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) \(\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\cos x.\sin x}{\cos^2x-\sin^2x}=\frac{2.\frac{\sin x}{\cos x}}{\frac{\cos^2x}{\cos^2x}-\frac{\sin^2x}{\cos^2x}}=\frac{2\tan x}{1-\tan^2x}\)

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

a: tan x(cot^2x-1)

\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)

=cotx-tanx/cotx=cotx(1-tan^2x)

b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)

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

c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)

\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)

\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)

=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)

=-cos^2x*cos^2x=-cos^4x

=>ĐPCM

Quên cách giải ptlg rồi nên lm câu 4 =.=

\(\cos3x=\cos\left(2x+x\right)=\cos2x.\cos x-\sin2x.\sin x\)

\(=\left(2\cos^2x-1\right)\cos x-2\sin^2x.\cos x\)

\(=2\cos^3x-\cos x-2\sin^2x.\cos x\)

\(\Rightarrow A=\frac{1+\cos x+2\cos^2x-1+2\cos^3x-\cos x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\left(1-\cos^2x\right).\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\cos x+2\cos^3x}{2\cos^2x-1+\cos x}\)

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

\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)

\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)

\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)

\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)

\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)

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

\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)

Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)

\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)

\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)

a.

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

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

\(\Leftrightarrow1-sin^2x=0\)

\(\Leftrightarrow cos^2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

\(\frac{\left(1+tanx\right)^2-2tan^2x}{1+tan^2x}=\frac{1+2tanx-tan^2x}{1+tan^2x}=\frac{cos^2x\left(1+2tanx-tan^2x\right)}{cos^2x\left(1+tan^2x\right)}\)

\(=\frac{cos^2x+2sinx.cosx-sin^2x}{cos^2x+sin^2x}=\frac{cos^2x-sin^2x+2sinx.cosx}{1}\)

\(=cos2x+sin2x\)

a, \(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)

\(=\dfrac{1+cos2x+cosx+cos3x}{2cos^2x+cosx-1}\)

\(=\dfrac{2cos^2x+2cos2x.cosx}{cos2x+cosx}\)

\(=\dfrac{2cosx\left(cos2x+cosx\right)}{cos2x+cosx}=2cosx\)

b) \(cos\dfrac{5x}{2}.cos\dfrac{3x}{2}+sin\dfrac{7x}{2}.sin\dfrac{x}{2}\)

\(=cos\dfrac{4x+x}{2}.cos\dfrac{4x-x}{2}+sin\dfrac{4x+3x}{2}.sin\dfrac{4x-3x}{2}\)

\(=\dfrac{1}{2}\left(cos4x+cosx\right)-\dfrac{1}{2}\left(cos4x-cos3x\right)\)

\(=\dfrac{1}{2}\left(cosx+cos3x\right)=\dfrac{1}{2}.2cos2x.cos\left(-x\right)\)\(=cosx.cos2x\)