tính số đo góc
3sina-căn 3 cos a=0
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)
\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)
b.
\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)
c.
\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)
\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)
\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)
\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)
\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)
\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\)
\(\cot=\frac{1}{\tan}=\frac{1}{\frac{2\sqrt{5}}{5}}=\frac{\sqrt{5}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.Ta có :
\(\cot41=\tan49\) ; \(\cot46=\tan44\)
sắp xếp :\(\tan27< \tan44< \tan47< \tan49\)\(\Rightarrow\tan27< \cot46< \tan47< \cot41\)
2.ta có
\(\cos28=\sin62;\cos41=\sin49\)
\(A=\cos^228+\cos^241+\cos^262+\cos^249\)
\(\Rightarrow A=\sin^262+\cos^262+\sin^249+\cos^249\)
\(\Rightarrow A=1+1=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`sin^2 α+cos^2 α =1`
`=> sinα =\sqrt(1-cos^2α)=\sqrt(1-(3/4)^2) = \sqrt7/4`
`=> tanα=(sinα)/(cosα)=(3\sqrt7)/7`
`=> cotα=1/(tanα)=\sqrt7/3`
Đề bài cho cos rồi tính cos làm gì nhỉ =))) Mình tính sin thay vào chỗ đấy nhé.
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\(cos\alpha=\dfrac{3}{4}\Rightarrow cos^2\alpha=\dfrac{9}{16}\)
Mà \(sin^2\alpha+cos^2\alpha=1\)
\(\Rightarrow sin^2\alpha=1-cos^2\alpha=1-\dfrac{9}{16}=\dfrac{7}{16}\)
\(\Rightarrow cos\alpha=\dfrac{\sqrt{7}}{4}\\ \Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{4}}{\dfrac{\sqrt{7}}{4}}=\dfrac{3\sqrt{7}}{7}\\ \Rightarrow cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{\sqrt{7}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(sina=\sqrt{3}cosa\)
\(\Rightarrow\dfrac{sina}{cosa}=\sqrt{3}\)
\(\Rightarrow tana=\sqrt{3}\)
\(\Rightarrow a=60^0\) (nếu góc nhọn)
![](https://rs.olm.vn/images/avt/0.png?1311)
`sin^2 α+cos^2α=1`
`<=> (2/3)^2+cos^2α=1`
`=> cosα= \sqrt5/3`
`=> tan α=(sinα)/(cosα) = (2\sqrt5)/5`
`=> cota = 1/(tanα)=sqrt5/2`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(3sina+4cosa\right)^2+\left(4sina-3cosa\right)^2\)
\(=9sin^2a+24sina.cosa+16cos^2a+16sin^2a-24sina.cosa+9cos^2a\)
\(=33sin^2a+33cos^2a=33\)
\(3sina-\sqrt{3}\cdot cosa=0\)
=>\(3\cdot sina=\sqrt{3}\cdot cosa\)
=>\(\dfrac{sina}{cosa}=\dfrac{\sqrt{3}}{3}=\dfrac{1}{\sqrt{3}}\)
=>\(tana=\dfrac{1}{\sqrt{3}}\)
=>\(a=30^0\)