Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/
\(tanx=\frac{sinx}{cosx}=\frac{sin^2x}{sinx.cosx}=\frac{2sin^2x}{2sinx.cosx}\)
\(=\frac{2\left(\frac{1-cos2x}{2}\right)}{sin2x}=\frac{1-cos2x}{sin2x}\)
2/
\(\frac{sin\left(60-x\right)cos\left(30-x\right)+cos\left(60-x\right)sin\left(30-x\right)}{sin4x}=\frac{sin\left(60-x+30-x\right)}{sin4x}=\frac{sin\left(90-2x\right)}{2sin2x.cos2x}\)
\(=\frac{cos2x}{2sin2x.cos2x}=\frac{1}{2sin2x}\)
3/
\(4cos\left(60+a\right)cos\left(60-a\right)+2sin^2a\)
\(=2\left(cos\left(60+a+60-a\right)+cos\left(60+a-60+a\right)\right)+2sin^2a\)
\(=2cos120+2cos2a+2\left(\frac{1-cos2a}{2}\right)\)
\(=-1+2cos2a+1-cos2a=cos2a\)
1) \(\dfrac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
\(VT=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)
\(VT=\dfrac{cosx\left(2cos-1\right)}{sinx\left(2cosx-1\right)}\)
\(VT=\dfrac{cosx}{sinx}=cotx=VP\) ( đpcm )
b) \(\dfrac{sinx+sin\dfrac{x}{2}}{1+cosx+cos\dfrac{x}{2}}=tan\dfrac{x}{2}\)
\(VT=\dfrac{sin\left(2.\dfrac{x}{2}\right)+sin\dfrac{x}{2}}{1+cos\left(2.\dfrac{x}{2}\right)+cos\dfrac{x}{2}}\)
\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{1+2cos^2\dfrac{x}{2}-1+cos\dfrac{x}{2}}\)
\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{2cos^2\dfrac{x}{2}+cos\dfrac{x}{2}}\)
\(VT=\dfrac{sin\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}{cos\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}\)
\(VT=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}=tan\dfrac{x}{2}=VP\) ( đpcm )
c) \(\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=tan^2\left(\dfrac{\pi}{4}-x\right)\)
\(VT=\dfrac{2cos2x-sin\left(2.2x\right)}{2cos2x+sin\left(2.2x\right)}\)
\(VT=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)
\(VT=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}\)
\(VT=\dfrac{1-sin2x}{1+sin2x}\)
\(VP=tan^2\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-cos2\left(\dfrac{\pi}{4}-x\right)}{1+cos2\left(\dfrac{\pi}{4}-x\right)}\)
\(VP=\dfrac{1-cos\left(\dfrac{\pi}{2}-2x\right)}{1+cos\left(\dfrac{\pi}{2}-2x\right)}\)
\(VP=\dfrac{1-sin2x}{1+cos2x}=VT\) ( đpcm )
d) \(tanx-tany=\dfrac{sin\left(x-y\right)}{cosx.cosy}\)
\(VP=\dfrac{sin\left(x-y\right)}{cosx.cosy}=\dfrac{sinx.cosy-cosx.siny}{cosx.cosy}\)
\(VP=\dfrac{sinx.cosy}{cosx.cosy}-\dfrac{cosx.siny}{cosx.cosy}\)
\(VP=\dfrac{sinx}{cosx}-\dfrac{siny}{cosy}=tanx-tany=VT\) ( đpcm )
a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).
\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).
ta có : \(\dfrac{sinx+sin\left(\dfrac{x}{2}\right)}{1+cosx+cos\left(\dfrac{x}{2}\right)}=\dfrac{2sin\left(\dfrac{x}{2}\right).cos\left(\dfrac{x}{2}\right)+sin\left(\dfrac{x}{2}\right)}{2cos^2\left(\dfrac{x}{2}\right)+cos\left(\dfrac{x}{2}\right)}\)
\(=\dfrac{sin\left(\dfrac{x}{2}\right)\left(2cos\left(\dfrac{x}{2}\right)+1\right)}{cos\left(\dfrac{x}{2}\right)\left(2cos\left(\dfrac{x}{2}\right)+1\right)}=\dfrac{sin\left(\dfrac{x}{2}\right)}{cos\left(\dfrac{x}{2}\right)}=tan\left(\dfrac{x}{2}\right)\left(đpcm\right)\)
Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý
Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)
Ta có các công thức cơ bản sau: \(cos\left(90^0+x\right)=-sinx;sin\left(90^0-x\right)=cosx\)
\(cot\left(90^0-x\right)=tanx;tan\left(90^0+x\right)=-cotx\)
Thay vào bài toán:
\(\dfrac{1-\left(-sinx\right)^2}{1-cos^2x}-tanx.\left(-cotx\right)=\dfrac{1-sin^2x}{1-cos^2x}+tanx.cotx\)
\(=\dfrac{cos^2x}{sin^2x}+1=\dfrac{cos^2x+sin^2x}{sin^2x}=\dfrac{1}{sin^2x}\)
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
\(=\dfrac{tan\left(\dfrac{pi}{2}+x\right)\cdot sin\left(-x\right)\cdot cos\left(x-pi\right)}{cos\left(\dfrac{pi}{2}-x\right)\cdot sin\left(x+pi\right)}\)
\(=\dfrac{-cotx\cdot sin\left(-x\right)\cdot\left(-cosx\right)}{sinx\cdot-sinx}\)
\(=\dfrac{cotx\cdot sinx\left(-1\right)\cdot cosx}{-sinx\cdot sinx}=\dfrac{\dfrac{cosx}{sinx}\cdot cosx}{sinx}=\dfrac{cos^2x}{sin^2x}=cot^2x\)
\(tử:=\dfrac{1}{2}\left[sin\left(60^o-x+30^o-x\right)+sin\left(60^o-x-30^2+x\right)\right]+\dfrac{1}{2}\left[sin\left(30^o-x+60^o-x\right)+sin\left(30^o-x-60^o+x\right)\right]\)
\(=\dfrac{1}{2}\left[2sin\left(\dfrac{\pi}{2}-2x\right)+sin\left(\dfrac{\pi}{6}\right)+sin\left(-\dfrac{\pi}{6}\right)\right]=\dfrac{1}{2}.\left[2sin\left(\dfrac{\pi}{2}-2x\right)+0\right]=sin\left(\dfrac{\pi}{2}-2x\right)=cos2x\)
\(VT=\dfrac{cos2x}{sin4x}=\dfrac{cos2x}{2sin2x.cos2x}=\dfrac{1}{2sin2x}=\dfrac{1}{4sinx.cosx}=\dfrac{\dfrac{1}{cos^2x}}{\dfrac{4sinx.cosx}{cos^2x}}=\dfrac{1+tan^2x}{\dfrac{4sĩnx}{cosx}}=\dfrac{1+tan^2x}{4tanx}=VP\)
thanks