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NV
14 tháng 6 2020

\(1+4\left(cosa+cos3a\right)+6cos2a+2cos^22a-1\)

\(=8cos2a.cosa+6cos2a+2cos^22a\)

\(=2cos2a\left(cos2a+4cosa+3\right)\)

\(=2cos2a\left(2cos^2a+4cosa+2\right)\)

\(=4cos2a\left(\left(2cos^2\frac{a}{2}-1\right)^2+2\left(2cos^2\frac{a}{2}-1\right)+1\right)\)

\(=4cos2a\left(4cos^4\frac{a}{2}-4cos^2\frac{a}{2}+1+4cos^2\frac{a}{2}-2+1\right)\)

\(=16cos2a.cos^4\frac{a}{2}\)

NV
8 tháng 6 2020

\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)

\(cos^2\left(a-\frac{\pi}{4}\right)-sin^2\left(a-\frac{\pi}{4}\right)=cos\left(2a-\frac{\pi}{2}\right)\)

\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)

NV
16 tháng 5 2019

\(2\left[\left(sinx+cosx+1\right)\left(sinx+cosx-1\right)\right]^2\)

\(=2\left[\left(sinx+cosx\right)^2-1\right]^2=2\left(sin^2x+cos^2x+2sinx.cosx-1\right)^2\)

\(=2\left(2sinx.cosx\right)^2=2sin^22x=1-cos4x\)

b/ \(\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a+1\right)}{2\left(cos^22a+2cos2a+1\right)}=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}\)

\(\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{4sin^4a}{4cos^4a}=tan^4a\)

c/ \(cos^22x+sin^22x-2sin2x.cos2x+2sin3x.cosx-2sinx.cosx-sin^2x\)

\(=1-sin4x+sin4x+sin2x-sin2x-sin^2x\)

\(=1-sin^2x=cos^2x\)

16 tháng 5 2019

Cảm ơn bạn nhiều lắm! khocroi

NV
14 tháng 6 2020

\(cot^2a+tan^2a=\frac{cos^2a}{sin^2a}+\frac{sin^2a}{cos^2a}=\frac{cos^4a+sin^4a}{sin^2a.cos^2a}=\frac{8\left(\frac{1+cos2a}{2}\right)^2+8\left(\frac{1-cos2a}{2}\right)^2}{2\left(2sina.cosa\right)^2}\)

\(=\frac{2+4cos2a+2cos^22a+2-4cos2a+2cos^22a}{2sin^22a}=\frac{4+4cos^22a}{2sin^22a}\)

\(=\frac{4+4\left(\frac{1+cos4a}{2}\right)}{2\left(\frac{1-cos4a}{2}\right)}=\frac{6+2cos4a}{1-cos4a}\)

NV
10 tháng 4 2019

\(cos^4a+sin^4a-6sin^2a.cos^2a\)

\(=cos^4a+sin^4a-2sin^2a.cos^2a-4sin^2a.cos^2a\)

\(=\left(cos^2a-sin^2a\right)^2-\left(2sina.cosa\right)^2\)

\(=cos^22a-sin^22a\)

\(=cos4a\)

NV
10 tháng 4 2019

1/

\(tanx=\frac{sinx}{cosx}=\frac{sin^2x}{sinx.cosx}=\frac{2sin^2x}{2sinx.cosx}\)

\(=\frac{2\left(\frac{1-cos2x}{2}\right)}{sin2x}=\frac{1-cos2x}{sin2x}\)

2/

\(\frac{sin\left(60-x\right)cos\left(30-x\right)+cos\left(60-x\right)sin\left(30-x\right)}{sin4x}=\frac{sin\left(60-x+30-x\right)}{sin4x}=\frac{sin\left(90-2x\right)}{2sin2x.cos2x}\)

\(=\frac{cos2x}{2sin2x.cos2x}=\frac{1}{2sin2x}\)

3/

\(4cos\left(60+a\right)cos\left(60-a\right)+2sin^2a\)

\(=2\left(cos\left(60+a+60-a\right)+cos\left(60+a-60+a\right)\right)+2sin^2a\)

\(=2cos120+2cos2a+2\left(\frac{1-cos2a}{2}\right)\)

\(=-1+2cos2a+1-cos2a=cos2a\)

NV
7 tháng 6 2020

\(A=\frac{\left(1+cos2x\right)}{cos2x}.tanx=\frac{\left(1+2cos^2x-1\right)}{cos2x}.\frac{sinx}{cosx}=\frac{2cos^2x.sinx}{cos2x.cosx}=\frac{2sinx.cosx}{cos2x}=\frac{sin2x}{cos2x}=tan2x\)

\(B=\frac{1+2sin2a.cos2a-1+2sin^22a}{1+2sin2a.cos2a+2cos^22a-1}=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)

\(C=\frac{2sina.cosa+sina}{1+2cos^2a-1+cosa}=\frac{sina\left(2cosa+1\right)}{cosa\left(2cosa+1\right)}=\frac{sina}{cosa}=tana\)