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\(\frac{2sin^2\frac{x}{2}+sin2x-1}{2sinx-1}+sinx=\frac{1-cosx+2sin2x.cosx-1}{2sinx-1}+sinx\)
\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx\)
\(=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)\)
\(=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(\frac{sin2x-cosx}{2sinx-1}+sinx=\frac{2sinx.cosx-cosx}{2sinx-1}+sinx\)
\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
Ta có 2 công thức: \(\left\{{}\begin{matrix}sinx+cosx=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\\sinx-cosx=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\end{matrix}\right.\)
\(\Rightarrow tan\left(\frac{\pi}{4}-x\right)=-tan\left(x-\frac{\pi}{4}\right)=-\frac{sin\left(x-\frac{\pi}{4}\right)}{cos\left(x-\frac{\pi}{4}\right)}=-\frac{sinx-cosx}{sinx+cosx}\)
\(=\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx-sinx\right)^2}{cos^2x-sin^2x}=\frac{1-2sinx.cosx}{cos2x}=\frac{1-sin2x}{cos2x}\)
Ta có:
\(2\sin^2\frac{x}{2}-1=-\cos x\)
Do đó: \(\frac{2\sin^2\frac{x}{2}+\sin2x-1}{2\sin x-1}+\sin x\)
\(=\frac{-\cos x+2\sin x.\cos x}{2\sin x-1}+\sin x\)
\(=\cos x+\sin x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
Đầu tiên bạn cần biết công thức \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
Ta có:
\(\frac{sinx+cosx+cos2x}{1-sin2x+cos2x+2cosx}=\frac{sinx+cosx+cos^2x-sin^2x}{1-2sinx.cosx+2cos^2x-1+2cosx}\)
\(=\frac{sinx+cosx+\left(cosx-sinx\right)\left(cosx+sinx\right)}{2cos^2x-2sinx.cosx+2cosx}=\frac{\left(sinx+cosx\right)\left(cosx-sinx+1\right)}{2cosx\left(cosx-sinx+1\right)}\)
\(=\frac{sinx+cosx}{2cosx}=\frac{sinx}{2cosx}+\frac{cosx}{2cosx}=\frac{1}{2}tanx+\frac{1}{2}\)
\(\frac{1-2sin^2x}{1-tanx}=\frac{cosx\left(1-2sin^2x\right)}{cosx-sinx}=\frac{cosx\left(cos^2x-sin^2x\right)}{cosx-sinx}=\frac{cosx\left(cosx+sinx\right)\left(cosx-sinx\right)}{cosx-sinx}\)
\(=cosx\left(cosx+sinx\right)=\frac{cosx\left(cosx+sinx\right)^2}{cosx+sinx}=\frac{cos^2x+sin^2x+2sinx.cosx}{1+\frac{sinx}{cosx}}=\frac{1+sin2x}{1+tanx}\)
\(\frac{x}{2}=a\Rightarrow\frac{cot^2a-cot^23a}{cos^2a.cos2a\left(1+cot^23a\right)}=\frac{sin^23a\left(cot^2a-cot^23a\right)}{cos^2a.cos2a}=\frac{sin^23a.cot^2a-cos^23a}{cos^2a.cos2a}\)
\(=\frac{sin^23a.cos^2a-cos^23a.sin^2a}{sin^2a.cos^2a.cos2a}=\frac{\left(sin3a.cosa-cos3a.sina\right)\left(sin3a.cosa+cos3a.sina\right)}{sin^2a.cos^2a.cos2a}\)
\(=\frac{sin\left(3a-a\right).sin\left(3a+a\right)}{sin^2a.cos^2a.cos2a}=\frac{sin2a.sin4a}{sin^2a.cos^2a.cos2a}=\frac{2sina.cosa.4sina.cosa.cos2a}{sin^2a.cos^2a.cos2a}\)
\(=\frac{8sin^2a.cos^2a.cos2a}{sin^2a.cos^2a.cos2a}=8\)
\(sin\left(a+b+a\right)=5sin\left(a+b-a\right)\)
\(\Leftrightarrow sin\left(a+b\right)cosa+cos\left(a+b\right).sina=5sin\left(a+b\right).cosa-5cos\left(a+b\right).sina\)
\(\Leftrightarrow6cos\left(a+b\right).sina=4sin\left(a+b\right).cosa\)
\(\Leftrightarrow\frac{2sin\left(a+b\right)cosa}{cos\left(a+b\right)sina}=3\Leftrightarrow\frac{2tan\left(a+b\right)}{tana}=3\)
\(\frac{1-sin2x}{1+sin2x}=\frac{sin^2x+cos^2x-2sinx.cosx}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(sinx-cosx\right)^2}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left[\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\right]^2}{\left[\sqrt{2}.sin\left(x+\frac{\pi}{4}\right)\right]^2}=tan^2\left(\frac{\pi}{4}-x\right)\)
Bạn coi lại đề, vế phải là tan chứ ko phải cot
\(\frac{sin2x-2sinx}{sin2x+2sinx}=\frac{2sinx.cosx-2sinx}{2sinx.cosx+2sinx}=\frac{2sinx\left(cosx-1\right)}{2sinx\left(cosx+1\right)}\)
\(=\frac{cosx-1}{cos+1}=\frac{1-2sin^2\frac{x}{2}-1}{2cos^2\frac{x}{2}-1+2}=\frac{-2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}=-tan^2\frac{x}{2}\)
Cảm ơn bạn, mình sẽ xem lại.