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NV
7 tháng 6 2020

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

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

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

14 tháng 6 2020

\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)

\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)

\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)

\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)

NV
25 tháng 5 2020

\(A=2sin2x.cos2x.cos4x=sin4x.cos4x=\frac{1}{2}sin8x\)

\(B=sin^4x+cos^6x-6sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x\)

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

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

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

\(D=\frac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}=\frac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\frac{cos3a}{sin3a}=cot3a\)

\(E=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)-\frac{1}{2}+\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)\)

\(=\frac{1}{2}\left[cos\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\right]=-sin\frac{\pi}{4}.sinx=-\frac{\sqrt{2}}{2}sinx\)

20 tháng 8 2018

A = 1

19 tháng 4 2021

Sao có \(cosb\) ở đây??

24 tháng 4 2021

đó là cosa đó anh,em xin lỗi em viết nhầm

b: \(=\left(\cos^2\alpha+\sin^2\alpha\right)^3-3\cos^2\alpha\sin^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)+3\cdot\sin^2\alpha\cdot\cos^2\alpha\)

=1

NV
5 tháng 12 2021

\(cos^4a-sin^4a+1=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1\)

\(=cos^2a-sin^2a+1=cos^2a-sin^2a+sin^2a+cos^2a\)

\(=2cos^2a\)

\(cos^6a+sin^6a+3sin^2a.cos^2a\)

\(=\left(cos^2a+sin^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)

\(=1-3sin^2a.cos^2a.1+3sin^2a.cos^2a\)

\(=1\)

NV
4 tháng 6 2020

\(\frac{cos7a+cos3x-2cos5a}{sin6x-sin4a}=2m\Leftrightarrow\frac{2cos5a.cos2a-2cos5a}{2cos5a.sina}=2m\)

\(\Leftrightarrow\frac{2cos5a\left(cos2a-1\right)}{2cos5a.sina}=2m\Leftrightarrow\frac{cos2a-1}{sina}=2m\)

\(\Leftrightarrow\frac{-2sin^2a}{sina}=2m\Leftrightarrow sina=-m\)

\(\Rightarrow cos2a=1-2sin^2a=1-2m^2\)

3 tháng 7 2021

\(\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)

\(=\left(1+\dfrac{sin^2a}{cos^2a}\right).cos^2a-\left(1+\dfrac{cos^2a}{sin^2a}\right).sin^2a\)

\(=cos^2a+sin^2a-sin^2a-cos^2a=\)\(0\)

Vậy B=0