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\(\frac{2sin^2\frac{x}{2}+sin2x-1}{2sinx-1}+sinx=\frac{1-cosx+2sin2x.cosx-1}{2sinx-1}+sinx\)
\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx\)
\(=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)\)
\(=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(\frac{sin2x-cosx}{2sinx-1}+sinx=\frac{2sinx.cosx-cosx}{2sinx-1}+sinx\)
\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
\(\cos^2=\frac{1}{1+tan^2x}=\frac{1}{1+25}\\ \Rightarrow cos=\frac{1}{\sqrt{26}}\left(6\pi< x< \frac{13}{2}\right)\)
\(\Rightarrow sin=\frac{5}{\sqrt{26}}\\ \Rightarrow sin2x=2sinxcosx=2\times\frac{5}{\sqrt{26}}\times\frac{1}{\sqrt{26}}=\frac{5}{13}\)
b) \(cos^2=1-sin^2x=\frac{16}{25}\\ \Rightarrow cos=-\frac{4}{5}\left(-\frac{3\pi}{2}< x< -\pi\right)\\\Rightarrow tanx=-\frac{3}{4} \\ tan\left(x-\frac{\pi}{4}\right)=\frac{tanx-tan\frac{\pi}{4}}{1+tanxtan\frac{\pi}{4}}=-7\)
6π là số chẵn nên viết được dưới dạng k2π nên nó quay về mức 0 còn \(\frac{13\pi}{2}=\frac{\pi}{2}+6\pi\) nên tóm lại nó lằm từ (0<x<\(\frac{\pi}{2}\))
Ta có 2 công thức: \(\left\{{}\begin{matrix}sinx+cosx=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\\sinx-cosx=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\end{matrix}\right.\)
\(\Rightarrow tan\left(\frac{\pi}{4}-x\right)=-tan\left(x-\frac{\pi}{4}\right)=-\frac{sin\left(x-\frac{\pi}{4}\right)}{cos\left(x-\frac{\pi}{4}\right)}=-\frac{sinx-cosx}{sinx+cosx}\)
\(=\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx-sinx\right)^2}{cos^2x-sin^2x}=\frac{1-2sinx.cosx}{cos2x}=\frac{1-sin2x}{cos2x}\)
\(\frac{1-2sin^2x}{1-tanx}=\frac{cosx\left(1-2sin^2x\right)}{cosx-sinx}=\frac{cosx\left(cos^2x-sin^2x\right)}{cosx-sinx}=\frac{cosx\left(cosx+sinx\right)\left(cosx-sinx\right)}{cosx-sinx}\)
\(=cosx\left(cosx+sinx\right)=\frac{cosx\left(cosx+sinx\right)^2}{cosx+sinx}=\frac{cos^2x+sin^2x+2sinx.cosx}{1+\frac{sinx}{cosx}}=\frac{1+sin2x}{1+tanx}\)
\(\frac{x}{2}=a\Rightarrow\frac{cot^2a-cot^23a}{cos^2a.cos2a\left(1+cot^23a\right)}=\frac{sin^23a\left(cot^2a-cot^23a\right)}{cos^2a.cos2a}=\frac{sin^23a.cot^2a-cos^23a}{cos^2a.cos2a}\)
\(=\frac{sin^23a.cos^2a-cos^23a.sin^2a}{sin^2a.cos^2a.cos2a}=\frac{\left(sin3a.cosa-cos3a.sina\right)\left(sin3a.cosa+cos3a.sina\right)}{sin^2a.cos^2a.cos2a}\)
\(=\frac{sin\left(3a-a\right).sin\left(3a+a\right)}{sin^2a.cos^2a.cos2a}=\frac{sin2a.sin4a}{sin^2a.cos^2a.cos2a}=\frac{2sina.cosa.4sina.cosa.cos2a}{sin^2a.cos^2a.cos2a}\)
\(=\frac{8sin^2a.cos^2a.cos2a}{sin^2a.cos^2a.cos2a}=8\)
\(sin\left(a+b+a\right)=5sin\left(a+b-a\right)\)
\(\Leftrightarrow sin\left(a+b\right)cosa+cos\left(a+b\right).sina=5sin\left(a+b\right).cosa-5cos\left(a+b\right).sina\)
\(\Leftrightarrow6cos\left(a+b\right).sina=4sin\left(a+b\right).cosa\)
\(\Leftrightarrow\frac{2sin\left(a+b\right)cosa}{cos\left(a+b\right)sina}=3\Leftrightarrow\frac{2tan\left(a+b\right)}{tana}=3\)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
\(\frac{1-sin2x}{1+sin2x}=\frac{sin^2x+cos^2x-2sinx.cosx}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(sinx-cosx\right)^2}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left[\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\right]^2}{\left[\sqrt{2}.sin\left(x+\frac{\pi}{4}\right)\right]^2}=tan^2\left(\frac{\pi}{4}-x\right)\)
Bạn coi lại đề, vế phải là tan chứ ko phải cot
\(\frac{sin2x-2sinx}{sin2x+2sinx}=\frac{2sinx.cosx-2sinx}{2sinx.cosx+2sinx}=\frac{2sinx\left(cosx-1\right)}{2sinx\left(cosx+1\right)}\)
\(=\frac{cosx-1}{cos+1}=\frac{1-2sin^2\frac{x}{2}-1}{2cos^2\frac{x}{2}-1+2}=\frac{-2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}=-tan^2\frac{x}{2}\)
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