Cho \(sin\)\(\alpha\) = \(-\dfrac{5}{9}\) và \(\dfrac{\pi}{2}< \alpha< \pi\) Tính \(sin\dfrac{\alpha}{2}\); \(cos\dfrac{\alpha}{2}\); \(tan\dfrac{\alpha}{2}\); \(cot\dfrac{\alpha}{2}\).
Những câu hỏi liên quan
tính F=\(\sin^2\dfrac{\pi}{6}+\sin^2\dfrac{2\pi}{6}+...+\sin^2\dfrac{5\pi}{6}+\sin^2\pi\)
2/ biết \(\sin\beta=\dfrac{4}{5},0< \beta< \dfrac{\pi}{2}\) giá trị của biểu thúc a=\(\dfrac{\sqrt{3}\sin\left(\alpha+\beta\right)-\dfrac{4\cos\left(\alpha+\beta\right)}{\sqrt{3}}}{\sin\alpha}\)
Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)
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Chứng minh đẳng thức: \(\dfrac{tan\left(\alpha-\dfrac{\pi}{2}\right).cos\left(\dfrac{3\pi}{2}+\alpha\right)-sin^3\left(\dfrac{7\pi}{2}-\alpha\right)}{cos\left(\alpha-\dfrac{\pi}{2}\right).tan\left(\dfrac{3\pi}{2}+\alpha\right)}=sin^2\alpha\)
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)
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Cho \(\tan\alpha=-3\) và \(\dfrac{\pi}{2}< \alpha< \pi.\)Tính \(\cos\alpha\),\(\sin\alpha\),\(\cot\alpha\)
Lời giải:
$\frac{\pi}{2}< a< \pi$ nên $\sin a>0; \cos a< 0$
$-3=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=-3\cos a$
$\Rightarrow \sin ^2a=9\cos ^2a$
$\Rightarrow 10\sin ^2a=9(\sin ^2a+\cos ^2a)=9$
$\Rightarrow \sin ^2a=\frac{9}{10}$
$\Rightarrow \sin a=\frac{3}{\sqrt{10}}$
$\cos a=\frac{\sin a}{-3}=\frac{-1}{\sqrt{10}}$
$\cot a=\frac{1}{\tan a}=\frac{-1}{3}$
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1. Cho tam giác $ABC$. Chứng minh rằng $\sin ^{2} A+\sin ^{2} B-\sin ^{2} C=2\sin A.\sin B.\cos C$.
2. Chứng minh rằng:
a. $\sin \alpha .\sin \left(\dfrac{\pi }{3} -\alpha \right).\sin \left(\dfrac{\pi }{3} +\alpha \right)=\dfrac{1}{4} \sin 3\alpha $
b. $\sin 5\alpha -2\sin \alpha \left({\rm cos} {\rm 4}\alpha +\cos 2\alpha \right)=\sin \alpha $
Cho \(\pi< \alpha< \dfrac{3\pi}{2}\) và sin a = \(\dfrac{-5}{13}\) . Tính cosa , sin2a , cos2a , và sin\(\dfrac{a}{2}\)
Lời giải:
$\sin ^2a+\cos ^2a=1$
$\cos ^2a=1-\sin ^2a=1-(\frac{-5}{13})^2=\frac{144}{169}$
Vì $\pi < a< \frac{3\pi}{2}$ nên $\cos a< 0$
Do đó: $\cos a=-\sqrt{\frac{144}{169}}=\frac{-12}{13}$
$\sin 2a=2\sin a\cos a=2.\frac{-5}{13}.\frac{-12}{13}=\frac{120}{169}$
$\cos 2a=\cos ^2a-\sin ^2a=2\cos ^2a-1=2.\frac{144}{169}-1=\frac{119}{169}$
$\cos a=\cos ^2\frac{a}{2}-\sin ^2\frac{a}{2}$
$=1-2\sin ^2\frac{a}{2}$
$\Leftrightarrow \frac{-12}{13}=1-2\sin ^2\frac{a}{2}$
$\Rightarrow \sin ^2\frac{a}{2}=\frac{25}{26}$
Vì $\pi < a< \frac{3\pi}{2}$ nên $\sin \frac{a}{2}>0$
$\Rightarrow \sin \frac{a}{2}=\frac{5}{\sqrt{26}}$
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Rút gọn cac biểu thức sau:Asinleft(dfrac{5pi}{2}-alpharight)+cosleft(13pi+alpharight)-3sinleft(alpha-5piright)Bsinleft(x+dfrac{85pi}{2}right)+cosleft(2017pi+xright)+sin^2left(33pi+xright)+sin^2left(x-dfrac{5pi}{2}right)+cosleft(x+dfrac{3pi}{2}right)Csinleft(x+dfrac{2017pi}{2}right)+2sin^2left(x-piright)+cosleft(x+2019piright)+cos2x+sinleft(x+dfrac{9pi}{2}right)
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Rút gọn cac biểu thức sau:
\(A=sin\left(\dfrac{5\pi}{2}-\alpha\right)+cos\left(13\pi+\alpha\right)-3sin\left(\alpha-5\pi\right)\)
\(B=sin\left(x+\dfrac{85\pi}{2}\right)+cos\left(2017\pi+x\right)+sin^2\left(33\pi+x\right)+sin^2\left(x-\dfrac{5\pi}{2}\right)+cos\left(x+\dfrac{3\pi}{2}\right)\)\(C=sin\left(x+\dfrac{2017\pi}{2}\right)+2sin^2\left(x-\pi\right)+cos\left(x+2019\pi\right)+cos2x+sin\left(x+\dfrac{9\pi}{2}\right)\)
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
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1.chứng minh hệ thức: dfrac{sinalpha+sin3alpha+sin5alpha}{cosalpha+cos3alpha+cos5alpha}tan3alpha2.rút gọn biểu thức: dfrac{1+sin4alpha-cos4alpha}{1+cos4alpha+sin4alpha}3. Tính 96sqrt{3}sindfrac{pi}{48}cosdfrac{pi}{48}cosdfrac{pi}{24}cosdfrac{pi}{12}cosdfrac{pi}{6}4. chứng minh rằng trong một △ABC ta có: tanA + tanB + tanC tanA tanB tanC (A,B,C cùng khác dfrac{pi}{2})
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1.\(\)chứng minh hệ thức: \(\dfrac{sin\alpha+sin3\alpha+sin5\alpha}{cos\alpha+cos3\alpha+cos5\alpha}=tan3\alpha\)
2.rút gọn biểu thức: \(\dfrac{1+sin4\alpha-cos4\alpha}{1+cos4\alpha+sin4\alpha}\)
3. Tính \(96\sqrt{3}sin\dfrac{\pi}{48}cos\dfrac{\pi}{48}cos\dfrac{\pi}{24}cos\dfrac{\pi}{12}cos\dfrac{\pi}{6}\)
4. chứng minh rằng trong một △ABC ta có:
tanA + tanB + tanC = tanA tanB tanC (A,B,C cùng khác \(\dfrac{\pi}{2}\))
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
\(\dfrac{1+sin4a-cos4a}{1+sin4a+cos4a}=\dfrac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin2a.cos2a+2cos^22a-1}=\dfrac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\dfrac{sin2a}{cos2a}=tan2a\)
\(96\sqrt{3}sin\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=48\sqrt{3}sin\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=24\sqrt{3}sin\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=12\sqrt{3}sin\left(\dfrac{\pi}{6}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=6\sqrt{3}sin\left(\dfrac{\pi}{3}\right)=6\sqrt{3}.\dfrac{\sqrt{3}}{2}=9\)
\(A+B+C=\pi\Rightarrow A+B=\pi-C\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
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Rút gọn các biểu thức :
a) dfrac{2sin2alpha-sin4alpha}{2sin2alpha+sin4alpha}
b) tanalphaleft(dfrac{1+cos^2alpha}{sinalpha}-sinalpharight)
c) dfrac{sinleft(dfrac{pi}{4}-alpharight)+cosleft(dfrac{pi}{4}-alpharight)}{sinleft(dfrac{pi}{4}-alpharight)-cosleft(dfrac{pi}{4}-alpharight)}
d) dfrac{sin5alpha-sin3alpha}{2cos4alpha}
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Rút gọn các biểu thức :
a) \(\dfrac{2\sin2\alpha-\sin4\alpha}{2\sin2\alpha+\sin4\alpha}\)
b) \(\tan\alpha\left(\dfrac{1+\cos^2\alpha}{\sin\alpha}-\sin\alpha\right)\)
c) \(\dfrac{\sin\left(\dfrac{\pi}{4}-\alpha\right)+\cos\left(\dfrac{\pi}{4}-\alpha\right)}{\sin\left(\dfrac{\pi}{4}-\alpha\right)-\cos\left(\dfrac{\pi}{4}-\alpha\right)}\)
d) \(\dfrac{\sin5\alpha-\sin3\alpha}{2\cos4\alpha}\)
Cho \(\alpha\) , \(\beta\in\left(0;\dfrac{\pi}{2}\right)\) và sin \(\alpha\) = \(\dfrac{1}{\sqrt{5}}\) ; Cos \(\alpha\) = \(\dfrac{1}{\sqrt{10}}\) . Tính Cos \(\left(\alpha+\beta\right)\)
Kiểm tra lại đề bài, \(cosa=\dfrac{1}{\sqrt{10}}\) hay \(cos\beta=\dfrac{1}{\sqrt{10}}\)?
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1. Cho 2cosleft(alpha+betaright)cosalphacosleft(pi+betaright)Tính Adfrac{1}{2sin^2alpha+3cos^2alpha}+dfrac{1}{2sin^2beta+3cos^2beta}2. Rút gọn: a) A4cosdfrac{2x}{3}cosdfrac{pi+2x}{3}cosdfrac{pi-2x}{3}b) Bdfrac{sinleft(a-bright).sinleft(a+bright)}{cos^2a.sin^2b}-tan^2a.cot^2b3. Chứng minh rằng: Nếu 2tan atanleft(a+bright) thì:a) sin bsin a.cosleft(a+bright)b) 3sin bsinleft(2a+bright)
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1. Cho \(2\cos\left(\alpha+\beta\right)=\cos\alpha\cos\left(\pi+\beta\right)\)
Tính \(A=\dfrac{1}{2\sin^2\alpha+3\cos^2\alpha}+\dfrac{1}{2\sin^2\beta+3\cos^2\beta}\)
2. Rút gọn: a) \(A=4\cos\dfrac{2x}{3}\cos\dfrac{\pi+2x}{3}\cos\dfrac{\pi-2x}{3}\)
b) \(B=\dfrac{\sin\left(a-b\right).\sin\left(a+b\right)}{\cos^2a.\sin^2b}-\tan^2a.\cot^2b\)
3. Chứng minh rằng: Nếu \(2\tan a=\tan\left(a+b\right)\) thì:
a) \(\sin b=\sin a.\cos\left(a+b\right)\)
b) \(3\sin b=\sin\left(2a+b\right)\)
1.
\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)
\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)
\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)
\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)
\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)
\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)
\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)
\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)
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2.
\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)
\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x\)
\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)
\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)
\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)
\(=tan^2a.cot^2b-2\)
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3.
\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b\right).cosa-cos\left(a+b\right)sina\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b-a\right)\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sinb\)
b.
\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)
\(\Leftrightarrow sin\left(2a+b\right)+sin\left(-b\right)=\dfrac{1}{2}sin\left(2a+b\right)+\dfrac{1}{2}sinb\)
\(\Leftrightarrow\dfrac{1}{2}sin\left(2a+b\right)=\dfrac{3}{2}sinb\)
\(\Leftrightarrow sin\left(2a+b\right)=3sinb\)
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