K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!
16 tháng 10 2023
a: \(VT=\dfrac{cot^2x}{1+cot^2x}\cdot\dfrac{1+tan^2x}{tan^2x}\)
\(=\dfrac{cot^2x}{\dfrac{1}{sin^2x}}\cdot\dfrac{\dfrac{1}{cos^2x}}{tan^2x}\)
\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{1}{cos^2x}:\dfrac{1}{sin^2x}\)
\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{sin^2x}{cos^2x}\)
\(=cot^2x\)
\(VP=\dfrac{tan^2x+cot^2x}{1+tan^4x}=\dfrac{\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}}{1+\dfrac{sin^4x}{cos^4x}}\)
\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}:\dfrac{cos^4x+sin^4x}{cos^4x}\)
\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}\cdot\dfrac{cos^4x}{cos^4x+sin^4x}=\dfrac{cos^2x}{sin^2x}=cot^2x\)
=>VT=VP
b:
\(\dfrac{tan^2x-cos^2x}{sin^2x}+\dfrac{cot^2x-sin^2x}{cos^2x}\)
\(=\dfrac{\left(\dfrac{sinx}{cosx}\right)^2-cos^2x}{sin^2x}+\dfrac{\left(\dfrac{cosx}{sinx}\right)^2-sin^2x}{cos^2x}\)
\(=\dfrac{sin^2x-cos^4x}{cos^2x\cdot sin^2x}+\dfrac{cos^2x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x+cos^2x-cos^4x-sin^4x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{1-\left(cos^2x+sin^2x\right)^2+2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}=2\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 6 2020
\(cos5x.cos3x+sin7x.sinx=\frac{1}{2}cos8x+\frac{1}{2}cos2x-\frac{1}{2}cos8x+\frac{1}{2}cos6x\)
\(=\frac{1}{2}\left(cos6x+cos2x\right)=cos4x.cos2x\)
\(\frac{1-2sin^22x}{1-sin4x}=\frac{cos^22x-sin^22x}{cos^22x+sin^22x-2sin2x.cos2x}\)
\(=\frac{\left(cos2x-sin2x\right)\left(cos2x+sin2x\right)}{\left(cos2x-sin2x\right)^2}=\frac{cos2x+sin2x}{cos2x-sin2x}=\frac{\frac{cos2x}{cos2x}+\frac{sin2x}{cos2x}}{\frac{cos2x}{cos2x}-\frac{sin2x}{cos2x}}=\frac{1+tan2x}{1-tan2x}\)
\(2cosx-3cos\left(\pi-x\right)+5sin\left(4\pi-\frac{\pi}{2}-x\right)+cot\left(\pi+\frac{\pi}{2}-x\right)\)
\(=2cosx+3cosx-5sin\left(\frac{\pi}{2}+x\right)+cot\left(\frac{\pi}{2}-x\right)\)
\(=5cosx-5cosx+tanx=tanx\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 6 2020
\(\frac{sin3x+sinx+sin4x}{cos4x+1+cosx+cos3x}=\frac{2sin2x.cosx+2sin2x.cos2x}{2cos^22x+2cos2x.cosx}=\frac{2sin2x\left(cosx+cos2x\right)}{2cos2x\left(cos2x+cosx\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\frac{sin^22x+2cos\left(2\pi+\pi+2x\right)-2}{-3+4cos2x+cos\left(\pi-4x\right)}=\frac{sin^22x-2cos2x-2}{-3+4cos2x-cos4x}=\frac{4sin^2x.cos^2x-2\left(2cos^2x-1\right)-2}{-3+4\left(1-2sin^2x\right)-\left(1-2sin^22x\right)}\)
\(=\frac{4cos^2x\left(sin^2x-1\right)}{-8sin^2x+2sin^22x}=\frac{2cos^2x.\left(-cos^2x\right)}{-4sin^2x+4sin^2x.cos^2x}=\frac{cos^4x}{2sin^2x\left(1-cos^2x\right)}\)
\(=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)
10 tháng 9 2023
a: Ta sẽ có hình vẽ sau:
Đặt \(x=\widehat{B}\)
sin x=sin B=AC/BC
cosx=cosB=AB/BC
\(tanx=tanB=\dfrac{AC}{AB}=\dfrac{sinx}{cosx}\)
=>\(tan^2x=\dfrac{sin^2x}{cos^2x}\)
b: \(cot^2x=\dfrac{1}{tan^2x}=1:\dfrac{sin^2x}{cos^2x}=\dfrac{cos^2x}{sin^2x}\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 4 2019
\(\frac{2}{sin4x}-tan2x=\frac{2}{2sin2x.cos2x}-\frac{sin2x}{cos2x}=\frac{1}{cos2x}\left(\frac{1}{sin2x}-sin2x\right)\)
\(=\frac{1}{cos2x}\left(\frac{1-sin^22x}{sin2x}\right)=\frac{1}{cos2x}\frac{cos^22x}{sin2x}=\frac{cos2x}{sin2x}=cot2x\)
HT
28 tháng 11 2019
Quên cách giải ptlg rồi nên lm câu 4 =.=
\(\cos3x=\cos\left(2x+x\right)=\cos2x.\cos x-\sin2x.\sin x\)
\(=\left(2\cos^2x-1\right)\cos x-2\sin^2x.\cos x\)
\(=2\cos^3x-\cos x-2\sin^2x.\cos x\)
\(\Rightarrow A=\frac{1+\cos x+2\cos^2x-1+2\cos^3x-\cos x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\left(1-\cos^2x\right).\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\cos x+2\cos^3x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos x\left(2\cos^2x+\cos x-1\right)}{2\cos^2x-1+\cos x}=2\cos x\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 5 2020
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
