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a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)
- \(cosx=\frac{1}{2}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)
- \(cosx=\frac{-1}{2}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)
b) Bạn làm tương tự câu a) nha.
a) \(\frac{1}{cos^2x}=1+tan^2x=1+\frac{9}{16}=\frac{25}{16}\)
\(\Leftrightarrow cos^2x=\frac{16}{25}\Leftrightarrow\orbr{\begin{cases}cosx=\frac{4}{5}\\cosx=\frac{-4}{5}\end{cases}}\)
- \(cosx=\frac{4}{5}\):
\(sinx=cosxtanx=\frac{4}{5}.\frac{3}{4}=\frac{3}{5}\)
\(cotx=\frac{1}{tanx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\).
- \(cosx=\frac{-4}{5}\):
\(sinx=cosxtanx=\frac{-4}{5}.\frac{3}{4}=\frac{-3}{5}\)
\(cotx=\frac{1}{tanx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\).
b) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{49}{625}=\frac{576}{625}\)
\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{24}{25}\\cosx=-\frac{24}{25}\end{cases}}\)
- \(cosx=\frac{24}{25}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{7}{24}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\frac{7}{24}}=\frac{24}{7}\)
- \(cosx=\frac{-24}{25}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{7}{25}}{\frac{-24}{25}}=-\frac{7}{24}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\frac{7}{24}}=\frac{-24}{7}\)
1:
a: sin a=căn 3/2
\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)
\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)
cot a=1/tan a=1/căn 3
b: \(tana=2\)
=>cot a=1/tan a=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=5\)
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)
c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)
tan a=5/13:12/13=5/12
cot a=1:5/12=12/5
Ta có CT: sin^2a+cos^2a=1 => cos^2a=1-sin^2a=0,64 => cosa=0,8
Áp dụng 2 công thức sau và làm tương tự:
tana=sina/cosa
cota=cosa/sina
A,xem lại đề
B\(=sin^6x+cos^6x+3sin^2x.cos^2x\)
\(=\left(sin^2x\right)^3+3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\left(cos^2x\right)^3\)
\(=\left(sin^2+cos^2x\right)^3\)
\(=1\)
a) Sửa đề: \(A=\cot48^0\cdot\cot42^0+\tan60^0\)
Ta có: \(A=\cot48^0\cdot\cot42^0+\tan60^0\)
\(=\cot48^0\cdot\tan48^0+\tan60^0\)
\(=1+\sqrt{3}\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)
ta có: tan x.cot x=1=>cot x=2/5
1+tan^2 x=1/cos^2 x=>cos=2/\(\sqrt{29}\)
tan x=sin x/cos x=>sin x=\(\frac{5\sqrt{29}}{4}\)