Câu 1 : chứng minh rằng : \(\frac{sina+sin2a+sin3a}{cosa+cos2a+cos3a}=tan2a\)
Câu 2 : chứng minh : \(cos^2\left(\alpha-\frac{\pi}{4}\right)-sin^2\left(\alpha-\frac{\pi}{4}\right)=sin2\alpha\)
Mọi người giúp em giải bài này ạ, em cảm ơn
Bài 1: Rút gọn biểu thức:
A=\(\frac{\sin2x+\sin x}{1+\cos2x+\cos x}\)
B=\(cota\left(\frac{1+\sin^2a}{\cos a}-cosa\right)\)
C=\(\frac{1+\cos x+\cos2x+\cos3x}{2\cos^2x+\cos x-1}\)
D=\(\frac{2\cos\left(\frac{\pi}{2}-x\right)\cdot\sin\left(\frac{\pi}{2}+x\right)\cdot\tan\left(\pi-x\right)}{\cot\left(\frac{\pi}{2}+x\right)\cdot\sin\left(\pi-x\right)}-2\cos x\)
E=\(\cos^2x\cdot\cot^2x+3\cos^2x-\cot^2x+2\sin^2x\)
\(F=\frac{\sin^2x+\sin^2x\tan^2x}{\cos^2x+\cos^2x\tan^2x}\)
\(G=\frac{1+cos2a-cosa}{2sina-sina}\)
H=\(sin^{^{ }4}\left(\frac{\pi}{2}+\alpha\right)-cos^4\left(\frac{3\pi}{2}-\alpha\right)+1\)
Bài 2: chứng minh
a) cho \(\Delta ABCchứngminhsin\frac{A+B}{2}=cos\frac{C}{2}\)
b) chứng minh biểu thức sau độc lập với biến x:
A=\(cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)\)
c)cho \(\Delta\) ABC chứng minh : sin A+sin B+ sin C= \(4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
d)CMR: \(\frac{cos2a}{1+sin2a}=\frac{cosa-sina}{cosa+sina}\)
e) CMR:\(E=\frac{sin\alpha+cos\alpha}{cos^3\alpha}=1+tan\alpha+tan^2\alpha+tan^3\alpha\)
f) CMR \(\Delta\)ABC cân khi và chỉ khi \(sinB=2cosAsinC\)
g) CM: \(\frac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
h)CM: \(\left(cos3x-cosx\right)^2+\left(sin3x-sinx\right)^2=4sin^2x\)
k) CMR trong tam giac ABC ta có: \(sin2A+sin2B+sin2C=4sinA\cdot sinB\cdot sinC\)
Bài 3: giải bất phương trình:
a)\(\frac{\left(1-3x\right)\left(2x^2+1\right)}{-2x^2-3x+5}>0\)
b)\(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\ge0\)
c)\(\frac{\left(3x-2\right)\left(x^2-9\right)}{x^2-4x+4}\le0\)
d)\(\frac{\left(2x^2+3x\right)\left(3-2x\right)}{1-x^2}\ge0\)
e)\(\frac{\left(x^2+2x+1\right)\left(x-1\right)}{3-x^2}\)
f)\(\frac{2x+1}{-x^2+x+6}\ge0\)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
Bài 2:
\(sin\frac{A+B}{2}=sin\left(\frac{180^0-C}{2}\right)=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)
b/
\(A=cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)=cosx+2cos\left(x+\pi\right).cos\frac{\pi}{3}\)
\(=cosx-2cosx.\frac{1}{2}=0\)
c/
\(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
d/ \(\frac{cos2a}{1+sin2a}=\frac{cos^2a-sin^2a}{cos^2a+sin^2a+2sina.cosa}=\frac{\left(cosa-sina\right)\left(cosa+sina\right)}{\left(cosa+sina\right)^2}=\frac{cosa-sina}{cosa+sina}\)
e/
\(E=\frac{sina+cosa}{cos^3a}=\frac{1}{cos^2a}\left(tana+1\right)=\left(1+tan^2a\right)\left(tana+1\right)\)
\(E=tan^3a+tan^2a+tana+1\)
Cho cos \(\alpha\)=\(-\frac{4}{5}\) và \(-\pi< \alpha< \frac{-3}{2}\pi\). Tính \(\sin2\alpha;\)\(\cos2a;\sin\left(\frac{5\pi}{2}-\alpha\right);\tan\left(\alpha+\frac{\pi}{4}\right);\cos\frac{\alpha}{2}\)
--.-- \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ
\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)
\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)
\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)
\(\cos2a=2\cos^2a-1=\frac{7}{25}\)
\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)
\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)
\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)
\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)
\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)
Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)
\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)
Câu 1 : Chọn đẳng thức đúng và chứng minh :
\(A.cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1-sin\alpha}{2}\) \(B.cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1+sin\alpha}{2}\)
\(C.cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1-cos\alpha}{2}\) \(D.cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1+cos\alpha}{2}\)
Theo mk là A đúng
ta có : cos2x = \(\frac{1+cos2x}{2}\)
=> cos2(\(\frac{\pi}{4}\)+\(\frac{\alpha}{2}\))= \(\frac{1+cos\left(\frac{\pi}{2}+\alpha\right)}{2}\) = \(\frac{1-sinx}{2}\)
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) \(\sin^2\left(45^0+\alpha\right)-\sin^2\left(30^0-\alpha\right)-\sin15^0\cos\left(15^0+2\alpha\right)=\sin2\alpha\)
Biết \(sin\alpha=\frac{3}{5}\) và \(\frac{\pi}{2}< \alpha< \pi\). Tính \(P=1-2sin^2\left(\frac{\pi}{4}-\alpha\right)+sin2\alpha+cos\left(\pi-2\alpha\right)-6tan\left(\frac{\pi}{2}-\alpha\right)\)
\(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\)
\(P=1-\left[1-cos\left(\frac{\pi}{2}-2a\right)\right]+sin2a-cos2a-6cota\)
\(=sin2a+sin2a-cos2a-6cota\)
\(=2sin2a-cos2a-6cota\)
\(=4sina.cosa-\left(cos^2a-sin^2a\right)-\frac{6cosa}{sina}\) (thay số và bấm máy)
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 \(0< \alpha,\beta< \frac{\pi}{2}\)và \(\left\{{}\begin{matrix}3\sin^2\alpha+2\sin^2\beta=1\\3\sin2\alpha-2\sin2\beta=0\end{matrix}\right.\). Chứng minh rằng: \(\alpha+2\beta=\frac{\pi}{2}\).
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 \(sin\alpha < 0 \Rightarrow \sin \alpha = - \frac{{2\sqrt 2 }}{3}\).
\(b)\;\,sin2\alpha = 2sin\alpha .cos\alpha = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} = - \frac{{4\sqrt 2 }}{9}\)
\(c)\;cos(\alpha + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6 + 1}}{6}\).
Tính \(\sin \left( {\alpha + \frac{\pi }{6}} \right),\cos \left( {\frac{\pi }{4} - \alpha } \right)\) biết \(\sin \alpha = - \frac{5}{{13}},\pi < \alpha < \frac{{3\pi }}{2}\)
\(\cos \alpha = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = - \frac{{12}}{{13}}\) (vì \(\pi < \alpha < \frac{{3\pi }}{2}\))
\(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)
\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha + \sin \frac{\pi }{4}sin\alpha = \frac{{ - 17\sqrt 2 }}{{26}}\)