Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)
\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)
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 \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)
\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)
mình làm r nha
https://hoc24.vn/cau-hoi/biet-cotadfrac12-gia-tri-bieu-thuc-adfrac4sinalpha5cosalpha2sinalpha-3cosalpha-bang-bao-nhieughi-ro-tung-loi-giai-nha.5724337531039
\(A=\dfrac{3sin\alpha-cos\alpha}{sin\alpha+cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-1}{\dfrac{sin\alpha}{cos\alpha}-1}=\dfrac{3tan\alpha-1}{tan\alpha-1}\)\(=\dfrac{3\sqrt{2}-1}{\sqrt{2}-1}=5+2\sqrt{2}\).
a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)
b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)
c/\(P=sin\left(30+60\right)=sin90=1\)
d/
\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)
\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)
e/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)
\(P=\dfrac{2sin\alpha-3cos\alpha}{3sin\alpha+2cos\alpha}\\ =\dfrac{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}\\ =\dfrac{2tan\alpha-3}{3tan\alpha+2}=\dfrac{2.3-3}{3.3+2}=\dfrac{3}{11}\)
Ta có: \(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha \ne {90^o})\)
\( \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = 1 + {3^2} = 10\)
\( \Leftrightarrow {\cos ^2}\alpha = \frac{1}{{10}} \Leftrightarrow \cos \alpha = \pm \frac{{\sqrt {10} }}{{10}}\)
Vì \({0^o} < \alpha < {180^o}\) nên \(\sin \alpha > 0\).
Mà \(\tan \alpha = 3 > 0 \Rightarrow \cos \alpha > 0 \Rightarrow \cos \alpha = \frac{{\sqrt {10} }}{{10}}\)
Lại có: \(\sin \alpha = \cos \alpha .\tan \alpha = \frac{{\sqrt {10} }}{{10}}.3 = \frac{{3\sqrt {10} }}{{10}}.\)
\( \Rightarrow P = \dfrac{{2.\frac{{3\sqrt {10} }}{{10}} - 3.\frac{{\sqrt {10} }}{{10}}}}{{3.\frac{{3\sqrt {10} }}{{10}} + 2.\frac{{\sqrt {10} }}{{10}}}} = \dfrac{{\frac{{\sqrt {10} }}{{10}}\left( {2.3 - 3} \right)}}{{\frac{{\sqrt {10} }}{{10}}\left( {3.3 + 2} \right)}} = \dfrac{3}{{11}}.\)
\(cot\alpha=\dfrac{1}{2}tacó:\)
\(cot\alpha=\dfrac{cos\alpha}{sin\alpha}\)
\(A=\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}\)
\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)
\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)
\(A=\dfrac{4+5\left(\dfrac{1}{2}\right)}{2-3\left(\dfrac{1}{2}\right)}\)
\(A=13\)
\(A=\dfrac{4\sin\alpha+5\cos\alpha}{2\sin\alpha-3\cos\alpha}\)
\(A=\dfrac{\dfrac{4\sin\alpha}{\sin\alpha}+\dfrac{5\cos\alpha}{\sin\alpha}}{\dfrac{2\sin\alpha}{\sin\alpha}-\dfrac{3\cos\alpha}{\sin\alpha}}\)
\(A=\dfrac{4+5\cot\alpha}{2-3\cot\alpha}\)
Thay cot α= \(\dfrac{1}{2}\) vào A, ta có:
\(A=\dfrac{4+5\times\dfrac{1}{2}}{2-3\times\dfrac{1}{2}}\)
\(A=\dfrac{4+\dfrac{5}{2}}{2-\dfrac{3}{2}}\)
\(A=\dfrac{13}{\dfrac{2}{\dfrac{1}{2}}}\)
A=13
\(C=\dfrac{sina\left(2sin^2a-3cos^2a\right)}{cosa\left(sin^2a-2\right)}\)
\(=\dfrac{sina\left(2sin^2a-3+3sin^2a\right)}{cosa\left(sin^2a-2\right)}\)
\(=\dfrac{sina\left(5sin^2a-3\right)}{cosa\left(-1-cos^2a\right)}\)
\(\dfrac{2sina+cosa}{2sin^3a-cos^3a}=\dfrac{\dfrac{2sina}{cos^3a}+\dfrac{cosa}{cos^3a}}{\dfrac{2sin^3a}{cos^3a}-\dfrac{cos^3a}{cos^3a}}=\dfrac{2tana.\dfrac{1}{cos^2a}+\dfrac{1}{cos^2a}}{2tan^3a-1}\)
\(=\dfrac{2tana\left(1+tan^2a\right)+1+tan^2a}{2tan^3a-1}=...\) (thay số và bấm máy)
Em vẫn ch hiểu tại sao cosa/cos3a lại ra 1/cos2a thầy giải thích giúp em vs ạ