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\(\frac{2sina+3cosa}{4sina-5cosa}=\frac{\frac{2sina}{cosa}+\frac{3cosa}{cosa}}{\frac{4sina}{cosa}-\frac{5cosa}{cosa}}=\frac{2tana+3}{4tana-5}=\frac{6+3}{12-5}=\frac{9}{7}\)
\(\frac{3sina-2cosa}{5sina+4cos^3a}=\frac{\frac{3sina}{cosa}-\frac{2cosa}{cosa}}{\frac{5sina}{cosa}+\frac{4cos^3a}{cosa}}=\frac{3tana-2}{5tana+4cos^2a}=\frac{3tana-2}{5tana+\frac{4}{1+tan^2a}}=\frac{9-2}{15+\frac{4}{10}}=\frac{5}{11}\)
\(tana-cota=3\Rightarrow\left(tana-cota\right)^2=9\)
\(\Rightarrow tan^2a+cot^2a-2=9\Rightarrow tan^2a+cot^2a=11\)
\(\frac{1}{tan^2a}+\frac{1}{cot^2a}=\frac{tan^2a+cot^2a}{\left(tana.cota\right)^2}=tan^2a+cot^2a=11\)
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: \(sin^2a+cos^2a=1\Leftrightarrow sin^2a+\frac{4}{9}\left(1-sina\right)^2=1\)
\(\Leftrightarrow9sin^2a+4sin^2a-8sina+4=9\)
\(\Leftrightarrow13sin^2a-8sina-5=0\Rightarrow\left[{}\begin{matrix}sina=1>0\left(l\right)\\sina=-\frac{5}{13}\end{matrix}\right.\)
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 -_-
Nhân cả tử và mẫu của phân số chứa tan với \(sina.cosa\)
\(A=\frac{sin^2x-cos^2x}{sin^2x+cos^2x}+cos2x=sin^2x-cos^2x+cos2x=-cos2x+cos2x=0\)
\(B=\frac{1+sin4a-cos4a}{1+sin4a+cos4a}=\frac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin4a.cos4a+2cos^22a-1}\)
\(B=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(C=\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a-1\right)}{2\left(cos^22a+2cos2a+1\right)}\)
\(C=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}=\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(D=\frac{sin^22a+4sin^4a-\left(2sina.cosa\right)^2}{4-4sin^2a-sin^22a}=\frac{sin^22a+4sin^4a-sin^22a}{4\left(1-sin^2a\right)-\left(2sina.cosa\right)^2}=\frac{4sin^4a}{4cos^2a-4sin^2a.cos^2a}\)
\(=\frac{sin^4a}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^2a.cos^2a}=\frac{sin^4a}{cos^4a}=tan^4a\)
Tham khảo:
\(E=\frac{cot\alpha+3tan\alpha}{2cot\alpha+tan\alpha}\\ E=\frac{1+3tan^2\alpha}{2+tan^2\alpha}\\ E=\frac{3\left(tan^2\alpha+1\right)-2}{1+\left(1+tan^2\alpha\right)}\\ E=\frac{\frac{3}{cos^2\alpha}-2}{\frac{1}{cos^2\alpha}+1}\\ E=\frac{3-2cos^2\alpha}{1+cos^2\alpha}\\ E=\frac{19}{13}\)
\(\left(sin^2a-1-2cos^2a\right)\frac{sin^2a}{cos^2a}=\left(-cos^2a-2cos^2a\right).\frac{sin^2a}{cos^2a}\)
\(=\frac{-3cos^2a.sin^2a}{cos^2a}=-3sin^2a\)
Đề sai hoặc bạn viết sai đề ở \(-2cos^2a\) trên tử số, phải là dấu "+" mới ra kết quả \(sin^2a\)
\(=\left(2cos^2a-\left(1-sin^2a\right)\right).\frac{sin^2a}{cos^2a}=\left(2cos^2a-cos^2a\right)\frac{sin^2a}{cos^2a}\)
\(=\frac{cos^2a.sin^2a}{cos^2a}=sin^2a\)
\(A=\frac{3sina-2cosa}{12sin^3a+4cos^3a}=\frac{\frac{3sina}{sin^3a}-\frac{2cosa}{sin^3a}}{12+\frac{4cos^3a}{sin^3a}}=\frac{3.\frac{1}{sin^2a}-2cota.\frac{1}{sin^2a}}{12+4cot^3a}\)
\(=\frac{3\left(1+cot^2a\right)-2cota\left(1+cot^2a\right)}{12+4cot^3a}=\frac{3\left(1+3^2\right)-2.3.\left(1+3^2\right)}{12+4.3^3}=...\)