Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)
bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)
\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\)
\(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
2.
ĐK: \(2x-y\ge0;y\ge0;y-x-1\ge0;y-3x+5\ge0\)
\(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\left(1\right)\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(1-y\right)\sqrt{2x-y}+y-1+2x-y-1-\left(2x-y-1\right)\sqrt{y}=0\)
\(\Leftrightarrow\left(1-y\right)\left(\sqrt{2x-y}-1\right)+\left(2x-y-1\right)\left(1-\sqrt{y}\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(1+\sqrt{y}\right)+\left(\sqrt{2x-y}-1\right)\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}+1\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(\sqrt{y}+\sqrt{2x-y}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2x-1\end{matrix}\right.\) (Vì \(\sqrt{y}+\sqrt{2x-y}+2>0\))
Nếu \(y=1\), khi đó:
\(\left(1\right)\Leftrightarrow x-5=\sqrt{-x}+\sqrt{-3x+6}\)
Phương trình này vô nghiệm
Nếu \(y=2x-1\), khi đó:
\(\left(1\right)\Leftrightarrow2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\) (Điều kiện: \(2\le x\le4\))
\(\Leftrightarrow2x\left(x-3\right)+x-3+1-\sqrt{x-2}+1-\sqrt{4-x}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1\right)=0\)
Ta thấy: \(1+\sqrt{x-2}\ge1\Rightarrow-\dfrac{1}{1+\sqrt{x-2}}\ge-1\Rightarrow1-\dfrac{1}{1+\sqrt{x-2}}\ge0\)
Lại có: \(\dfrac{1}{1+\sqrt{4-x}}>0\); \(2x>0\)
\(\Rightarrow\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1>0\)
Nên phương trình \(\left(1\right)\) tương đương \(x-3=0\Leftrightarrow x=3\Rightarrow y=5\)
Ta thấy \(\left(x;y\right)=\left(3;5\right)\) thỏa mãn điều kiện ban đầu.
Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)=\left(3;5\right)\)
a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).
b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).
1.
\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)
\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)
\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)
\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)
\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)
\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)
\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)
\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)
2.
\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)
\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x\)
\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)
\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)
\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)
\(=tan^2a.cot^2b-2\)
a)\(sin\left(\alpha+\dfrac{\pi}{2}\right)=cos\left[\dfrac{\pi}{2}-\left(\alpha+\dfrac{\pi}{2}\right)\right]=cos\left(-\alpha\right)=cos\alpha\).
b) \(cos\left(x+\dfrac{\pi}{2}\right)=sin\left[\dfrac{\pi}{2}-\left(x+\dfrac{\pi}{2}\right)\right]=sin\left(-x\right)=-sinx\).
c) \(tan\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{sin\left(\alpha+\dfrac{\pi}{2}\right)}{cos\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{cos\alpha}{-sin\alpha}=-cot\alpha\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{cos\left(\alpha+\dfrac{\pi}{2}\right)}{sin\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{-sin\alpha}{cos\alpha}=-tan\alpha\).