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\(\dfrac{sin2\alpha+sin5\alpha-sin3\alpha}{1+cos\alpha-2sin^22\alpha}\)
\(=\dfrac{2cos\dfrac{5\alpha}{2}.sin\left(-\dfrac{\alpha}{2}\right)+2sin\dfrac{5\alpha}{2}.cos\dfrac{5\alpha}{2}}{cos4\alpha+cos\alpha}\)
\(=\dfrac{2cos\dfrac{5\alpha}{2}.\left(sin\dfrac{5\alpha}{2}-sin\dfrac{\alpha}{2}\right)}{2cos\dfrac{5\alpha}{2}.cos\dfrac{3\alpha}{2}}\)
\(=\dfrac{4cos\dfrac{5\alpha}{2}.cos\dfrac{3\alpha}{2}.sin\alpha}{2cos\dfrac{5\alpha}{2}.cos\dfrac{3\alpha}{2}}\)
\(=2sin\alpha\)
a) \(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=\dfrac{sin2\alpha}{cos2\alpha}:\left(\dfrac{sin4\alpha}{cos4\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\right)\)
\(=\dfrac{sin2\alpha}{cos2\alpha}:\dfrac{sin4\alpha cos2\alpha-sin2\alpha cos4\alpha}{cos4\alpha cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos2\alpha}.\dfrac{cos4\alpha.cos2\alpha}{sin2\alpha}=cos4\alpha\).
b) \(\sqrt{1+sin\alpha}-\sqrt{1-sin\alpha}=\sqrt{sin^2\dfrac{\alpha}{2}+2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)\(-\sqrt{sin^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)
\(=\sqrt{\left(sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right)^2}-\sqrt{\left(sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right)^2}\)
\(=\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên \(0< \alpha< \dfrac{\pi}{4}\).
Trong \(\left(0;\dfrac{\pi}{4}\right)\) thì \(sin\dfrac{\alpha}{2}\) tăng dần từ 0 tới \(\dfrac{\sqrt{2}}{2}\) và \(cos\dfrac{\alpha}{2}\) giảm dần từ 1 tới \(\dfrac{\sqrt{2}}{2}\) nên \(\left|sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right|=-\left(sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right)=cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\).
Vì vậy:
\(\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
\(=sin\dfrac{\alpha}{4}+cos\dfrac{\alpha}{4}-\left(cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\right)=2sin\dfrac{\alpha}{4}\).
\(\dfrac{cosa+cos5a+cos3a}{sina+sin5a+sin3a}=\dfrac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}\)
\(=\dfrac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\dfrac{cos3a}{sin3a}=cot3a\)
\(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\left(\dfrac{1-cos4b}{cos\left(a-b\right)}\right)=\dfrac{\left(cosa.cosb+sina.sinb\right)}{sinb.cosb}.\dfrac{2sin^22b}{cos\left(a-b\right)}\)
\(=\dfrac{cos\left(a-b\right)}{\dfrac{1}{2}sin2b}.\dfrac{2sin^22b}{cos\left(a-b\right)}=4sin2b\)
\(\frac{1-cosa}{1-cos^2a}-\frac{1}{1+cosa}=\frac{1-cosa}{\left(1-cosa\right)\left(1+cosa\right)}-\frac{1}{1+cosa}=\frac{1}{1+cosa}-\frac{1}{1+cosa}=0\)
\(\frac{1-sin^2a.cos^2a}{cos^2a}-cos^2a=\frac{1}{cos^2a}-\frac{sin^2a.cos^2a}{cos^2a}-cos^2a\)
\(=\frac{1}{cos^2a}-\left(sin^2a+cos^2a\right)=\frac{1}{cos^2a}-1\)
\(=\frac{1-cos^2a}{cos^2a}=\frac{sin^2a}{cos^2a}=tan^2a\)
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
\(P=\frac{sin2a+2cos4a.sina}{cos4a+cosa}=\frac{2sina.cosa+2sina.cos4a}{cos4a+cosa}=\frac{2sina\left(cosa+cos4a\right)}{cos4a+cosa}=2sina\)