Cho \(\cos\alpha\)=\(\frac{1}{3}\).Tính P= \(\frac{sin3a-sina}{sin2a}\)
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NV
Nguyễn Việt Lâm
Giáo viên
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
Nguyễn Việt Lâm
Giáo viên
13 tháng 6 2020
\(A=\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(B=\frac{sin^2a\left(1+tan^2a\right)}{cos^2a\left(1+cot^2a\right)}=\frac{sin^2a.\frac{1}{cos^2a}}{cos^2a.\frac{1}{sin^2a}}=\frac{sin^4a}{cos^4a}=tan^4a\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2019
\(Q=2\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+2=2\left(\frac{sina.cosa+sina}{cosa\left(cosa+1\right)}\right)^2+2\)
\(=2\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+2=2tan^2a+2=2\left(1+tan^2a\right)=\frac{2}{cos^2a}\)
\(A=\frac{sin2a+2cos4a.sina}{cos4a+cosa}=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)
NQ
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 5 2020
\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)
\(=\frac{1-2sin^2\frac{a}{2}-1}{2cos^2\frac{a}{2}-1+1}=\frac{-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=-tan^2\frac{a}{2}\)
\(\frac{sin^4x-sin^2x+cos^2x}{cos^4x-cos^2x+sin^2x}=\frac{sin^2x\left(sin^2x-1\right)+cos^2x}{cos^2x\left(cos^2x-1\right)+sin^2x}=\frac{-sin^2x.cos^2x+cos^2x}{-cos^2x.sin^2x+sin^2x}\)
\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 5 2019
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 5 2019
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 5 2020
\(A=2sin2x.cos2x.cos4x=sin4x.cos4x=\frac{1}{2}sin8x\)
\(B=sin^4x+cos^6x-6sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x\)
\(=1-2\left(2sinx.cosx\right)^2=1-2sin^22x=cos4x\)
\(C=\frac{cos2a+1-2cos^22a}{2sin2a.cos2a+sin2a}=\frac{\left(1-cos2a\right)\left(2cos2a+1\right)}{sin2a\left(2cos2a+1\right)}=\frac{1-cos2a}{sin2a}\)
\(=\frac{1-\left(1-2sin^2a\right)}{2sina.cosa}=\frac{2sin^2a}{2sina.cosa}=\frac{sina}{cosa}=tana\)
\(D=\frac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}=\frac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\frac{cos3a}{sin3a}=cot3a\)
\(E=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)-\frac{1}{2}+\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)\)
\(=\frac{1}{2}\left[cos\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\right]=-sin\frac{\pi}{4}.sinx=-\frac{\sqrt{2}}{2}sinx\)