CMR
\(1+\cos2\alpha=2\cos^2\alpha\)
K
Khách
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QL
0
4 tháng 8 2023
Ta có xn luôn dương
Ta có \(2x_n+1=\) \(2\times\dfrac{\left(2+cos\alpha\right)x_n+cos^2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}+1=\)
\(=\dfrac{6x_n+2cos^2\alpha+2-cos2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}\)
\(=\dfrac{6x_n+2cos^2\alpha+2sin^2a+1}{\left(2x_n+1\right)\left(1-cos2\alpha\right)+1}\)
\(=\dfrac{3\left(2x_n+1\right)}{2\sin^2\alpha\left(2x_n+1\right)+1}\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}=\dfrac{2\sin^2\alpha\left(2x_n+1\right)+1}{3\left(2x_n+1\right)}\)
\(=\dfrac{1}{3}\left(2\sin^2\alpha+\dfrac{1}{2x_n+1}\right)\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\dfrac{1}{3}\left(\dfrac{1}{2x_n+1}-\sin^2\alpha\right)\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{2x_1+1}-\sin^2\alpha\right)\)
\(=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{3}-\sin^2\alpha\right)\)
\(\Rightarrow y_n=\sum\limits^{n-1}_{i=0}\left(\dfrac{1}{3}\right)^i\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)
\(=\dfrac{1-\left(\dfrac{1}{3}\right)^n}{1-\dfrac{1}{3}}\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)
NT
0
PT
1
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 5 2019
\(A=\frac{4sin2a}{1-2cos^2\frac{a}{2}}=\frac{4\left(2sina.cosa\right)}{1-\left(1+cosa\right)}=\frac{8sina.cosa}{-cosa}=-8sina\)
\(B=\frac{1+cosa-sina}{1-cosa-sina}=\frac{1+2cos^2\frac{a}{2}-1-2sin\frac{a}{2}cos\frac{a}{2}}{1-\left(1-2sin^2\frac{a}{2}\right)-2sin\frac{a}{2}cos\frac{a}{2}}=\frac{2cos^2\frac{a}{2}-2sin\frac{a}{2}cos\frac{a}{2}}{2sin^2\frac{a}{2}-2sin\frac{a}{2}cos\frac{a}{2}}\)
\(=\frac{-cos\frac{a}{2}\left(2sin\frac{a}{2}-2cos\frac{a}{2}\right)}{sin\frac{a}{2}\left(2sin\frac{a}{2}-2cos\frac{a}{2}\right)}=\frac{-cos\frac{a}{2}}{sin\frac{a}{2}}=-cot\frac{a}{2}\)
Ta có hình vẽ, với \(\Delta\)ABC vuông ở A; đường cao AH; trung tuyến AM và \(\alpha=\widehat{ACB}\):
\(\Delta\)ABC vuông tại A có đường trung tuyến AM nên \(\Delta\)ACM cân ở M => ^AMB = 2.^ACM = 2.^ACB = 2\(\alpha\)
Ta có: \(\cos2\alpha=\frac{HM}{AM}=\frac{HM}{CM}\Rightarrow1+\cos2\alpha=\frac{HM+CM}{CM}=\frac{CH}{CM}\)
\(\Rightarrow1+\cos2\alpha=2.\frac{CH}{BC}=2.\frac{CH}{AC}.\frac{AC}{BC}=2.\cos\alpha.\cos\alpha=2\cos^2\alpha\)(ĐPCM).