Cho \(sinA=\dfrac{3}{5}\).Vậy \(cosA=\)? (biết\(0^o\le A\le90^0\))
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Bài 2:
\(\cos\alpha=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}\)
\(\tan\alpha=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)
\(\cot\alpha=\dfrac{\sqrt{5}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Có: `1+tan^2a=1/(cos^2a)`
`<=> 1+(3/5)^2=1/(cos^2a)`
`=> cosa=\sqrt10/4`
`=> sina = \sqrt(1-cos^2a) = \sqrt6/4`
b) Có: `sin^2a + cos^2a=1`
`<=> sin^2a + (1/4)^2=1`
`=> sina=\sqrt15/4`
`=> tana = (sina)/(cosa) = \sqrt15`
Má ơi,tính sai:
a)\(\left[{}\begin{matrix}cos\alpha=\dfrac{5\sqrt{34}}{34}\\cos\alpha=\dfrac{-5\sqrt{34}}{34}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}sin\alpha=cos\alpha.tan\alpha=\dfrac{3\sqrt{34}}{34}\\sin\alpha=cos\alpha.tan\alpha=\dfrac{-3\sqrt{34}}{34}\end{matrix}\right.\)
b)\(\left[{}\begin{matrix}sin\alpha=\dfrac{\sqrt{15}}{4}\\sin\alpha=\dfrac{-\sqrt{15}}{4}\end{matrix}\right.\)\(\left[{}\begin{matrix}tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\sqrt{15}\\tatn\alpha=-\sqrt{15}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow2\cdot sin\left(\dfrac{a}{2}\right)\cdot cos\left(\dfrac{a}{2}\right)+2\cdot cos^2\left(\dfrac{a}{2}\right)-1-\dfrac{cos\left(\dfrac{a}{2}\right)}{sin\left(\dfrac{a}{2}\right)}=0\)
=>\(2\cdot cos\left(\dfrac{a}{2}\right)\left(sin\left(\dfrac{a}{2}\right)+cos\left(\dfrac{a}{2}\right)\right)=\dfrac{cos\left(\dfrac{a}{2}\right)+sin\left(\dfrac{a}{2}\right)}{sin\left(\dfrac{a}{2}\right)}\)
=>\(\left(cos\left(\dfrac{a}{2}\right)+sin\left(\dfrac{a}{2}\right)\right)\left(sin\left(a\right)-1\right)=0\)
=>cos(a/2)=-sin(a/2) hoặc sin a-1=0
=>cot(a/2)=-1 hoặc sina =1
=>a=-pi/2(loại) hoặc a=pi/2
\(tan\left(a+\dfrac{2013pi}{2}\right)=tan\left(a+\dfrac{pi}{2}\right)=tan\left(\dfrac{pi}{2}+\dfrac{pi}{2}\right)=tanpi=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\cos a-\sin a=\dfrac{1}{5}\\ \Leftrightarrow\left(\cos a-\sin a\right)^2=\dfrac{1}{25}\\ \Leftrightarrow1-2\sin a\cos a=\dfrac{1}{25}\\ \Leftrightarrow2\sin a\cos a=\dfrac{24}{25}\)
Mà \(\cos a=\dfrac{1}{5}+\sin a\)
\(\Leftrightarrow2\sin a\left(\dfrac{1}{5}+\sin a\right)=\dfrac{24}{25}\\ \Leftrightarrow\dfrac{2}{5}\sin a+2\sin^2a-\dfrac{24}{25}=0\\ \Leftrightarrow\left[{}\begin{matrix}\sin a=\dfrac{3}{5}\\\sin a=-\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\cos a=\dfrac{4}{5}\\\cos a=-\dfrac{3}{5}\end{matrix}\right.\\ \Leftrightarrow\cot a=\dfrac{4}{5}\cdot\dfrac{5}{3}=\dfrac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\sin A=1,4-\cos A\)
Thế vào \(\sin^2A+\cos^2A=1\)ta được
\(25\cos^2A-35\cos A+12=0\)
\(\Leftrightarrow\orbr{\begin{cases}\cos A=0,8\\\cos A=0,6\end{cases}\Rightarrow\orbr{\begin{cases}\sin A=0,6\\\sin A=0,8\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}\cot A=\frac{4}{3}\\\cot A=\frac{3}{5}\end{cases}}\)
giả sử tam giác ABC vuông tại A
đặt Ab=c; AC=b; BC=a, \(\widehat{B}\)=A
ta có:
\(sinA+cosA=\frac{b}{a}+\frac{c}{a}=\frac{b+c}{a}=\frac{7}{5}\)
=>b+c=7
=>(b+c)2=b2+2bc+c2=49
=>\(sin^2A+cos^2A=\left(\frac{b}{a}\right)^2+\left(\frac{c}{a}\right)^2=\frac{b^2+c^2}{a^2}=\frac{a^2}{a^2}=\frac{25}{25}\)
=>b2+c2=25
ta có:
(b+c)2-b2-c2=49-25
2bc=24
bc=12
ta có: b.c=12; b+c=7
=> 3.4=4.3=1.12=12.1=2.6=6.2
mà b+c=7=> b=4,c=3 hoặc b=3,c=4
=> cot A= 4/3 hoặc 3/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{sina+cosa}{sina-cosa}=3=>sina+cosa=3sina-3cosa\)
\(=>2sina=4cosa=>sina=2cosa\)
\(=>tana=\dfrac{sina}{cosa}=\dfrac{2cosa}{cosa}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(sin^2\alpha+cos^2\alpha=1\Rightarrow sin^2\alpha+\left(sin\alpha+\dfrac{1}{5}\right)^2=1\)
\(\Rightarrow25sin^2\alpha+5sin\alpha-12=0\\\Rightarrow\left(5sin\alpha-3\right)\left(5sin\alpha+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}sin\alpha=\dfrac{3}{5}\Rightarrow cos\alpha=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\Rightarrow cot\alpha=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\\sin\alpha=-\dfrac{4}{5}\left(loại\right)\end{matrix}\right. \)
Ta có: \(sin^2a+cos^2a=1\)
\(\Rightarrow cos^2a=1-sin^2a=1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\Rightarrow cosa=\dfrac{4}{5}\)(vì \(0^o\le a\le90^o\))