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\(A=\dfrac{\sqrt{2}.cosx-2cos\left(\dfrac{\pi}{4}+x\right)}{-\sqrt{2}.sinx+2sin\left(\dfrac{\pi}{4}+x\right)}\)
\(=\dfrac{\sqrt{2}.cosx-2\left(cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\right)}{-\sqrt{2}.sinx+2\left(sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx\right)}\)
\(=\dfrac{\sqrt{2}.cosx-\sqrt{2}.cosx+\sqrt{2}.sinx}{-\sqrt{2}.sinx+\sqrt{2}.cosx+\sqrt{2}.sinx}\)
\(=\dfrac{\sqrt{2}.sinx}{\sqrt{2}.cosx}=tanx\)
\(sinx+cosx=\frac{1}{2}\Rightarrow\left(sinx+cosx\right)^2=\frac{1}{4}\Rightarrow sin^2x+cos^2x+2sinx.cosx=\frac{1}{4}\)
\(\Rightarrow2sinx.cosx=\frac{1}{4}-1=-\frac{3}{4}\Rightarrow sinx.cosx=-\frac{3}{8}\)
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:
\(tanx+cotx=2\Rightarrow\left(tanx+cotx\right)^2=4\)
\(\Rightarrow tan^2x+2tanx.cotx+cot^2x=4\Rightarrow tan^2x+cot^2x+2=4\)
\(\Rightarrow tan^2x+cot^2x=2\)
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(A=-2cosx+2cosx+tan^2x-\frac{1}{cos^2x}\)
\(=tan^2x-\left(1+tan^2x\right)=-1\)
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
\(\frac{2sin^2\frac{x}{2}+sin2x-1}{2sinx-1}+sinx=\frac{1-cosx+2sin2x.cosx-1}{2sinx-1}+sinx\)
\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx\)
\(=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)\)
\(=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(2sin\left(\frac{\pi}{2}+x\right)+sin\left(3\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)\)
\(=2cosx+sinx-cosx-sinx\)
\(=cosx\)
\(\frac{sin3x+sinx+sin4x}{cos4x+1+cosx+cos3x}=\frac{2sin2x.cosx+2sin2x.cos2x}{2cos^22x+2cos2x.cosx}=\frac{2sin2x\left(cosx+cos2x\right)}{2cos2x\left(cos2x+cosx\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\frac{sin^22x+2cos\left(2\pi+\pi+2x\right)-2}{-3+4cos2x+cos\left(\pi-4x\right)}=\frac{sin^22x-2cos2x-2}{-3+4cos2x-cos4x}=\frac{4sin^2x.cos^2x-2\left(2cos^2x-1\right)-2}{-3+4\left(1-2sin^2x\right)-\left(1-2sin^22x\right)}\)
\(=\frac{4cos^2x\left(sin^2x-1\right)}{-8sin^2x+2sin^22x}=\frac{2cos^2x.\left(-cos^2x\right)}{-4sin^2x+4sin^2x.cos^2x}=\frac{cos^4x}{2sin^2x\left(1-cos^2x\right)}\)
\(=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)
\(P=\frac{sin^2x+cos^2x+2sinx.cosx-1}{\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right).cotx}-\frac{1}{cosx-sinx}\)
\(=\frac{2sinx.cosx}{\left(cosx-sinx\right).\frac{cosx}{sinx}}-\frac{1}{cosx-sinx}=\frac{2sin^2x}{cosx-sinx}-\frac{1}{cosx-sinx}\)
\(=\frac{2sin^2x-1}{cosx-sinx}=\frac{2sin^2x-\left(sin^2x+cos^2x\right)}{cosx-sinx}=\frac{sin^2x-cos^2x}{cosx-sinx}\)
\(=\frac{\left(sinx-cosx\right)\left(sinx+cosx\right)}{cosx-sinx}=-\left(sinx+cosx\right)\)
\(\frac{\sqrt{2}cosx-2cos\left(\frac{\pi}{4}+x\right)}{2sin\left(\frac{\pi}{4}+x\right)-\sqrt{2}sinx}\\ =\frac{cosx-\sqrt{2}cos\left(\frac{\pi}{4}+x\right)}{\sqrt{2}sin\left(\frac{\pi}{4}+x\right)-sinx}\\ =\frac{cosx-\sqrt{2}\left(\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx\right)}{\sqrt{2}\left(\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx\right)-sinx}\\ =\frac{cosx-cosx+sinx}{cosx+sinx-sinx}\\ =\frac{sinx}{cosx}=tanx\)