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\(A=\dfrac{3sin\alpha-cos\alpha}{sin\alpha+cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-1}{\dfrac{sin\alpha}{cos\alpha}-1}=\dfrac{3tan\alpha-1}{tan\alpha-1}\)\(=\dfrac{3\sqrt{2}-1}{\sqrt{2}-1}=5+2\sqrt{2}\).
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 = \(\frac{2016}{\sin^2\alpha-\sin\alpha\cos\alpha-\cos^2\alpha}=\frac{2016}{9\cos^2\alpha-3\cos^2\alpha-\cos^2\alpha}\) \(=\frac{2016}{5\cos^2\alpha}=\frac{2016}{5}\times\frac{1}{\cos^2\alpha}=\frac{2016}{5}\times\left(1+\tan^2\alpha\right)\) \(=\frac{2016}{5}\left(1+9\right)=4032\)
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
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
\(B=cos^273^0+cos^287^0+cos^23^0+cos^217^0\)
\(\Rightarrow B=cos^273^0+cos^287^0+cos^2\left(90^0-87^0\right)+cos^2\left(90^0-73^0\right)\)
\(\Rightarrow B=cos^273^0+cos^287^0+sin^287^0+sin^273^0\)
\(\Rightarrow B=\left(cos^273^0+sin^273^0\right)+\left(cos^287^0+sin^287^0\right)\)
\(\Rightarrow B=1+1=2\)
\(A=cos3a+2cos\left(\pi-3a\right)sin^2\left(\dfrac{\pi}{4}-1,5a\right)\)
\(=cos3a-2cos3a\dfrac{1-cos\left(\dfrac{\pi}{2}-3a\right)}{2}\)
\(=cos3a-cos3a\left(1-sin3a\right)\)
\(=cos3a-cos3a+cos3asin3a=\dfrac{1}{2}sin6a\)
\(=\dfrac{1}{2}sin\left(6\dfrac{5\pi}{6}\right)=\dfrac{1}{2}sin\left(4\pi+\pi\right)=\dfrac{1}{2}sin\pi=0\)
Vì a=\(\dfrac{5\pi}{6}\) nên: \(3a=\dfrac{5\pi}{2}\) => \(\cos3a=0\)
\(\pi-3a=\pi-\dfrac{5\pi}{2}=\dfrac{-3\pi}{2}\)
=> \(\cos\left(\pi-3a\right)=0\)
a/ \(\frac{\pi}{6}< x< \frac{\pi}{3}\Rightarrow cosx>0\)
\(cos^2x=\frac{1}{1+tan^2x}=\frac{1}{10}\)
\(cotx=\frac{1}{tanx}=\frac{1}{3}\)
Thay số và bấm máy
b/ \(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\tana< 0\end{matrix}\right.\)
\(sina=\sqrt{1-cos^2a}=\frac{3}{5}\)
\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)
\(A=\frac{6sina.cosa-\frac{2tana}{1-tan^2a}}{cosa-\left(2cos^2a-1\right)}\)
Thay số và bấm máy
c/ \(\frac{3\pi}{2}< x< 2\pi\Rightarrow\left\{{}\begin{matrix}cosx>0\\sinx< 0\end{matrix}\right.\)
\(cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{1}{\sqrt{5}}\)
\(sinx=cosx.tanx=-\frac{2}{\sqrt{5}}\)
\(B=\frac{cos^2x+2sinx.cosx}{\frac{2tanx}{1-tan^2x}-\left(2cos^2x-1\right)}\)
Thay số
Lời giải:
Ta có:
\(P=\frac{2\sin \alpha+3\cos \alpha}{4\sin \alpha-5\cos \alpha}=\frac{2+\frac{3\cos \alpha}{\sin \alpha}}{4-\frac{5\cos \alpha}{\sin \alpha}}\)
\(=\frac{2+3\cot \alpha}{4-5\cot\alpha}=\frac{2+3.3}{4-5.3}=-1\)
\(E=\)\(cos^273+1-sin^247+cos73\left(cos120.cos73+sin120.sin73\right)\)
\(=cos^273+1-\left(sin120.cos73-cos120.sin73\right)^2-\dfrac{1}{2}.cos^273+\dfrac{\sqrt{3}}{2}cos73.sin73\)
\(=cos^273+1-\left(\dfrac{\sqrt{3}}{2}.cos73+\dfrac{1}{2}.sin73\right)^2-\dfrac{1}{2}.cos73^2+\dfrac{\sqrt{3}}{2}cos73.sin73\)
\(=\dfrac{1}{2}cos^273+1-\left(\dfrac{3}{4}cos^273+\dfrac{\sqrt{3}}{2}.cos73.sin73+\dfrac{1}{4}sin^273\right)+\dfrac{\sqrt{3}}{2}.cos73.sin73\)
\(=1-\dfrac{1}{4}.cos^273-\dfrac{1}{4}.sin^273\)
\(=1-\dfrac{1}{4}=\dfrac{3}{4}\)
Đề có sai không bạn nhỉ?