Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(tanx=3\) \(\Leftrightarrow sinx=3cosx\)
\(A=\dfrac{2.3.cosx-3cosx}{4cosx+5.3cosx}=\dfrac{3cosx}{19cosx}=\dfrac{3}{19}\)
\(B=\dfrac{sin^2x-4sinxcosx+3cos^2x}{5-2sin^2x}\)
\(=\dfrac{\left(3cosx\right)^2-4.3cosx.cosx+3cos^2x}{5-2\left(3cosx\right)^2}\)
\(=\dfrac{9cos^2x-12cos^2x+3cos^2x}{5-18cos^2x}=0\)
a: \(A=\sqrt{3}\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\dfrac{\sqrt{3}}{2}sinx-\dfrac{3}{2}cosx+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\sqrt{3}sinx-cosx\)
c: \(=2\left[\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right]+4sinx+1\)
\(=\sqrt{3}sin2x-cos2x+4sinx+1\)
d: \(D=\sqrt{3}cos2x+sin2x+2\cdot\left(\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right)\)
\(=\sqrt{3}\cdot cos2x+sin2x+\sqrt{3}\cdot sin2x-cos2x\)
\(=cos2x\left(\sqrt{3}-1\right)+sin2x\left(1+\sqrt{3}\right)\)
Câu 2:
\(A=2\cdot\dfrac{1}{2}+3\cdot\dfrac{1}{2}+1=1+1+1=3\)
Bài 3:
\(cos^2a=1-\left(\dfrac{12}{13}\right)^2=\dfrac{25}{169}\)
mà cosa>0
nên cosa=5/13
=>tan a=12/5; cot a=5/12
Câu 4: \(sin^2a=1-\dfrac{1}{4}=\dfrac{3}{4}\)
mà sina <0
nên sin a=-căn 3/2
=>tan a=-căn 3
\(A=-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\cdot\left(-\sqrt{3}\right)=-\sqrt{3}\)
\(\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}=\dfrac{\dfrac{4sin\alpha}{cos\alpha}+\dfrac{5cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{4tan\alpha+5}{2tan\alpha-3}\)
Biết \(tan\)=\(\dfrac{1}{3}\) nên ta có:
\(\dfrac{4\times\dfrac{1}{2}+5}{2\times\dfrac{1}{2}-3}=\dfrac{2+5}{2-3}=\dfrac{7}{-2}=\dfrac{-7}{2}\)
\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)
\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)
Biết cotα=\(\dfrac{1}{2}\) nên ta có:
\(A=\dfrac{4+5\cdot\dfrac{1}{2}}{2-3\cdot\dfrac{1}{2}}\)
\(A=\dfrac{4+\dfrac{5}{2}}{2-\dfrac{3}{2}}\)
A= 13
\(2sin\left(x+y\right)=sinx+siny\)
\(\Leftrightarrow2.2.sin\dfrac{x+y}{2}.cos\dfrac{x+y}{2}=2.sin\dfrac{x+y}{2}.cos\dfrac{x-y}{2}\)
\(\Leftrightarrow2cos\dfrac{x+y}{2}=cos\dfrac{x-y}{2}\)
\(\Leftrightarrow2\left(cos\dfrac{x}{2}.cos\dfrac{y}{2}-sin\dfrac{x}{2}.sin\dfrac{y}{2}\right)=cos\dfrac{x}{2}.cos\dfrac{y}{2}+sin\dfrac{x}{2}.sin\dfrac{y}{2}\)
\(\Leftrightarrow cos\dfrac{x}{2}.cos\dfrac{y}{2}=3.sin\dfrac{x}{2}.sin\dfrac{y}{2}\)
\(\Leftrightarrow\left(sin\dfrac{x}{2}:cos\dfrac{x}{2}\right).\left(sin\dfrac{y}{2}:cos\dfrac{y}{2}\right)=\dfrac{1}{3}\)
\(\Leftrightarrow tan\dfrac{x}{2}.tan\dfrac{y}{2}=\dfrac{1}{3}\)
2sin(x+y)=sinx+siny2sin(x+y)=sinx+siny
⇔2.2.sinx+y2.cosx+y2=2.sinx+y2.cosx−y2⇔2.2.sinx+y2.cosx+y2=2.sinx+y2.cosx−y2
⇔2cosx+y2=cosx−y2⇔2cosx+y2=cosx−y2
⇔2(cosx2.cosy2−sinx2.siny2)=cosx2.cosy2+sinx2.siny2⇔2(cosx2.cosy2−sinx2.siny2)=cosx2.cosy2+sinx2.siny2
⇔cosx2.cosy2=3.sinx2.siny2⇔cosx2.cosy2=3.sinx2.siny2
⇔(sinx2:cosx2).(siny2:cosy2)=13⇔(sinx2:cosx2).(siny2:cosy2)=13
⇔tanx2.tany2=13⇔tanx2.tany2=13
\(\dfrac{3sin\alpha-4cos\alpha}{2sin\alpha+3cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}+\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{3tan\alpha-4}{2tan\alpha+3}\)
Biết tanα=\(-\dfrac{1}{4}\) nên ta có:
\(\dfrac{3\cdot\dfrac{-1}{4}-4}{2\cdot\dfrac{-1}{4}+3}=\dfrac{-\dfrac{3}{4}-4}{-\dfrac{1}{2}+3}=\dfrac{-19}{10}\)
Đề thiếu. Bạn xem lại đề.