Cho \(5\sin2\alpha-6\cos\alpha=0\) và \(0< \alpha< \dfrac{\pi}{2}\)
Tính A = \(\cos(\dfrac{\pi}{2}-\alpha)+\sin\left(2017\pi-\alpha\right)-\cot(2018\pi+\alpha)\)
Đơn giản các biểu thức sau:
G = \(cos\left(\alpha-5\pi\right)+sin\left(-\dfrac{3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
H = \(cot\left(\alpha-2\pi\right).cos\left(\alpha-\dfrac{3\pi}{2}\right)+cos\left(\alpha-6\pi\right)-2sin\left(\alpha-\pi\right)\)
bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)
\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\) \(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)
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) \(\dfrac{\sin5\alpha-\sin3\alpha}{2\cos4\alpha}\)
Chứng minh các hệ thức sau :
a) \(\sin\alpha+\sin\left(\alpha+\dfrac{14}{3}\pi\right)+\sin\left(\alpha-\dfrac{8}{3}\pi\right)=0\)
b) \(\dfrac{\sin4a}{1+\cos4a}.\dfrac{\cos2a}{1+\cos2a}=\cot\left(\dfrac{3}{2}\pi-a\right)\)
c) \(\left(\cos a-\cos b\right)^2-\left(\sin a-\sin b\right)^2=-4\sin^2\dfrac{a-b}{2}\cos\left(a+b\right)\)
d) \(\sin^2\left(45^0+\alpha\right)-\sin^2\left(30^0-\alpha\right)-\sin15^0\cos\left(15^0+2\alpha\right)=\sin2\alpha\)
Chung minh rang voi moi goc luong giac α lam cho bieu thuc xac dinh thi
a) \(\dfrac{1-sin2\alpha}{1+sin2\alpha}\)=cot\(^2\)(\(\dfrac{\pi}{4}\)+α) b) \(\dfrac{sin\alpha+sin\beta cos\left(\alpha+\beta\right)}{cos\alpha-sin\beta sin\left(\alpha+\beta\right)}\)=tan\(\left(\alpha+\beta\right)\).
a, \(\dfrac{1-sin2a}{1+sin2a}\)
\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)
\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)
\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)
b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)
\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)
\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)
\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)
\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)
\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)
Cho \(cos\alpha = \frac{1}{3}\) và \( - \frac{\pi }{2} < \alpha < 0\). Tính
\(\begin{array}{l}a)\;sin\alpha \\b)\;sin2\alpha \\c)\;cos\left( {\alpha + \frac{\pi }{3}} \right)\end{array}\)
a, Ta có: \({\sin ^2}x + co{s^2}x = 1\)
\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha + {\left( {\frac{1}{3}} \right)^2} = 1\\ \Leftrightarrow \sin \alpha = \pm \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} = \pm \frac{{2\sqrt 2 }}{3}\end{array}\)
Vì \( - \frac{\pi }{2} < \alpha < 0\) nên \(sin\alpha < 0 \Rightarrow \sin \alpha = - \frac{{2\sqrt 2 }}{3}\).
\(b)\;\,sin2\alpha = 2sin\alpha .cos\alpha = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} = - \frac{{4\sqrt 2 }}{9}\)
\(c)\;cos(\alpha + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6 + 1}}{6}\).
Rút gọn cac biểu thức sau:
\(A=sin\left(\dfrac{5\pi}{2}-\alpha\right)+cos\left(13\pi+\alpha\right)-3sin\left(\alpha-5\pi\right)\)
\(B=sin\left(x+\dfrac{85\pi}{2}\right)+cos\left(2017\pi+x\right)+sin^2\left(33\pi+x\right)+sin^2\left(x-\dfrac{5\pi}{2}\right)+cos\left(x+\dfrac{3\pi}{2}\right)\)\(C=sin\left(x+\dfrac{2017\pi}{2}\right)+2sin^2\left(x-\pi\right)+cos\left(x+2019\pi\right)+cos2x+sin\left(x+\dfrac{9\pi}{2}\right)\)
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
Chứng minh đẳng thức: \(\dfrac{tan\left(\alpha-\dfrac{\pi}{2}\right).cos\left(\dfrac{3\pi}{2}+\alpha\right)-sin^3\left(\dfrac{7\pi}{2}-\alpha\right)}{cos\left(\alpha-\dfrac{\pi}{2}\right).tan\left(\dfrac{3\pi}{2}+\alpha\right)}=sin^2\alpha\)
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)
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 \(cos\beta=\dfrac{1}{\sqrt{10}}\)?
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 $\sin a>0; \cos a< 0$
$-3=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=-3\cos a$
$\Rightarrow \sin ^2a=9\cos ^2a$
$\Rightarrow 10\sin ^2a=9(\sin ^2a+\cos ^2a)=9$
$\Rightarrow \sin ^2a=\frac{9}{10}$
$\Rightarrow \sin a=\frac{3}{\sqrt{10}}$
$\cos a=\frac{\sin a}{-3}=\frac{-1}{\sqrt{10}}$
$\cot a=\frac{1}{\tan a}=\frac{-1}{3}$