nếu cos α=\(\dfrac{1}{3}\)và 0<α<\(\dfrac{\Pi}{2}\) thì sin α bằng bao nhiêu?
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5 tháng 7 2023
a: VT=sin^2a(sin^2a+cos^2a)+cos^2a
=sin^2a+cos^2a
=1=VP
b: \(VT=\dfrac{sina+sina\cdot cosa+sina-sina\cdot cosa}{1-cos^2a}=\dfrac{2sina}{sin^2a}=\dfrac{2}{sina}=VP\)
c: \(VT=\dfrac{sin^2a+1+2cosa+cos^2a}{sina\left(1+cosa\right)}\)
\(=\dfrac{2\left(cosa+1\right)}{sina\left(1+cosa\right)}=\dfrac{2}{sina}=VP\)
2 tháng 10 2021
a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)
\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)
2 tháng 10 2021
b: Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)
\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)
=4
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 5 2021
\(\dfrac{3\pi}{2}\le a\le2\pi\Rightarrow3\pi\le2a\le4\pi\)
\(\Rightarrow sin2a\le0\)
\(cos^2a-sin^2a=\dfrac{1}{2}\Leftrightarrow cos2a=\dfrac{1}{2}\)
\(\Rightarrow sin2a=-\sqrt{1-cos^22a}=-\dfrac{\sqrt{3}}{2}\)
Y
24 tháng 8 2023
\(a,VT=cot\alpha+\dfrac{sin\alpha}{1+cos\alpha}\\ =\dfrac{cos\alpha}{sin\alpha}+\dfrac{sin\alpha}{1+cos\alpha}\\ =\dfrac{cos\alpha\left(1+cos\alpha\right)+sin^2\alpha}{sin\alpha\left(1+cos\alpha\right)}\\ =\dfrac{cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha\left(1+cos\alpha\right)}\\ =\dfrac{cos\alpha+1}{sin\alpha\left(1+cos\alpha\right)}\\ =\dfrac{1}{sin\alpha}=VP\left(dpcm\right)\)
\(b,VT=\dfrac{1}{1-sin\alpha}+\dfrac{1}{1+sin\alpha}\\ =\dfrac{1+sin\alpha+1-sin\alpha}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}\\ =\dfrac{2}{1-sin^2\alpha}\\ =\dfrac{2}{cos^2\alpha}=VP\left(dpcm\right)\)
21 tháng 3 2022
\(sin\left(\text{α}-\dfrac{\Pi}{4}\right)-cos\left(\text{α}-\dfrac{\Pi}{4}\right)\)
\(=sin\text{α}.cos\dfrac{\Pi}{4}-cos\text{α}-sin\dfrac{\Pi}{4}-\left(cos\text{α}.cos\dfrac{\Pi}{4}+sin\text{α}.sin\dfrac{\Pi}{4}\right)\)
\(=sin\text{α}.\dfrac{\sqrt{2}}{2}-\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-sin\text{α}.\dfrac{\sqrt{2}}{2}\)
\(=\dfrac{-2\sqrt{2}}{6}\)
\(=\dfrac{-\sqrt{2}}{3}\)
26 tháng 7 2023
a: pi/2<a<pi
=>sin a>0
\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)
\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)
\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)
b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
c: \(sin\left(a-\dfrac{pi}{3}\right)\)
\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)
\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)
d: \(cos\left(a-\dfrac{pi}{6}\right)\)
\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)
\(sin^2\text{α}=1-cos^2\text{α}=1-\left(\dfrac{1}{3}\right)^2=\dfrac{8}{9}\)
vì π<α<\(\dfrac{\Pi}{2}\)⇒sin α=\(\dfrac{2\sqrt{2}}{3}\)