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Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha,cos\alpha< 0;tan\alpha,cot\alpha< 0\).
\(cos\left(\alpha-\dfrac{\pi}{2}\right)=cos\left(\dfrac{\pi}{2}-\alpha\right)=sin\alpha< 0\).
\(sin\left(\dfrac{\pi}{2}+\alpha\right)=cos\alpha< 0\).
\(tan\left(\dfrac{3\pi}{2}-\alpha\right)=tan\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)\)\(=tan\left(-\dfrac{\pi}{2}-\alpha\right)\)\(=-tan\left(\dfrac{\pi}{2}+\alpha\right)=cot\left(\alpha\right)>0\).
\(cot\left(\alpha+\pi\right)=cot\left(\alpha\right)>0\).
Câu 2:
\(A=2\cdot\dfrac{1}{2}+3\cdot\dfrac{1}{2}+1=1+1+1=3\)
Bài 3:
\(cos^2a=1-\left(\dfrac{12}{13}\right)^2=\dfrac{25}{169}\)
mà cosa>0
nên cosa=5/13
=>tan a=12/5; cot a=5/12
Câu 4: \(sin^2a=1-\dfrac{1}{4}=\dfrac{3}{4}\)
mà sina <0
nên sin a=-căn 3/2
=>tan a=-căn 3
\(A=-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\cdot\left(-\sqrt{3}\right)=-\sqrt{3}\)
Với 0 < α < :
a) sin(α - π) < 0; b) cos( - α) < 0;
c) tan(α + π) > 0; d) cot(α + ) < 0
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
a) \(sin\left(\alpha-\pi\right)=-sin\left(\pi-\alpha\right)=-sin\alpha< 0\).
b) \(cos\left(\dfrac{3\pi}{2}-\alpha\right)=cos\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)=cos\left(-\dfrac{\pi}{2}-\alpha\right)\)
\(=cos\left(\dfrac{\pi}{2}+\alpha\right)=-sin\alpha< 0\).
c) \(tan\left(\alpha+\pi\right)=tan\alpha>0\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=-tan\alpha< 0\).
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) 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 =
\(A=cos3a+2cos\left(\pi-3a\right)sin^2\left(\dfrac{\pi}{4}-1,5a\right)\)
\(=cos3a-2cos3a\dfrac{1-cos\left(\dfrac{\pi}{2}-3a\right)}{2}\)
\(=cos3a-cos3a\left(1-sin3a\right)\)
\(=cos3a-cos3a+cos3asin3a=\dfrac{1}{2}sin6a\)
\(=\dfrac{1}{2}sin\left(6\dfrac{5\pi}{6}\right)=\dfrac{1}{2}sin\left(4\pi+\pi\right)=\dfrac{1}{2}sin\pi=0\)
Vì a=\(\dfrac{5\pi}{6}\) nên: \(3a=\dfrac{5\pi}{2}\) => \(\cos3a=0\)
\(\pi-3a=\pi-\dfrac{5\pi}{2}=\dfrac{-3\pi}{2}\)
=> \(\cos\left(\pi-3a\right)=0\)
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: \(tan\alpha=\dfrac{1}{2}\).
\(sin\alpha=\sqrt{\dfrac{1}{1+cot^2\alpha}}=\dfrac{1}{\sqrt{3}}\).
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{\dfrac{2}{3}}\).
b) Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
Vì vậy:
\(cos\alpha=\sqrt{1-0,6^2}=\dfrac{4}{5}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=0,6:\dfrac{4}{5}=0,75;cot\alpha=1:tan\alpha=\dfrac{4}{3}\).
Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(sin\alpha>0;tan\alpha< 0;cot\alpha< 0\).
\(sin\alpha=\sqrt{1-cos^2\alpha}=\dfrac{\sqrt{51}}{10}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{51}}{10}:\left(-0,7\right)=-\dfrac{\sqrt{51}}{7}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-7}{\sqrt{51}}\).
a) \(A=sin\left(\dfrac{\pi}{4}+x\right)-cos\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-\left(cos\dfrac{\pi}{4}.cosx+sin\dfrac{\pi}{4}.sinx\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\)
\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2}.cosx+\dfrac{\sqrt{2}}{2}.sinx-\dfrac{\sqrt{2}}{2}.cosx-\dfrac{\sqrt{2}}{2}.sinx\)
\(\Leftrightarrow A=0\)
b) \(B=cos\left(\dfrac{\pi}{6}-x\right)-sin\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-\left(sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-sin\dfrac{\pi}{3}.cosx-cos\dfrac{\pi}{3}.sinx\)
\(\Leftrightarrow B=\dfrac{\sqrt{3}}{2}.cosx+\dfrac{1}{2}.sinx-\dfrac{\sqrt{3}}{2}.cosx-\dfrac{1}{2}.sinx\)
\(\Leftrightarrow B=0\)
c) \(C=sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow C=sin^2x+\left(cos\dfrac{\pi}{3}.cosx+sin\dfrac{\pi}{3}.sinx\right).\left(cos\dfrac{\pi}{3}.cosx-sin\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\left(\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right).\left(\dfrac{1}{2}.cosx-\dfrac{\sqrt{3}}{2}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\dfrac{1}{4}.cos^2x-\dfrac{3}{4}.sin^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}.sin^2x+\dfrac{1}{4}.cos^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow C=\dfrac{1}{4}\)
d) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)
\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)
\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)
\(\Leftrightarrow D=tanx.cotx\)
\(\Leftrightarrow D=1\)
a.Ta có : \(x\in\left(\pi;\dfrac{3}{2}\pi\right)\Rightarrow cosx< 0\)
\(cosx=-\sqrt{1-sin^2x}=-\sqrt{1-0,8^2}=-0,6\)
\(tanx=\dfrac{4}{3};cotx=\dfrac{3}{4}\)
b. cos 2x = \(cos^2x-sin^2x=0,6^2-0,8^2=-0,28\)
\(P=2.cos2x=-0,56\)
\(Q=tan\left(2x+\dfrac{\pi}{3}\right)=\dfrac{tan2x+tan\dfrac{\pi}{3}}{1-tan2x.tan\dfrac{\pi}{3}}=\dfrac{tan2x+\sqrt{3}}{1-tan2x.\sqrt{3}}\)
tan 2x = \(\dfrac{2tanx}{1-tan^2x}=\dfrac{\dfrac{2.4}{3}}{1-\left(\dfrac{4}{3}\right)^2}=\dfrac{-24}{7}\)
\(Q=\dfrac{-\dfrac{24}{7}+\sqrt{3}}{1+\dfrac{24}{7}.\sqrt{3}}\) \(=\dfrac{-24+7\sqrt{3}}{7+24\sqrt{3}}\)