Rút gọn
A=cot\(a\) +\(\dfrac{sin\alpha}{1+cos\alpha}\) (0o<\(\alpha\)<900)
K
Khách
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18 tháng 7 2022
a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alphaA=tanα+tanα1tanα+3tanα1=tan2α+1tan2α+3=cos2α1cos2α1+2=1+2cos2α Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}A=1+2⋅169=817.
b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}B=cos3αsin3α+cos3α3cos3α+cos3α2sinαcos3αsinα−cos3αcosα=tan3α+3+2tanα(tan2α+1)tanα(tan2α+1)−(tan2α+1).
Suy ra B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}B=22+3+22(2+1)2(2+1)−(2+1)=3+823(2−1).
BT
10 tháng 5 2017
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.
BT
10 tháng 5 2017
Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(tan\alpha< 0,cot\alpha< 0;cos\alpha< 0\).
Vì vậy: \(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{7}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{4}:\dfrac{-\sqrt{7}}{4}=\dfrac{-3}{\sqrt{7}}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-\sqrt{7}}{3}\).
\(A=\dfrac{2tan\alpha-3cot\alpha}{cos\alpha+tan\alpha}\)\(=\dfrac{2.\dfrac{-3}{\sqrt{7}}-3.\dfrac{-\sqrt{7}}{3}}{\dfrac{-\sqrt{7}}{4}+\dfrac{-3}{\sqrt{7}}}\)
\(=\dfrac{\dfrac{-6}{\sqrt{7}}+\sqrt{7}}{\dfrac{-7-12}{4\sqrt{7}}}\)\(=\dfrac{\dfrac{-6+7}{\sqrt{7}}.4\sqrt{7}}{-19}\)\(=\dfrac{\dfrac{1}{\sqrt{7}}.4\sqrt{7}}{-19}=-\dfrac{4}{19}\).
BT
10 tháng 5 2017
b) \(\dfrac{cos^2\alpha+cot^2\alpha}{tan\alpha-cot\alpha}=\dfrac{\left(-\dfrac{\sqrt{7}}{4}\right)^2+\left(\dfrac{-\sqrt{7}}{3}\right)^2}{\dfrac{-3}{\sqrt{7}}+\dfrac{\sqrt{7}}{3}}\)
\(=\dfrac{\dfrac{7}{16}+\dfrac{7}{9}}{\dfrac{-9+7}{3\sqrt{7}}}=\dfrac{\dfrac{175}{144}}{\dfrac{-2}{3\sqrt{7}}}=\dfrac{-175}{96\sqrt{7}}\).
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 1 2021
\(A=\dfrac{cos^2a-sin^2a}{\dfrac{cos^2a}{sin^2a}-\dfrac{sin^2a}{cos^2a}}-cos^2a=\dfrac{cos^2a.sin^2a\left(cos^2a-sin^2a\right)}{\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)}-cos^2a\)
\(=cos^2a.sin^2a-cos^2a=cos^2a\left(sin^2a-1\right)=-cos^4a\)
\(B=\sqrt{\left(1-cos^2a\right)^2+6cos^2a+3cos^4a}+\sqrt{\left(1-sin^2a\right)^2+6sin^2a+3sin^4a}\)
\(=\sqrt{4cos^4a+4cos^2a+1}+\sqrt{4sin^4a+4sin^2a+1}\)
\(=\sqrt{\left(2cos^2a+1\right)^2}+\sqrt{\left(2sin^2a+1\right)^2}\)
\(=2\left(sin^2a+cos^2a\right)+2=4\)
8 tháng 6 2020
vậy thì kết quả là
\(\sin2\alpha=-0.96\)
\(\)còn \(\cos\left(\alpha+\frac{\pi}{6}\right)\) thì đúng vì -(-0.8) mà sorry thiếu ngủ hôm qua -_-
BT
11 tháng 5 2017
a) \(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=\dfrac{sin2\alpha}{cos2\alpha}:\left(\dfrac{sin4\alpha}{cos4\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\right)\)
\(=\dfrac{sin2\alpha}{cos2\alpha}:\dfrac{sin4\alpha cos2\alpha-sin2\alpha cos4\alpha}{cos4\alpha cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos2\alpha}.\dfrac{cos4\alpha.cos2\alpha}{sin2\alpha}=cos4\alpha\).
BT
11 tháng 5 2017
b) \(\sqrt{1+sin\alpha}-\sqrt{1-sin\alpha}=\sqrt{sin^2\dfrac{\alpha}{2}+2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)\(-\sqrt{sin^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)
\(=\sqrt{\left(sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right)^2}-\sqrt{\left(sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right)^2}\)
\(=\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên \(0< \alpha< \dfrac{\pi}{4}\).
Trong \(\left(0;\dfrac{\pi}{4}\right)\) thì \(sin\dfrac{\alpha}{2}\) tăng dần từ 0 tới \(\dfrac{\sqrt{2}}{2}\) và \(cos\dfrac{\alpha}{2}\) giảm dần từ 1 tới \(\dfrac{\sqrt{2}}{2}\) nên \(\left|sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right|=-\left(sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right)=cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\).
Vì vậy:
\(\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
\(=sin\dfrac{\alpha}{4}+cos\dfrac{\alpha}{4}-\left(cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\right)=2sin\dfrac{\alpha}{4}\).
AH
Akai Haruma
Giáo viên
1 tháng 10 2018
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
AH
Akai Haruma
Giáo viên
1 tháng 10 2018
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
18 tháng 7 2022
a) Vì 90^{\circ}<\alpha<180^{\circ}90∘<α<180∘ nên \cos \alpha<0cosα<0 mặt khác \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 suy ra \cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-\sqrt{1-\dfrac{1}{9}}=-\dfrac{2 \sqrt{2}}{3}cosα=−1−sin2α=−1−91=−322.
Do đó \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{\dfrac{1}{3}}{-\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}tanα=cosαsinα=−32231=−221.
b) Vì \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 nên \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}sinα=1−cos2α=1−94=35 và \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{2}{3}}{\dfrac{\sqrt{5}}{3}}=-\dfrac{2}{\sqrt{5}}cotα=sinαcosα=35−32=−52.
c) Vì \tan \gamma=-2 \sqrt{2}<0 \Rightarrow \cos \alpha<0tanγ=−22<0⇒cosα<0 mặt khác \tan ^{2} \alpha+1=\dfrac{1}{\cos ^{2} \alpha}tan2α+1=cos2α1 nên \cos \alpha=-\sqrt{\dfrac{1}{\tan ^{2}+1}}=-\sqrt{\dfrac{1}{8+1}}=-\dfrac{1}{3}cosα=−tan2+11=−8+11=−31.
Ta có \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \alpha=\tan \alpha \cdot \cos \alpha=-2 \sqrt{2} \cdot\left(-\dfrac{1}{3}\right)=\dfrac{2 \sqrt{2}}{3}tanα=cosαsinα⇒sinα=tanα⋅cosα=−22⋅(−31)=322 \Rightarrow \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{1}{3}}{\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}⇒cotα=sinαcosα=322−31=−221.
ta có : \(A=cot\alpha+\dfrac{sin\alpha}{1+cos\alpha}=\dfrac{cos\alpha}{sin\alpha}+\dfrac{sin\alpha}{1+cos\alpha}\)
\(=\dfrac{cos\alpha\left(1+cos\alpha\right)+sin^2\alpha}{sin\alpha\left(1+cos\alpha\right)}=\dfrac{cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha\left(1+cos\alpha\right)}\)
\(=\dfrac{1+cos\alpha}{sin\alpha\left(1+cos\alpha\right)}=\dfrac{1}{sin\alpha}\)