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tiếc quá em 2011

Ta có \(2\sin x\cos x=\left(\sin x+\cos x\right)^2-\left(\sin^2x+\cos^2x\right)\) 

\(=\left(\dfrac{3}{4}\right)^2-1=-\dfrac{7}{16}\)  

Từ đó \(A=\left|\sin x-\cos x\right|\)

\(\Rightarrow A^2=\left(\sin x-\cos x\right)^2\)

\(A^2=\sin^2x+\cos^2x-2\sin x\cos x\)

\(A^2=1+\dfrac{7}{16}=\dfrac{23}{16}\)

\(\Rightarrow A=\dfrac{\sqrt{23}}{4}\) (do \(A\ge0\))

Có \(\cos x+\sin x=\dfrac{3}{4}\)

\(\Leftrightarrow\left(\cos x+\sin x\right)^2=\dfrac{9}{16}\)

\(\Leftrightarrow2.\sin x.\cos x+1=\dfrac{9}{16}\)

\(\Leftrightarrow\sin x.\cos x=-\dfrac{7}{32}\)

Lại có \(\left(\cos x+\sin x\right)^2=\left(\cos x-\sin x\right)^2+4.\sin x.\cos x=\dfrac{9}{16}\)

\(\Leftrightarrow\left(\cos x-\sin x\right)^2=\dfrac{23}{16}\)

\(\Leftrightarrow\left|\sin x-\cos x\right|=\dfrac{\sqrt{23}}{4}\)

 Mình sửa lại đề bài là AB cắt CD tại T chứ không phải là AD cắt BC đâu. 

\(a,cos\alpha=\dfrac{5}{13}\)

\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{5}{13}\right)^2}\Leftrightarrow tan^2\alpha=\dfrac{144}{25}\Leftrightarrow tan\alpha=\dfrac{12}{5}\)

\(cot\alpha=\dfrac{1}{tan\alpha}=1:\dfrac{12}{5}=\dfrac{5}{12}\)

\(b,sin\alpha=\dfrac{7}{12}\)

\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\left(\dfrac{7}{12}\right)^2}=\dfrac{\sqrt{95}}{12}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{\sqrt{95}}{12}\right)^2}\Leftrightarrow tan\alpha=\dfrac{49}{95}\)

\(cot\alpha=1:\dfrac{49}{95}=\dfrac{95}{49}\)

\(c,tan\alpha=\dfrac{15}{4}\)

\(cot\alpha=1:\dfrac{15}{4}=\dfrac{4}{15}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(\dfrac{15}{4}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\sqrt{\dfrac{16}{241}}\)

\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\sqrt{\dfrac{16}{241}}\right)^2}\approx0,97\)

\(d,cot\alpha=-\dfrac{1}{\sqrt{3}}\\ tan\alpha=1:\left(-\dfrac{1}{\sqrt{3}}\right)=-\sqrt{3}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(-\sqrt{3}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\dfrac{1}{2}\)

\(sin\alpha=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{3}}{2}\)

1e+84937

Ta có x_n luôn dương

Ta có \(2x_n+1=\) \(2\times\dfrac{\left(2+cos\alpha\right)x_n+cos^2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}+1=\)

\(=\dfrac{6x_n+2cos^2\alpha+2-cos2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}\)

\(=\dfrac{6x_n+2cos^2\alpha+2sin^2a+1}{\left(2x_n+1\right)\left(1-cos2\alpha\right)+1}\)

\(=\dfrac{3\left(2x_n+1\right)}{2\sin^2\alpha\left(2x_n+1\right)+1}\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}=\dfrac{2\sin^2\alpha\left(2x_n+1\right)+1}{3\left(2x_n+1\right)}\)

\(=\dfrac{1}{3}\left(2\sin^2\alpha+\dfrac{1}{2x_n+1}\right)\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\dfrac{1}{3}\left(\dfrac{1}{2x_n+1}-\sin^2\alpha\right)\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{2x_1+1}-\sin^2\alpha\right)\)

\(=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{3}-\sin^2\alpha\right)\)

\(\Rightarrow y_n=\sum\limits^{n-1}_{i=0}\left(\dfrac{1}{3}\right)^i\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)

\(=\dfrac{1-\left(\dfrac{1}{3}\right)^n}{1-\dfrac{1}{3}}\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)

Ta có \(\overrightarrow{BH}=\overrightarrow{BC}+\overrightarrow{CH}=-\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{CA}\)

\(\overrightarrow{MH}=\dfrac{1}{2}\overrightarrow{BA}=\dfrac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CA}\right)=\dfrac{1}{2}\overrightarrow{CA}-\dfrac{1}{2}\overrightarrow{CB}\)