![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(D=sin^2a-cosa-1=-cos^2a-cosa=-\left(cos^2a+cosa+\frac{1}{4}\right)+\frac{1}{4}\le\frac{1}{4}\)
mình đang học onl nên là rep muộn chút
Đặt \(sina=x;cosa=y\)ta có : \(x^2+y^2=1\)
Khi đó : \(-E=x^2+y^2-x-y-1=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)
\(< =>E\le\frac{3}{2}\)
sai thì thôi nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(cos\alpha=\frac{1}{2}\Leftrightarrow\alpha=\frac{-\pi}{3}\)(vì \(\frac{-\pi}{2}< \alpha< 0\))
\(cot\left(\frac{\pi}{3}-\alpha\right)=cot\left(\frac{2\pi}{3}\right)=\frac{-\sqrt{3}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lơ giải:
\(\frac{25}{16}=(\sin a+\cos a)^2=\sin ^2a+\cos ^2a+2\sin a\cos a=1+2\sin a\cos a\)
\(\Rightarrow \sin a\cos a=\frac{9}{32}\)
\((\sin a-\cos a)^2=(\sin a+\cos a)^2-4\sin a\cos a=\frac{25}{16}-4.\frac{9}{32}=\frac{7}{16}\)
\(\Rightarrow \sin a-\cos a=\pm \frac{\sqrt{7}}{4}\)
Do đó:
\(D=\sin ^3a-\cos ^3a=(\sin a-\cos a)(\sin ^2a+\sin a\cos a+\cos ^2a)\)
\(=(\sin a-\cos a)(1+\sin a\cos a)\)
\(=\pm \frac{\sqrt{7}}{4}(1+\frac{9}{32})=\pm \frac{41\sqrt{7}}{128}\)
Lơ giải:
\(\frac{25}{16}=(\sin a+\cos a)^2=\sin ^2a+\cos ^2a+2\sin a\cos a=1+2\sin a\cos a\)
\(\Rightarrow \sin a\cos a=\frac{9}{32}\)
\((\sin a-\cos a)^2=(\sin a+\cos a)^2-4\sin a\cos a=\frac{25}{16}-4.\frac{9}{32}=\frac{7}{16}\)
\(\Rightarrow \sin a-\cos a=\pm \frac{\sqrt{7}}{4}\)
Do đó:
\(D=\sin ^3a-\cos ^3a=(\sin a-\cos a)(\sin ^2a+\sin a\cos a+\cos ^2a)\)
\(=(\sin a-\cos a)(1+\sin a\cos a)\)
\(=\pm \frac{\sqrt{7}}{4}(1+\frac{9}{32})=\pm \frac{41\sqrt{7}}{128}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(sin^2b+cos^2b=1\)
=>\(sin^2b=1-\dfrac{1}{5}=\dfrac{4}{5}\)
=>\(sinb=\dfrac{2}{\sqrt{5}}\) hoặc \(sinb=-\dfrac{2}{\sqrt{5}}\)
TH1: \(sinb=\dfrac{2}{\sqrt{5}}\)
\(tanb=\dfrac{2}{\sqrt{5}}:\dfrac{1}{\sqrt{5}}=2\)
cot b=1/tanb=1/2
TH2: \(sinb=-\dfrac{2}{\sqrt{5}}\)
\(tanb=\dfrac{-2}{\sqrt{5}}:\dfrac{1}{\sqrt{5}}=-2\)
cot b=1/tan b=-1/2
c: \(1+cot^2y=\dfrac{1}{sin^2y}\)
=>\(\dfrac{1}{sin^2y}=1+2=3\)
=>\(sin^2y=\dfrac{1}{3}\)
=>\(siny=\dfrac{1}{\sqrt{3}}\) hoặc \(siny=-\dfrac{1}{\sqrt{3}}\)
TH1: \(siny=\dfrac{1}{\sqrt{3}}\)
\(coty=\dfrac{cosy}{siny}\)
=>\(cosy=\dfrac{1}{\sqrt{3}}\cdot\left(-\sqrt{2}\right)=\dfrac{-\sqrt{2}}{\sqrt{3}}\)
\(tany=\dfrac{1}{coty}=\dfrac{-1}{\sqrt{2}}\)
TH2: \(siny=-\dfrac{1}{\sqrt{3}}\)
\(cosy=coty\cdot siny=\left(-\sqrt{2}\right)\cdot\dfrac{-1}{\sqrt{3}}=\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{\sqrt{6}}{3}\)
$tany=\frac{1}{coty}=\frac{-1}{\sqrt{2}}$