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\(\frac{2sina+3cosa}{4sina-5cosa}=\frac{\frac{2sina}{cosa}+\frac{3cosa}{cosa}}{\frac{4sina}{cosa}-\frac{5cosa}{cosa}}=\frac{2tana+3}{4tana-5}=\frac{6+3}{12-5}=\frac{9}{7}\)
\(\frac{3sina-2cosa}{5sina+4cos^3a}=\frac{\frac{3sina}{cosa}-\frac{2cosa}{cosa}}{\frac{5sina}{cosa}+\frac{4cos^3a}{cosa}}=\frac{3tana-2}{5tana+4cos^2a}=\frac{3tana-2}{5tana+\frac{4}{1+tan^2a}}=\frac{9-2}{15+\frac{4}{10}}=\frac{5}{11}\)
\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)
\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)
a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)
b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)
Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)
\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)
\(tana-cota=3\Rightarrow\left(tana-cota\right)^2=9\)
\(\Rightarrow tan^2a+cot^2a-2=9\Rightarrow tan^2a+cot^2a=11\)
\(\frac{1}{tan^2a}+\frac{1}{cot^2a}=\frac{tan^2a+cot^2a}{\left(tana.cota\right)^2}=tan^2a+cot^2a=11\)
\(sin\alpha=sin\left(180-\alpha\right)=\dfrac{3}{5}\Rightarrow cos\left(180-a\right)=\sqrt{1-sin^2\alpha}=\dfrac{4}{5}\Rightarrow cos\alpha=-\dfrac{4}{5}\)
\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{5}}{-\dfrac{4}{5}}=-\dfrac{3}{4}\Rightarrow cot\alpha=-\dfrac{4}{3}\)
\(\Rightarrow A=\dfrac{3.\dfrac{3}{5}-\dfrac{4}{5}}{-\dfrac{3}{4}+\dfrac{4}{3}}=\dfrac{12}{7}\)
Câu 1:
\(sina+cosa=\frac{\sqrt{2}}{2}\Leftrightarrow\left(sina+cosa\right)^2=\frac{1}{2}\)
Chia 2 vế cho \(cos^2a:\) :
\(\left(\frac{sina+cosa}{cosa}\right)^2=\frac{1}{2}.\frac{1}{cos^2a}\Leftrightarrow\left(tana+1\right)^2=\frac{1}{2}\left(1+tan^2a\right)\)
\(\Leftrightarrow tan^2a+4tana+1=0\)
Tiếp tục chia 2 vế cho \(tana\): :
\(\Rightarrow tana+4+cota=0\Rightarrow tana+cota=-4\)
\(P=tan^2a+cot^2a=tan^2a+2+cot^2a-2=\left(tana+cota\right)^2-2=\left(-4\right)^2-2=14\)
Câu 2:
\(3cosa+2sina=2\Rightarrow cosa=\frac{2-2sina}{3}=\frac{2}{3}\left(1-sina\right)\)
Mặt khác ta luôn có: \(sin^2a+cos^2a=1\Leftrightarrow sin^2a+\frac{4}{9}\left(1-sina\right)^2=1\)
\(\Leftrightarrow9sin^2a+4sin^2a-8sina+4=9\)
\(\Leftrightarrow13sin^2a-8sina-5=0\Rightarrow\left[{}\begin{matrix}sina=1>0\left(l\right)\\sina=-\frac{5}{13}\end{matrix}\right.\)
\(\left(sin^2a-1-2cos^2a\right)\frac{sin^2a}{cos^2a}=\left(-cos^2a-2cos^2a\right).\frac{sin^2a}{cos^2a}\)
\(=\frac{-3cos^2a.sin^2a}{cos^2a}=-3sin^2a\)
Đề sai hoặc bạn viết sai đề ở \(-2cos^2a\) trên tử số, phải là dấu "+" mới ra kết quả \(sin^2a\)
\(=\left(2cos^2a-\left(1-sin^2a\right)\right).\frac{sin^2a}{cos^2a}=\left(2cos^2a-cos^2a\right)\frac{sin^2a}{cos^2a}\)
\(=\frac{cos^2a.sin^2a}{cos^2a}=sin^2a\)
vậy thì kết quả là
\(\sin2\alpha=-0.96\)
\(\)còn \(\cos\left(\alpha+\frac{\pi}{6}\right)\) thì đúng vì -(-0.8) mà sorry thiếu ngủ hôm qua -_-
\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)
\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)
\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)
\(A=\frac{3sina-2cosa}{12sin^3a+4cos^3a}=\frac{\frac{3sina}{sin^3a}-\frac{2cosa}{sin^3a}}{12+\frac{4cos^3a}{sin^3a}}=\frac{3.\frac{1}{sin^2a}-2cota.\frac{1}{sin^2a}}{12+4cot^3a}\)
\(=\frac{3\left(1+cot^2a\right)-2cota\left(1+cot^2a\right)}{12+4cot^3a}=\frac{3\left(1+3^2\right)-2.3.\left(1+3^2\right)}{12+4.3^3}=...\)