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Ta có: với 0 ° < α < 90 ° thì sinx < 1, suy ra sinx – 1 < 0
Ta có: *nếu x = 45 ° thì sinx = cosx, suy ra: sinx – cosx = 0
*nếu x < 45 ° thì cosx = sin( 90 ° – x)
Vì x < 45 ° nên 90 ° – x > 45 ° , suy ra: sinx < sin( 90 ° – x)
Vậy sinx – cosx < 0
*nếu x > 45 ° thì cosx = sin( 90 ° – x)
Vì x > 45 ° nên 90 ° – x < 45 ° , suy ra: sinx > sin( 90 ° – x)
Vậy sinx – cosx > 0.
a) sin = đối / huyền => sinx < 1 => sinx - 1 < 0
b) cos = kề / huyền => cosx < 1 => 1 - cosx > 0
c) sinx - cosx = sinx - sin(90-x)
Nếu x > 90-x hay x > 45 thì sinx - sin(90-x) > 0 hay sinx - cosx > 0
Nếu x < 90-x hay x < 45 thì sinx - sin(90-x) < 0 hay sinx - cosx < 0
d) Tương tự câu c)
1.
\(sin^2x+cos^2x=1\Rightarrow\left(\dfrac{1}{4}\right)^2+cos^2x=1\)
\(\Rightarrow cos^2x=\dfrac{15}{16}\Rightarrow cosx=\dfrac{\sqrt{15}}{4}\)
2.
\(tanx=\dfrac{1}{3}\Rightarrow tan^2x=\dfrac{1}{9}\Rightarrow\dfrac{sin^2x}{cos^2x}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{sin^2x}{1-sin^2x}=\dfrac{1}{9}\Rightarrow9sin^2x=1-sin^2x\)
\(\Rightarrow sin^2x=\dfrac{1}{10}\Rightarrow sinx=\dfrac{\sqrt{10}}{10}\)
ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)
Ta có:
\(\left\{{}\begin{matrix}tanx=3\\sin^2x+cos^2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\9cos^2x+cos^2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\cos^2x=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\cosx=\pm\dfrac{1}{\sqrt{10}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}sinx=\dfrac{3}{\sqrt{10}}\\cosx=\dfrac{1}{\sqrt{10}}\end{matrix}\right.\\\left\{{}\begin{matrix}sinx=-\dfrac{3}{\sqrt{10}}\\cosx=-\dfrac{1}{\sqrt{10}}\end{matrix}\right.\end{matrix}\right.\)
$\sin x=0,6\\\Leftrightarrow \sin^2 x=0,36\\\Rightarrow \cos^2 x=0,64\\\Leftrightarrow \cos x=0,8(x>0)$
\(\tan x=\frac{\sin x}{\cos x}=\frac{3}{5}\Rightarrow\sin x=\frac{3}{5}\cos x\)
\(\Rightarrow N=\frac{\sin x.\cos x}{\sin^2x-\cos^2x}=\frac{\sin x.\cos x}{\left(\sin x-\cos x\right)\left(\sin x+\cos x\right)}\)
\(=\frac{\frac{3}{5}.\cos^2x}{\left(\frac{3}{5}\cos x-\cos x\right)\left(\frac{3}{5}\cos x+\cos x\right)}=\frac{\frac{3}{5}\cos^2x}{\frac{-16}{25}.\cos^2x}=\frac{-15}{16}\)