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\(2sin\left(\frac{\pi}{4}+a\right)sin\left(\frac{\pi}{4}-a\right)=cos2a-cos\left(\frac{\pi}{2}\right)=cos2a\)
\(tanx-\frac{1}{tanx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}=\frac{sin^2x-cos^2x}{sinx.cosx}=-\frac{2\left(cos^2x-sin^2x\right)}{2sinx.cosx}=\frac{2cos2x}{sin2x}=-2cot2x=-\frac{2}{tan2x}\)
\(\pi< a< \frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\) \(\Rightarrow sin2a=2sina.cosa>0\)
\(\Rightarrow sin2a=\sqrt{1-cos^22a}=\frac{3\sqrt{7}}{8}\)
\(cos2a=1-2sin^2a=\frac{1}{8}\)
\(\Leftrightarrow sin^2a=\frac{7}{16}\Rightarrow sina=-\frac{\sqrt{7}}{4}\)
\(\Rightarrow M=\frac{-\frac{\sqrt{7}}{4}-\frac{3\sqrt{7}}{8}}{-\frac{\sqrt{7}}{4}+\frac{3\sqrt{7}}{8}}=...\)
\(sinx\left(1-tan^2\frac{x}{2}\right)=sinx\left(1-\frac{sin^2\frac{x}{2}}{cos^2\frac{x}{2}}\right)=sinx\left(1-\frac{1-cosx}{1+cosx}\right)\)
\(=sinx\left(\frac{1+cosx-\left(1-cosx\right)}{1+cosx}\right)=\frac{2sinx.cosx}{1+cosx}\)
\(1-sin2x.sin3x-cos2x.cos3x=1-\left(cos3x.cos2x+sin3x.sin2x\right)=1-cos\left(3x-2x\right)=1-cosx\)
\(\Rightarrow\frac{1-sin2x.sin3x-cos2x.cos3x}{sinx\left(1-tan^2\frac{x}{2}\right)}=\frac{1-cosx}{\frac{2sinx.cosx}{1+cosx}}=\frac{\left(1-cosx\right)\left(1+cosx\right)}{2sinx.cosx}\)
\(=\frac{1-cos^2x}{2sinx.cosx}=\frac{sin^2x}{2sinx.cosx}=\frac{sinx}{2cosx}=\frac{1}{2}tanx\)
\(\frac{\pi}{2}< a< \pi\Rightarrow\pi< 2a< 2\pi\)
Mà \(tan2a< 0\) \(\Rightarrow\frac{3\pi}{2}< 2a< 2\pi\Rightarrow cos2a>0\)
\(\Rightarrow cos2a=\frac{1}{\sqrt{1+tan^22a}}=\frac{3}{5}\)
\(tan\left(2a+\frac{\pi}{4}\right)=\frac{tan2a+tan\frac{\pi}{4}}{1-tan2a.tan\frac{\pi}{4}}=\frac{-\frac{4}{3}+1}{1+\frac{4}{3}}=...\)
--.-- \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ
\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)
\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)
\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)
\(\cos2a=2\cos^2a-1=\frac{7}{25}\)
\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)
\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)
\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)
\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)
\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)
Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)
\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)
\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(cos^2\left(a-\frac{\pi}{4}\right)-sin^2\left(a-\frac{\pi}{4}\right)=cos\left(2a-\frac{\pi}{2}\right)\)
\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)
\(tan^2\left(x-a\right)+tan^2\left(x+a\right)=\frac{sin^2\left(x-a\right)}{cos^2\left(x-a\right)}+\frac{sin^2\left(x+a\right)}{cos^2\left(x+a\right)}\)
\(=\frac{sin^2\left(x-a\right).cos^2\left(x+a\right)+sin^2\left(x+a\right).cos^2\left(x-a\right)}{cos^2\left(x-a\right).cos^2\left(x+a\right)}\)
\(=\frac{\left(sin2x-sin2a\right)^2+\left(sin2x+sin2a\right)^2}{\left(cos2x+cos2a\right)^2}\)
\(=\frac{sin^22x-2sin2x.sin2a+sin^22a+sin^22x+2sin2x.sin2a+sin^22a}{\left(cos2x+cos2a\right)^2}\)
\(=\frac{2\left(sin^22x+sin^22a\right)}{\left(cos2x+cos2a\right)^2}\)
\(P=sin^22a+cos^22a+sin^22b+cos^22b+2sin2a.sin2b+2cos2a.cos2b\)
\(P=2+2\left(sin2a.sin2b+cos2a.cos2b\right)=2+2cos\left(2a-2b\right)\)
\(P=2+2cos\frac{\pi}{3}=3\)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
\(A=\frac{cos^2a-sin^2a}{2sin^2a+3sina.cosa}=\frac{\frac{cos^2a}{cos^2a}-\frac{sin^2a}{sin^2a}}{\frac{2sin^2a}{cos^2a}+\frac{3sina.cosa}{cos^2a}}=\frac{1-tan^2a}{2tan^2a+3tana}=\frac{1-2^2}{2.2^2+3.2}=...\)
\(\frac{1-2sin2a+cos2a}{1+2sin2a+cos2a}=\frac{1-4sina.cosa+2cos^2a-1}{1+4sina.cosa+2cos^2a-1}=\frac{2cosa\left(cosa-2sina\right)}{2cosa\left(cosa+2sina\right)}\)
\(=\frac{cosa-2sina}{cosa+2sina}\)
Bạn coi lại đề, muốn ra được biểu thức vế phải thì trước sin2a không được có số 2