}{7}\right)\)
\(\Leftrightarrow P.sin\dfrac{\pi}{7}=-\dfrac{1}{8}sin\dfrac{\pi}{7}\)
\(\Rightarrow P=-\dfrac{1}{8}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 1 2019
\(B=cos^273^0+cos^287^0+cos^23^0+cos^217^0\)
\(\Rightarrow B=cos^273^0+cos^287^0+cos^2\left(90^0-87^0\right)+cos^2\left(90^0-73^0\right)\)
\(\Rightarrow B=cos^273^0+cos^287^0+sin^287^0+sin^273^0\)
\(\Rightarrow B=\left(cos^273^0+sin^273^0\right)+\left(cos^287^0+sin^287^0\right)\)
\(\Rightarrow B=1+1=2\)
30 tháng 3 2017
\(A=cos3a+2cos\left(\pi-3a\right)sin^2\left(\dfrac{\pi}{4}-1,5a\right)\)
\(=cos3a-2cos3a\dfrac{1-cos\left(\dfrac{\pi}{2}-3a\right)}{2}\)
\(=cos3a-cos3a\left(1-sin3a\right)\)
\(=cos3a-cos3a+cos3asin3a=\dfrac{1}{2}sin6a\)
\(=\dfrac{1}{2}sin\left(6\dfrac{5\pi}{6}\right)=\dfrac{1}{2}sin\left(4\pi+\pi\right)=\dfrac{1}{2}sin\pi=0\)
NK
30 tháng 3 2017
Vì a=\(\dfrac{5\pi}{6}\) nên: \(3a=\dfrac{5\pi}{2}\) => \(\cos3a=0\)
\(\pi-3a=\pi-\dfrac{5\pi}{2}=\dfrac{-3\pi}{2}\)
=> \(\cos\left(\pi-3a\right)=0\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2021
a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)
b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)
c/\(P=sin\left(30+60\right)=sin90=1\)
d/
\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)
\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)
e/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)
\(E=\)\(cos^273+1-sin^247+cos73\left(cos120.cos73+sin120.sin73\right)\)
\(=cos^273+1-\left(sin120.cos73-cos120.sin73\right)^2-\dfrac{1}{2}.cos^273+\dfrac{\sqrt{3}}{2}cos73.sin73\)
\(=cos^273+1-\left(\dfrac{\sqrt{3}}{2}.cos73+\dfrac{1}{2}.sin73\right)^2-\dfrac{1}{2}.cos73^2+\dfrac{\sqrt{3}}{2}cos73.sin73\)
\(=\dfrac{1}{2}cos^273+1-\left(\dfrac{3}{4}cos^273+\dfrac{\sqrt{3}}{2}.cos73.sin73+\dfrac{1}{4}sin^273\right)+\dfrac{\sqrt{3}}{2}.cos73.sin73\)
\(=1-\dfrac{1}{4}.cos^273-\dfrac{1}{4}.sin^273\)
\(=1-\dfrac{1}{4}=\dfrac{3}{4}\)
Đề có sai không bạn nhỉ?