Thu gọn biểu thức:
A=sin2x+sin4x+sin6x+sin8x
B=\(\frac{sin2a-2sin4a+sin6a}{1+cos2a+cos4a}\)
C=\(\frac{cos5a.cos3a+sin7a.sina}{sin6a+sin2a}\)
Những câu hỏi liên quan
7 tháng 6 2020 lúc 18:12
Rút gọn
\(A=\left(\frac{1}{cos2x}+1\right).tanx\)
\(B=\frac{1+sin4a-cos4a}{1+sin4a+cos4a}\)
\(C=\frac{sin2a+sina}{1+cos2a+cosa}\)
\(A=\frac{\left(1+cos2x\right)}{cos2x}.tanx=\frac{\left(1+2cos^2x-1\right)}{cos2x}.\frac{sinx}{cosx}=\frac{2cos^2x.sinx}{cos2x.cosx}=\frac{2sinx.cosx}{cos2x}=\frac{sin2x}{cos2x}=tan2x\)
\(B=\frac{1+2sin2a.cos2a-1+2sin^22a}{1+2sin2a.cos2a+2cos^22a-1}=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(C=\frac{2sina.cosa+sina}{1+2cos^2a-1+cosa}=\frac{sina\left(2cosa+1\right)}{cosa\left(2cosa+1\right)}=\frac{sina}{cosa}=tana\)
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chứng minh tam giác ABC đều
a) sin2A+sin2B+sin2C=sinA+sinB+sinC
b) sin6A + sin6B + sin 6C = 0
c) sin A + sinB + sinC = \(cos\frac{A}{2}+cos\frac{B}{2}+cos\frac{C}{2}\)
d) \(sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}=\frac{1}{8}\)
9. Rút gọn các biểu thức sau
A= cos7x - cos8x - cos9x + cos10x / sin7x - sin8x - sin9x + sin10x
B = sin2x + 2sin3x + sin4x / sin3x +2sin4x + sin5x
C= 1+cosx + cos2x + cos3x / cosx + 2cos^2 . x -1
D = sin4x + sin5x + sin6x / cos4x + cos5x + cos6x
\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)
\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)
\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)
\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)
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Rút gọn biểu thức sau:
A4sinx*cosx*cos2x*cos4x
Bcos^4x -6cos^x*sin^2x+sim^4x
Cfrac{text{cos2a-cos4a}}{sin4a+sin2a}
Dfrac{text{cosa+cos3a+cos5a}}{sina+sin3a+sin5a}
Esin^2(frac{pi}{8}+frac{x}{2})-sin^2(frac{pi}{8}-frac{x}{2})
Ffrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}
Đọc tiếp
Rút gọn biểu thức sau:
A=4sinx*cosx*cos2x*cos4x
B=cos^4x -6cos^x*sin^2x+sim^4x
C=\(\frac{\text{cos2a-cos4a}}{sin4a+sin2a}\)
D=\(\frac{\text{cosa+cos3a+cos5a}}{sina+sin3a+sin5a}\)
E=sin^2(\(\frac{\pi}{8}\)+\(\frac{x}{2}\))-sin^2(\(\frac{\pi}{8}\)-\(\frac{x}{2}\))
F=\(\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)
\(A=2sin2x.cos2x.cos4x=sin4x.cos4x=\frac{1}{2}sin8x\)
\(B=sin^4x+cos^6x-6sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x\)
\(=1-2\left(2sinx.cosx\right)^2=1-2sin^22x=cos4x\)
\(C=\frac{cos2a+1-2cos^22a}{2sin2a.cos2a+sin2a}=\frac{\left(1-cos2a\right)\left(2cos2a+1\right)}{sin2a\left(2cos2a+1\right)}=\frac{1-cos2a}{sin2a}\)
\(=\frac{1-\left(1-2sin^2a\right)}{2sina.cosa}=\frac{2sin^2a}{2sina.cosa}=\frac{sina}{cosa}=tana\)
\(D=\frac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}=\frac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\frac{cos3a}{sin3a}=cot3a\)
\(E=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)-\frac{1}{2}+\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)\)
\(=\frac{1}{2}\left[cos\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\right]=-sin\frac{\pi}{4}.sinx=-\frac{\sqrt{2}}{2}sinx\)
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Biểu thức A = cos 4 a + cos 2 a . sin 2 a + sin 2 a bằng :
rút gọn
\(\dfrac{sin2a+1}{cos2a}-\dfrac{1-sin2a}{sina-cosb}\)
a) cos4a - sin4a +1 = 2cos2a
b) cos6a + sin6a + 3sin2a.cos2a = 1
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
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\(cos^4a-sin^4a+1=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1\)
\(=cos^2a-sin^2a+1=cos^2a-sin^2a+sin^2a+cos^2a\)
\(=2cos^2a\)
\(cos^6a+sin^6a+3sin^2a.cos^2a\)
\(=\left(cos^2a+sin^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a.1+3sin^2a.cos^2a\)
\(=1\)
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Tính cos2a theo m biết a thỏa: \(\frac{cos7a-2cos5a+cos3a}{sin6a-sin4a}=2m\)
\(\frac{cos7a+cos3x-2cos5a}{sin6x-sin4a}=2m\Leftrightarrow\frac{2cos5a.cos2a-2cos5a}{2cos5a.sina}=2m\)
\(\Leftrightarrow\frac{2cos5a\left(cos2a-1\right)}{2cos5a.sina}=2m\Leftrightarrow\frac{cos2a-1}{sina}=2m\)
\(\Leftrightarrow\frac{-2sin^2a}{sina}=2m\Leftrightarrow sina=-m\)
\(\Rightarrow cos2a=1-2sin^2a=1-2m^2\)
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Rút gọn biểu thức:
B = (1+ tan2a).(1- sin2a) \(-\)(1+ cotg2a).(1- cos2a)
\(\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)
\(=\left(1+\dfrac{sin^2a}{cos^2a}\right).cos^2a-\left(1+\dfrac{cos^2a}{sin^2a}\right).sin^2a\)
\(=cos^2a+sin^2a-sin^2a-cos^2a=\)\(0\)
Vậy B=0
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