Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)
b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)
\(=\frac{1+tanx}{1+tanx}=1\)
c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)
\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)
Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)
\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)
\(cos^2x-cos^2y=\left(1-sin^2x\right)-\left(1-sin^2y\right)=sin^2y-sin^2x\)
d) \(\frac{1+sin^2x}{1-sin^2x}=\frac{cos^2x+sin^2x+sin^2x}{cos^2x+sin^2x-sin^2x}=\frac{cos^2x+2sin^2x}{cos^2x}=1+2\left(\frac{sinx}{cosx}\right)^2=1+2tan^2x\)
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) \(\dfrac{1}{1+tan\alpha}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{1}{1+\dfrac{1}{cot\alpha}}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{1}{\dfrac{cot\alpha+1}{cot\alpha}}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{cot\alpha}{cot\alpha+1}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{cot\alpha+1}{cot\alpha+1}=1\) (đpcm)
b) \(tan^2x+cot^2x+2\)
\(=\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}+2\)
\(=\dfrac{sin^2x}{cos^2x}+1+\dfrac{cos^2x}{sin^2x}+1\)
\(=\dfrac{sin^2x+cos^2x}{cos^2x}+\dfrac{cos^2x+sin^2x}{sin^2x}\)
\(=\dfrac{1}{cos^2x}+\dfrac{1}{sin^2x}\) (đpcm)
c) \(sinx.cosx.\left(1+tanx\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sinx.cosx.tanx\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sinx.cosx.\dfrac{sinx}{cosx}\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sin^2x\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sin^2x\right)\left(1+\dfrac{cosx}{sinx}\right)\)
\(=sinx.cosx+cos^2x+sin^2x+sinx.cosx\)
\(=1+sin^2x.cos^2x\)
Câu cuối không biết chỗ sai, mong mọi người chỉ bảo ạ ^^
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.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)
\(\Leftrightarrow16-12.sin^22x=7\)
\(\Leftrightarrow3-4sin^22x=0\)
\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
Chứng minh đẳng thức:
\(\frac{\sin^2x}{\sin x-\cos x}-\frac{\sin x+\cos x}{\tan^2x-1}=\sin x+\cos x\)
\(\frac{\sin^2x}{\sin x-\cos x}-\frac{\sin x+\cos x}{\tan^2x-1}\)
\(=\frac{\sin^2x}{\sin x-\cos x}-\frac{\sin x+\cos x}{\frac{\sin^2x-\cos^2x}{\cos^2x}}\)
\(=\frac{\sin^2x}{\sin x-\cos x}-\frac{\cos^2x}{\sin x-\cos x}=\sin x+\cos x\)
Xong
a. ĐKXĐ: ...
\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)
b.
\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)
\(\Leftrightarrow4cos^32x-2cos2x-1=0\)
Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề
c. ĐKXĐ: ...
\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)
\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Rightarrow x=...\\\frac{cosx+sinx}{sinx.cosx}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cosx+sinx=sinx.cosx\)
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)
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
Giả sử tất cả các biểu thức đều xác định
a/
\(tan^2x-sin^2x=\frac{sin^2x}{cos^2x}-sin^2x=sin^2x\left(\frac{1}{cos^2x}-1\right)\)
\(=sin^2x\left(\frac{1-cos^2x}{cos^2x}\right)=sin^2x.\frac{sin^2x}{cos^2x}=sin^2x.tan^2x\)
b/
\(tanx+cotx=\frac{sinx}{cosx}+\frac{cosx}{sinx}=\frac{sin^2x+cos^2x}{sinx.cosx}=\frac{1}{sinx.cosx}\)
c/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
d/
\(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{1}{1+\frac{1}{tanx}}=\frac{1}{1+tanx}+\frac{tanx}{1+tanx}=\frac{1+tanx}{1+tanx}=1\)
e/
\(\left(1-\frac{1}{cosx}\right)\left(1+\frac{1}{cosx}\right)+tan^2x=1-\frac{1}{cos^2x}+tan^2x\)
\(=\frac{cos^2x-1}{cos^2x}+tan^2x=\frac{-sin^2x}{cos^2x}+tan^2x=-tan^2x+tan^2x=0\)