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a) Do 0 < α < nên sinα > 0, tanα > 0, cotα > 0
sinα =
cotα = ; tanα =
b) π < α < nên sinα < 0, cosα < 0, tanα > 0, cotα > 0
cosα = -√(1 - sin2 α) = -√(1 - 0,49) = -√0,51 ≈ -0,7141
tanα ≈ 0,9802; cotα ≈ 1,0202.
c) < α < π nên sinα > 0, cosα < 0, tanα < 0, cotα < 0
cosα = ≈ -0,4229.
sinα =
cotα = -
d) Vì < α < 2π nên sinα < 0, cosα > 0, tanα < 0, cotα < 0
Ta có: tanα =
cosα =
a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)
a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).
a) \(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=\dfrac{sin2\alpha}{cos2\alpha}:\left(\dfrac{sin4\alpha}{cos4\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\right)\)
\(=\dfrac{sin2\alpha}{cos2\alpha}:\dfrac{sin4\alpha cos2\alpha-sin2\alpha cos4\alpha}{cos4\alpha cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos2\alpha}.\dfrac{cos4\alpha.cos2\alpha}{sin2\alpha}=cos4\alpha\).
b) \(\sqrt{1+sin\alpha}-\sqrt{1-sin\alpha}=\sqrt{sin^2\dfrac{\alpha}{2}+2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)\(-\sqrt{sin^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)
\(=\sqrt{\left(sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right)^2}-\sqrt{\left(sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right)^2}\)
\(=\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên \(0< \alpha< \dfrac{\pi}{4}\).
Trong \(\left(0;\dfrac{\pi}{4}\right)\) thì \(sin\dfrac{\alpha}{2}\) tăng dần từ 0 tới \(\dfrac{\sqrt{2}}{2}\) và \(cos\dfrac{\alpha}{2}\) giảm dần từ 1 tới \(\dfrac{\sqrt{2}}{2}\) nên \(\left|sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right|=-\left(sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right)=cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\).
Vì vậy:
\(\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
\(=sin\dfrac{\alpha}{4}+cos\dfrac{\alpha}{4}-\left(cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\right)=2sin\dfrac{\alpha}{4}\).
a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cot\alpha>0;tan\alpha>0\).
Vì vậy: \(sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{15}}{4}:\dfrac{-1}{4}=\sqrt{15}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\sqrt{15}}\).
b) Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).
a)\(sin\left(\alpha+\dfrac{\pi}{2}\right)=cos\left[\dfrac{\pi}{2}-\left(\alpha+\dfrac{\pi}{2}\right)\right]=cos\left(-\alpha\right)=cos\alpha\).
b) \(cos\left(x+\dfrac{\pi}{2}\right)=sin\left[\dfrac{\pi}{2}-\left(x+\dfrac{\pi}{2}\right)\right]=sin\left(-x\right)=-sinx\).
c) \(tan\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{sin\left(\alpha+\dfrac{\pi}{2}\right)}{cos\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{cos\alpha}{-sin\alpha}=-cot\alpha\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{cos\left(\alpha+\dfrac{\pi}{2}\right)}{sin\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{-sin\alpha}{cos\alpha}=-tan\alpha\).
Lời giải:
Vì \(\frac{\pi}{2}< \alpha< \pi \Rightarrow \frac{\pi}{4}< \frac{\alpha}{2}< \frac{\pi}{2}\)
\(\Rightarrow 1> \sin \frac{\alpha}{2}, \cos \frac{\alpha}{2}>0(*)\)
Mà \(\frac{-5}{9}=\sin \alpha=2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}<0\), điều này hoàn toàn mâu thuẫn với $(*)$
Bạn xem lại đề bài.
a)\(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\sin^2\alpha=1-\cos^2\alpha\)
\(\Rightarrow1-2^2=-3\) \(\Rightarrow\cos=-\sqrt{3}\left(0< \alpha< \dfrac{\pi}{2}\right)\)
b) \(\tan\alpha\times\cot\alpha=1\Rightarrow\tan\alpha=\dfrac{1}{\cot\alpha}\Rightarrow\tan=\dfrac{1}{4}\)
a)Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
\(cos\alpha=2sin\alpha\)(1)
Nếu \(sin\alpha=0\Rightarrow cos\alpha\) (vô lý).
Vì vậy \(sin\alpha\ne0\) . Từ (1) \(\Rightarrow\dfrac{cos\alpha}{sin\alpha}=2\)\(\Leftrightarrow cot\alpha=2\).
Suy ra: \(tan\alpha=\dfrac{1}{2}\).
\(sin\alpha=\sqrt{\dfrac{1}{1+cot^2\alpha}}=\dfrac{1}{\sqrt{3}}\).
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{\dfrac{2}{3}}\).