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)\(\frac{2x+1}{-x^2+x+6}\ge0\)
\(\frac{1+sin^2x}{1-sin^2x}=\frac{1+sin^2x}{cos^2x}=\frac{1}{cos^2x}+\frac{sin^2x}{cos^2x}=1+tan^2x+tan^2x=1+2tan^2x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}-sina.cosa=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}-sina.cosa\)
\(=sin^2a+cos^2a+sina.cosa-sina.cosa=1\)
\(\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cosx.cos2x}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(\frac{1-2sin^2a}{cosa+sina}+\frac{2cos^2a-1}{cosa-sina}=\frac{cos^2a-sin^2a}{cosa+sina}+\frac{cos^2a-sin^2a}{cosa-sina}\)
\(=\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa+sina}+\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa-sina}=cosa-sina+cosa+sina=2cosa\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)