Rút gọn biểu thức M = 1 + cos α + cos 2 α + cos 3 α 2 cos 2 α + cos α - 1 ta được
A. M = - 2 cos α
B. M = 2 cos α
C. M = cos α
D. M = 2 sin α
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A = sin6α+ 3sin2α .cos2α + cos6α = sin6α + 3sin2α .cos2α ( sin2α + cos2α ) + cos6α = sin6α + 3sin4 α .cos2α + 3sin4α .cos4α + cos6α = (sin2α + cos2α )2 |
= 1
Ta có: \(A=\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
\(=2\left(\sin^2\alpha+\cos^2\alpha\right)\)
=2
bài 1: ta có : \(cos^220+cos^240+cos^250+cos^270\)
\(=cos^220+cos^270+cos^240+cos^250\)
\(=cos^220+cos^2\left(90-20\right)+cos^240+cos^2\left(90-40\right)\)
\(=cos^220+sin^220+cos^240+sin^240=1+1=2\)
bài 2: a) ta có : \(cot^2\alpha-cos^2\alpha=cos^2\alpha\left(\dfrac{1}{sin^2\alpha}-1\right)=cos^2\alpha.\left(\dfrac{1-sin^2\alpha}{sin^2\alpha}\right)\)
\(=cos^2\alpha.\left(\dfrac{cos^2\alpha}{sin^2\alpha}\right)=cos^2\alpha.cot^2\alpha\left(đpcm\right)\)
b) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Leftrightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Leftrightarrow\dfrac{1+cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1-cos\alpha}\left(đpcm\right)\)
a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)
\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)
b: Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)
\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)
=4
A = \(\left(sin^2a+cos^2a\right)^2=1^2=1\)
D = \(sin^2\left(sin^2B+cos^2B\right)+cos^2a=sin^2a+cos^2a=1\)
a: VT=sin^2a(sin^2a+cos^2a)+cos^2a
=sin^2a+cos^2a
=1=VP
b: \(VT=\dfrac{sina+sina\cdot cosa+sina-sina\cdot cosa}{1-cos^2a}=\dfrac{2sina}{sin^2a}=\dfrac{2}{sina}=VP\)
c: \(VT=\dfrac{sin^2a+1+2cosa+cos^2a}{sina\left(1+cosa\right)}\)
\(=\dfrac{2\left(cosa+1\right)}{sina\left(1+cosa\right)}=\dfrac{2}{sina}=VP\)
Chọn B