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 )

c: \(\cot\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

\(=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+\frac{x}{2}\right)=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{\tan\left(\frac{\pi}{4}\right)+\tan\left(\frac{x}{2}\right)}{1-\tan\left(\frac{\pi}{4}\right)\cdot\tan\left(\frac{x}{2}\right)}=\frac{\tan\left(\frac{x}{2}\right)+1}{1-\tan\left(\frac{x}{2}\right)}\)

\(=\left(\frac{\sin\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)}+1\right):\left(1-\frac{\sin\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)}\right)=\frac{\sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)}\) (1)

\(\frac{cosx}{1-\sin x}=\frac{cos\left(2\cdot\frac{x}{2}\right)}{1-\sin\left(2\cdot\frac{x}{2}\right)}=\frac{cos^2\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)}{\left(cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)^2}\) (2)

Từ (1),(2) suy ra \(\frac{cosx}{1-\sin x}=\cot\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

b: \(\tan\left(x+\frac{\pi}{4}\right)=\frac{\tan x+\tan\left(\frac{\pi}{4}\right)}{1-tanx\cdot\tan\left(\frac{\pi}{4}\right)}=\frac{\tan x+1}{1-\tan x}\)

\(=\left(\frac{\sin x}{cosx}+1\right):\left(1-\frac{\sin x}{cosx}\right)=\frac{\sin x+cosx}{cosx-\sin x}\)

\(\frac{1+\sin2x}{cos2x}=\frac{1+2\cdot\sin x\cdot cosx}{cos^2x-\sin^2x}=\frac{\left(cosx+\sin x\right)^2}{\left(cosx-\sin x\right)\left(cosx+\sin x\right)}=\frac{cosx+\sin x}{cosx-\sin x}\)

Do đó: \(\tan\left(x+\frac{\pi}{4}\right)=\frac{1+\sin2x}{cos2x}\)

mai đăng lại bài này nhé t làm cho h đi ngủ

ừ

Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)

2.

\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)

\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)

3.

\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)

4.

\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)

5.

\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)

\(=tan^2x+1+tan^2x=1+2tan^2x\)

a: \(\sin3x+cos2x=1+2\cdot\sin x\cdot cos2x\)

=>sin3x+cos2x=1+sin(x+2x)+sin(x-2x)

=>sin3x+cos2x=1+sin3x-sin x

=>cos2x-1+sin x=0

=>\(1-2\cdot\sin^2x-1+\sin x=0\)

=>\(-2\cdot\sin^2x+\sin x=0\)

=>sin x(2sin x-1)=0

TH1: sin x=0

=>\(x=k\pi\)

TH2: 2sin x-1=0

=>\(\sin x=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)

b: \(\sin^3x+cos^3x=2\cdot\left(\sin^5x+cos^5x\right)\)

=>\(\sin^3x-2\cdot\sin^5x+cos^3x-2\cdot cos^5x=0\)

=>\(\sin^3x\left(1-2\cdot\sin^2x\right)+cos^3x\left(1-2\cdot cos^2x\right)=0\)

=>\(\sin^3x\cdot cos2x-cos^3x\cdot cos2x=0\)

=>\(cos2x\left(\sin^3x-cos^3x\right)=0\)

TH1: cos2x=0

=>\(2x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

TH2: \(\sin^3x-cos^3x=0\)

=>\(\sin^3x=cos^3x\)

=>sin x=cosx

=>\(\sin x-cosx=0\)

=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x-\frac{\pi}{4}\right)=0\)

=>\(x-\frac{\pi}{4}=k\pi\)

=>\(x=\frac{\pi}{4}+k\pi\)

f: ĐKXĐ: \(\begin{cases}\sin x<>0\\ cosx<>0\end{cases}\Rightarrow\begin{cases}x<>k\pi\\ x<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow x<>\frac{k\pi}{2}\)

\(\frac{\tan x-\sin x}{\sin^3x}=\frac{1}{cosx}\)

=>\(\frac{\frac{\sin x}{cosx}-\sin x}{\sin^3x}=\frac{1}{cosx}\)

=>\(\frac{\frac{1}{cosx}-1}{\sin^2x}=\frac{1}{cosx}\)

=>\(\sin^2x=cosx\cdot\left(\frac{1}{cosx}-1\right)=1-cosx\)

=>\(1-cos^2x=1-cosx\)

=>\(cos^2x-cosx=0\)

=>cosx(cosx-1)=0

TH1: cosx=0

=>\(x=\frac{\pi}{2}+k\pi\) (loại)

TH2: cosx-1=0

=>cosx=1

=>\(x=k2\pi\)

=>sin x=0

=>Loại

a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)

\(\Leftrightarrow3cos4x-cos6x-2=0\)

Đặt \(t=2x\)

Pttt:\(3cos2t-cos3t-2=0\)

\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)

\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))

\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)

\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)

\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))

Vậy...

a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)

Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)

\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)

\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)

\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)

\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))

\(\Rightarrow sinx=1\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)

Vậy...

Xét tam giác ABC vuông tại A có AH là đường cao và AM là trung tuyến

Đặt \(\widehat{MAC}=\widehat{MCA}=x\)thì \(\widehat{BMA}=2x\)(theo tính chất đường trung tuyến ứng với cạnh huyền của tam giác vuông)

a) Ta có: \(\sin2x=\frac{AH}{AM}=2.\frac{AH}{BC}=2.\frac{AH}{AC}.\frac{AC}{BC}=2.\sin ACH.\cos ACB=2\cos x.\sin x\)

b) \(\cos2x=\frac{HM}{AM}=\frac{2HM}{BC}=\frac{2HC-2CM}{BC}=2.\frac{HC}{BC}-1=2.\frac{HC}{ AC}.\frac{AC}{BC}-1=2.\cos ACH.\cos ACB-1=2\cos^2x-1=2\cos^2x-\left(\sin^2x+\cos^2x\right)=\cos^2x-\sin^2x\)c) \(\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\cos x.\sin x}{\cos^2x-\sin^2x}=\frac{2.\frac{\sin x}{\cos x}}{\frac{\cos^2x}{\cos^2x}-\frac{\sin^2x}{\cos^2x}}=\frac{2\tan x}{1-\tan^2x}\)