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a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(tan\alpha,cot\alpha>0\) và \(sin\alpha,cos\alpha< 0\).
\(\left\{{}\begin{matrix}tan\alpha-3cot\alpha=6\\tan\alpha cot\alpha=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\\left(6+3cot\alpha\right)cot\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\3cot^2\alpha+6cot\alpha-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\cot\alpha=\dfrac{-3+2\sqrt{3}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=3+2\sqrt{3}\\cot\alpha=\dfrac{-3+2\sqrt{3}}{3}\end{matrix}\right.\).
Có \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{1}{tan^2\alpha+1}\).
Có thể đề sai.
a/ Ta có: \(tan\alpha=5\Rightarrow cot\alpha=\frac{1}{5}\) . Đề: \(\frac{sin\alpha}{sin^3\alpha+cos^3\alpha}=\frac{\frac{1}{sin^2\alpha}}{1+\frac{cos^3\alpha}{sin^3\alpha}}=\frac{1+cot^2\alpha}{1+cot^3\alpha}=\frac{1+\left(\frac{1}{5}\right)^2}{1+\left(\frac{1}{5}\right)^3}=\frac{65}{63}\)
b/ Ta có vế trái \(=\frac{sin^2x+cos^2x+cos^2x-sin^2x+\left(sinx+sin3x\right)}{1+2sinx}=\frac{2cos^2x+2.sin2x.cosx}{1+2sinx}=\frac{2cos^2x+4.sinx.cos^2x}{1+2sinx}=\frac{2cos^2x.\left(1+2sinx\right)}{1+2sinx}=2cos^2x\) ( = vế phải)
Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(tan\alpha< 0,cot\alpha< 0;cos\alpha< 0\).
Vì vậy: \(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{7}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{4}:\dfrac{-\sqrt{7}}{4}=\dfrac{-3}{\sqrt{7}}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-\sqrt{7}}{3}\).
\(A=\dfrac{2tan\alpha-3cot\alpha}{cos\alpha+tan\alpha}\)\(=\dfrac{2.\dfrac{-3}{\sqrt{7}}-3.\dfrac{-\sqrt{7}}{3}}{\dfrac{-\sqrt{7}}{4}+\dfrac{-3}{\sqrt{7}}}\)
\(=\dfrac{\dfrac{-6}{\sqrt{7}}+\sqrt{7}}{\dfrac{-7-12}{4\sqrt{7}}}\)\(=\dfrac{\dfrac{-6+7}{\sqrt{7}}.4\sqrt{7}}{-19}\)\(=\dfrac{\dfrac{1}{\sqrt{7}}.4\sqrt{7}}{-19}=-\dfrac{4}{19}\).
b) \(\dfrac{cos^2\alpha+cot^2\alpha}{tan\alpha-cot\alpha}=\dfrac{\left(-\dfrac{\sqrt{7}}{4}\right)^2+\left(\dfrac{-\sqrt{7}}{3}\right)^2}{\dfrac{-3}{\sqrt{7}}+\dfrac{\sqrt{7}}{3}}\)
\(=\dfrac{\dfrac{7}{16}+\dfrac{7}{9}}{\dfrac{-9+7}{3\sqrt{7}}}=\dfrac{\dfrac{175}{144}}{\dfrac{-2}{3\sqrt{7}}}=\dfrac{-175}{96\sqrt{7}}\).
a)\(sin\left(\alpha+\dfrac{\pi}{2}\right)=cos\left[\dfrac{\pi}{2}-\left(\alpha+\dfrac{\pi}{2}\right)\right]=cos\left(-\alpha\right)=cos\alpha\).
b) \(cos\left(x+\dfrac{\pi}{2}\right)=sin\left[\dfrac{\pi}{2}-\left(x+\dfrac{\pi}{2}\right)\right]=sin\left(-x\right)=-sinx\).
c) \(tan\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{sin\left(\alpha+\dfrac{\pi}{2}\right)}{cos\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{cos\alpha}{-sin\alpha}=-cot\alpha\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{cos\left(\alpha+\dfrac{\pi}{2}\right)}{sin\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{-sin\alpha}{cos\alpha}=-tan\alpha\).
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 = \(\frac{2016}{\sin^2\alpha-\sin\alpha\cos\alpha-\cos^2\alpha}=\frac{2016}{9\cos^2\alpha-3\cos^2\alpha-\cos^2\alpha}\) \(=\frac{2016}{5\cos^2\alpha}=\frac{2016}{5}\times\frac{1}{\cos^2\alpha}=\frac{2016}{5}\times\left(1+\tan^2\alpha\right)\) \(=\frac{2016}{5}\left(1+9\right)=4032\)
a) P = sin2α + sin2α.\(\frac{cos\text{α}}{sin\text{α}}\) + cos2α - cos2α.\(\frac{sin\text{α}}{cos\text{α}}\)
=sin2α + sinα.cosα + cos2α - cosα.sinα
=sin2α + cos2α
=1
Vậy P không phụ thuộc vào α
b) Q= -cos4α(2cos2α -1 -2) +sin4α(1 -2sin2α+2)
= -cos4α(cos2α -2) +sin4α(cos2α +2)
=-cos4α.cos2α +2cos4α +sin4α.cos2α +2sin4α
=cos2α(sin4α -cos4α) +2(sin4α +cos4α)
=cos2α [\(\left(\frac{1-cos^22\text{α}}{2}\right)^2-\left(\frac{1+cos^22\text{α}}{2}\right)^2\)]+2.[\(\left(\frac{1-cos^22\text{α}}{2}\right)^2+ \left(\frac{1+cos^22\text{α}}{2}\right)^2\)]
= -cos2α.cos2α +1+cos22α
= -cos22α +1+cos22α
=1
Vậy Q không phụ thuộc vào α
a) \(0< \alpha< 90^o\)
b) \(90^o< \alpha< 180^o\)
c) \(0< \alpha< 90^o\)
d) \(90^o< \alpha< 180^o\)
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).
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).
Ta có: \(\frac{\tan^2\alpha-\sin^2\alpha}{\cot^2\alpha-\cos^2\alpha}=\frac{\frac{\sin^2\alpha}{\cos^2\alpha}-\sin^2\alpha}{\frac{\cos^2\alpha}{\sin^2\alpha}-\cos^2\alpha}=\frac{\sin^2\alpha\left(\frac{1-\cos^2\alpha}{\cos^2\alpha}\right)}{\cos^2\alpha\left(\frac{1-\sin^2\alpha}{\sin^2\alpha}\right)}=\frac{\sin^2\alpha\left(\frac{\sin^2\alpha}{\cos^2\alpha}\right)}{\cos^2\alpha\left(\frac{\cos^2\alpha}{\sin^2\alpha}\right)}=\frac{\frac{\sin^4\alpha}{\cos^2\alpha}}{\frac{\cos^4\alpha}{\sin^2\alpha}}=\frac{\sin^4\alpha}{\cos^2\alpha}.\frac{\sin^2\alpha}{\cos^4\alpha}=\frac{\sin^6\alpha}{\cos^6\alpha}=\tan^6\alpha\)
P/s: Áp dụng công thức lượng giác cơ bản và liên hệ toán học giữa các hàm là đc :) Em lp 5 vậy nên sai thì thông cảm ạ