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\(A=1-cos^2x+2cosx+1=3-\left(cosx-1\right)^2\le3\)
\(A_{max}=3\) khi \(cosx=1\)
\(B=1-sin^2x-2sin^2x-3=-1-\left(sinx+1\right)^2\le-1\)
\(B_{max}=-1\) khi \(sinx=-1\)
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{2}-1\right)}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{2}}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{2}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{4}-1\right)}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{4}}}=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{4}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{8}-1\right)}=\sqrt{cos^2\frac{x}{8}}=cos\frac{x}{8}\)
\(B=\sqrt{2+\sqrt{2+\sqrt{2+2\left(2cos^2\frac{a}{2}-1\right)}}}\)
\(=\sqrt{2+\sqrt{2+\sqrt{4cos^2\frac{a}{2}}}}=\sqrt{2+\sqrt{2+2cos\frac{a}{2}}}\)
\(=\sqrt{2+\sqrt{2+2\left(cos^2\frac{a}{4}-1\right)}}=\sqrt{2+\sqrt{4cos^2\frac{a}{4}}}\)
\(=\sqrt{2+2cos\frac{a}{4}}=\sqrt{2+2\left(2cos^2\frac{a}{8}-1\right)}=2cos\frac{a}{8}\)
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosa}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+cos^2\frac{a}{2}-\frac{1}{2}}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{2}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{4}-1\right)}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{4}}=\sqrt{\frac{1}{2}+\frac{1}{2}\left(cos^2\frac{a}{8}-1\right)}\)
\(=cos\frac{a}{8}\Rightarrow n=8\)
a/
\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)
b/
\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)
c/
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)
\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)
d/
\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)
e/
\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)
Các câu c, e đều sử dụng kết quả từ câu b
f/
\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)
\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)
\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)
\(=2.\left(-2sin^2x\right)^2=8sin^4x\)
g/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
h/
\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
i/
\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
j/
\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosa}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{a}{2}}}=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{2}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{4}-1\right)}=\sqrt{cos^2\frac{a}{4}}\)
\(=cos\frac{a}{4}\)
\(\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}\)
\(=\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{\frac{1}{2}+cos^2\frac{a}{2}-\frac{1}{2}}}\)
\(=\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{cos^2\frac{a}{2}}}=\sqrt{\frac{1}{2}-\frac{1}{2}cos\frac{a}{2}}\)
\(=\sqrt{\frac{1}{2}-\frac{1}{2}\left(1-2sin^2\frac{a}{4}\right)}=\sqrt{\frac{1}{2}-\frac{1}{2}+sin^2\frac{a}{4}}\)
\(=\sqrt{sin^2\frac{a}{4}}=sin\frac{a}{4}\)
\(\pi< a< \frac{3\pi}{2}\Rightarrow2\pi< 2a< 3\pi\Rightarrow sin2a>0\)
\(cot2a=\frac{1}{2}\Rightarrow sin2a=\frac{1}{\sqrt{1+cot^22a}}=\frac{2\sqrt{5}}{5}\)
\(cos\left(a+\frac{\pi}{3}\right)+cos\left(a-\frac{\pi}{3}\right)=2cosa.cos\frac{\pi}{3}=cosa\)
\(tan\left(\frac{\pi}{2}-a\right)+tan\left(\frac{\pi}{2}+\frac{a}{2}\right)=\frac{-sin\frac{a}{2}}{cos\left(\frac{\pi}{2}-a\right).cos\left(\frac{\pi}{2}+\frac{a}{2}\right)}=\frac{sin\frac{a}{2}}{sina.sin\frac{a}{2}}=\frac{1}{sina}\)
\(\Rightarrow M=sina.cosa=\frac{1}{2}sin2a=\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow2a+b=7\)
\(\frac{1}{2}+\frac{1}{2}cosx=\frac{1}{2}\left(1+cosx\right)=\frac{1}{2}\left(1+2cos^2\frac{x}{2}-1\right)=cos^2\frac{x}{2}\)
Do \(0< x< \frac{\pi}{2}\Rightarrow cos\frac{x}{k}>0\) \(\forall k\) nguyên dương
\(\Rightarrow A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosx}}}\)
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{2}}}\)
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{4}}\)
\(A=cos\frac{x}{8}\)
\(\Rightarrow\) Với \(n=\pm8\) thì đẳng thức luôn đúng