Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(A=-2cosx+2cosx+tan^2x-\frac{1}{cos^2x}\)
\(=tan^2x-\left(1+tan^2x\right)=-1\)
2sin(π2+x)+sin(3π−x)+sin(3π2+x)+cos(π2+x)2sin(π2+x)+sin(3π−x)+sin(3π2+x)+cos(π2+x)
=2cosx+sinx−cosx−sinx=2cosx+sinx−cosx−sinx
=cosx
\(sinx+cosx=\frac{1}{2}\Rightarrow\left(sinx+cosx\right)^2=\frac{1}{4}\Rightarrow sin^2x+cos^2x+2sinx.cosx=\frac{1}{4}\)
\(\Rightarrow2sinx.cosx=\frac{1}{4}-1=-\frac{3}{4}\Rightarrow sinx.cosx=-\frac{3}{8}\)
Vậy ta có:
\(sin^3x+cos^3x=\left(sinx+cosx\right)\left[\left(sinx+cosx\right)^2-3sinx.cosx\right]\)
\(=\frac{1}{2}\left(\frac{1}{4}+\frac{9}{8}\right)=\frac{11}{16}\)
Bài 2: Mục đích của bài này là gì bạn? Ko thấy yêu cầu?
Bài 3:
\(tanx+cotx=2\Rightarrow\left(tanx+cotx\right)^2=4\)
\(\Rightarrow tan^2x+2tanx.cotx+cot^2x=4\Rightarrow tan^2x+cot^2x+2=4\)
\(\Rightarrow tan^2x+cot^2x=2\)
\(A=\dfrac{\sqrt{2}.cosx-2cos\left(\dfrac{\pi}{4}+x\right)}{-\sqrt{2}.sinx+2sin\left(\dfrac{\pi}{4}+x\right)}\)
\(=\dfrac{\sqrt{2}.cosx-2\left(cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\right)}{-\sqrt{2}.sinx+2\left(sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx\right)}\)
\(=\dfrac{\sqrt{2}.cosx-\sqrt{2}.cosx+\sqrt{2}.sinx}{-\sqrt{2}.sinx+\sqrt{2}.cosx+\sqrt{2}.sinx}\)
\(=\dfrac{\sqrt{2}.sinx}{\sqrt{2}.cosx}=tanx\)
a/ \(\pi< x< \frac{3\pi}{2}\Rightarrow sinx< 0\)
\(\Rightarrow sinx=-\sqrt{1-cos^2x}=-\frac{5}{13}\)
\(sin\left(\frac{\pi}{3}-x\right)=sin\frac{\pi}{3}cosx-cos\frac{\pi}{3}sinx=\frac{\sqrt{3}}{2}.\left(-\frac{12}{13}\right)-\frac{1}{2}.\left(-\frac{5}{13}\right)=\frac{5-12\sqrt{3}}{26}\)
b/ \(\pi< x< \frac{3\pi}{2}\Rightarrow cosx< 0\)
\(\Rightarrow cosx=-\sqrt{1-sin^2x}=-\frac{3}{5}\)
\(cot\left(x-\frac{\pi}{4}\right)=\frac{cos\left(x-\frac{\pi}{4}\right)}{sin\left(x-\frac{\pi}{4}\right)}=\frac{sinx+cosx}{sinx-cosx}=7\)
c/ \(cot\left(\frac{5\pi}{2}-x\right)=cot\left(2\pi+\frac{\pi}{2}-x\right)=tanx=2\)
\(\Rightarrow tan\left(x+\frac{\pi}{4}\right)=\frac{tanx+tan\frac{\pi}{4}}{1-tanx.tan\frac{\pi}{4}}=\frac{2+1}{1-2.1}=-3\)
\(\frac{2sin^2\frac{x}{2}+sin2x-1}{2sinx-1}+sinx=\frac{1-cosx+2sin2x.cosx-1}{2sinx-1}+sinx\)
\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx\)
\(=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)\)
\(=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(\pi< x< \frac{3\pi}{2}\Rightarrow sinx< 0;cosx< 0;tanx>0;cotx>0\)
\(tanx-3cotx=6\Leftrightarrow tanx-\frac{3}{tanx}=6\)
\(\Leftrightarrow tan^2x-6tanx-3=0\Rightarrow\left[{}\begin{matrix}tanx=3+2\sqrt{3}\\tanx=3-2\sqrt{3}< 0\left(l\right)\end{matrix}\right.\)
\(\frac{1}{cos^2x}=1+tan^2x\Rightarrow cos^2x=\frac{1}{1+tan^2x}\Rightarrow cosx=\frac{-1}{\sqrt{1+tan^2x}}\) (do \(cosx< 0\))
\(\Rightarrow cosx=\frac{-1}{\sqrt{22+12\sqrt{3}}}\Rightarrow sinx=-\sqrt{1-cos^2x}=-\sqrt{\frac{15+6\sqrt{3}}{26}}\)
\(cotx=\frac{1}{tanx}=\frac{1}{3+2\sqrt{3}}\)
Số xấu dữ dội, bạn tự thay vào kết quả :(
\(\frac{\sqrt{2}cosx-2cos\left(\frac{\pi}{4}+x\right)}{2sin\left(\frac{\pi}{4}+x\right)-\sqrt{2}sinx}\\ =\frac{cosx-\sqrt{2}cos\left(\frac{\pi}{4}+x\right)}{\sqrt{2}sin\left(\frac{\pi}{4}+x\right)-sinx}\\ =\frac{cosx-\sqrt{2}\left(\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx\right)}{\sqrt{2}\left(\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx\right)-sinx}\\ =\frac{cosx-cosx+sinx}{cosx+sinx-sinx}\\ =\frac{sinx}{cosx}=tanx\)