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\(Q=2\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+2=2\left(\frac{sina.cosa+sina}{cosa\left(cosa+1\right)}\right)^2+2\)
\(=2\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+2=2tan^2a+2=2\left(1+tan^2a\right)=\frac{2}{cos^2a}\)
\(A=\frac{sin2a+2cos4a.sina}{cos4a+cosa}=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)
\(A=\frac{sin2a+sin5a-sin3a}{1+cosa-2sin^22a}=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)
\(A=2sin2x.cos2x.cos4x=sin4x.cos4x=\frac{1}{2}sin8x\)
\(B=sin^4x+cos^6x-6sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x\)
\(=1-2\left(2sinx.cosx\right)^2=1-2sin^22x=cos4x\)
\(C=\frac{cos2a+1-2cos^22a}{2sin2a.cos2a+sin2a}=\frac{\left(1-cos2a\right)\left(2cos2a+1\right)}{sin2a\left(2cos2a+1\right)}=\frac{1-cos2a}{sin2a}\)
\(=\frac{1-\left(1-2sin^2a\right)}{2sina.cosa}=\frac{2sin^2a}{2sina.cosa}=\frac{sina}{cosa}=tana\)
\(D=\frac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}=\frac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\frac{cos3a}{sin3a}=cot3a\)
\(E=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)-\frac{1}{2}+\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)\)
\(=\frac{1}{2}\left[cos\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\right]=-sin\frac{\pi}{4}.sinx=-\frac{\sqrt{2}}{2}sinx\)
\(A=\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(B=\frac{sin^2a\left(1+tan^2a\right)}{cos^2a\left(1+cot^2a\right)}=\frac{sin^2a.\frac{1}{cos^2a}}{cos^2a.\frac{1}{sin^2a}}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(\left(\frac{1}{cos2x}+1\right)tanx=\left(\frac{cos2x+1}{cos2x}\right).\frac{sinx}{cosx}=\frac{2cos^2x}{cos2x}.\frac{sinx}{cosx}\)
\(=\frac{2sinx.cosx}{cos2x}=\frac{sin2x}{cos2x}=tan2x\)
\(\frac{cos7a+cosa+cos5a+cos3a}{sin7a+sina+sin5a+sin3a}=\frac{2cos4a.cos3a+2cos4a.cosa}{2sin4a.cos3a+2sin4a.cosa}\)
\(=\frac{cos4a\left(2cos3a+2cosa\right)}{sin4a\left(2cos3a+2cosa\right)}=\frac{cos4a}{sin4a}=cot4a\)
a, \(\frac{sin2a+cosa}{2sina+1}=\frac{2sinacosa+cóa}{2sina+1}\)= \(\frac{cosa\left(2sina+1\right)}{2sina+1}\)= cos a (đpcm)
b, P= \(\frac{\left(sin^2x-cos^2x\right)\left(sin^2+cos^2x\right).\left(sin^2+2sinx.cosx+cos^2x-1\right)}{1+2cos2x-1}\)
= \(\frac{\left(sin^2x-cos^2x\right).2sinx.cosx}{2cos2x}\)
= \(\frac{-cos2x.sin2x}{2.cos2x}\)= -1/2 sin 2x
#mã mã#
\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(cos^2\left(a-\frac{\pi}{4}\right)-sin^2\left(a-\frac{\pi}{4}\right)=cos\left(2a-\frac{\pi}{2}\right)\)
\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)
\(A=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)