\(A=\dfrac{sina+sin3a+sin5a}{cosa+cos3a+cos5a}\)
\(A=\dfrac{2sin3a.cosa+sin3a}{2cos3a.cosa+cos3a}\)
\(A=\dfrac{sin3a\left(2cosa+1\right)}{cos3a\left(2cosa+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
8B = 8sinx . cosx . cos2x . cos4x
⇔ 8B = 4sin2x . cos2x . cos4x
⇔ 8B = 2sin4x . cos4x
⇔ 8B = sin8x
⇔ B = \(\dfrac{1}{8}.sin8x\)
\(C=\dfrac{tana-sina}{tana+sina}=\dfrac{\dfrac{tana}{tana}-\dfrac{sina}{tana}}{\dfrac{tana}{tana}+\dfrac{sina}{tana}}=\dfrac{1-cosa}{1+cosa}\)
\(tan^2\dfrac{a}{2}=\dfrac{sin^2\dfrac{a}{2}}{cos^2\dfrac{a}{2}}=\dfrac{\dfrac{1-cosa}{2}}{\dfrac{1+cosa}{2}}=\dfrac{1-cosa}{1+cosa}\)
Vậy ta có \(C=\dfrac{tana-sina}{tana+sina}=tan^2\dfrac{a}{2}\)
(4) \(D=sin\left(x+\dfrac{\pi}{4}\right)-sin\left(\dfrac{\pi}{4}-x\right)\)
\(D=2cos\left(\dfrac{x+\dfrac{\pi}{4}+\dfrac{\pi}{4}-x}{2}\right).sin\left(\dfrac{x+\dfrac{\pi}{4}-\dfrac{\pi}{4}+x}{2}\right)\)
\(D=2cos\dfrac{\pi}{4}.sinx=\sqrt{2}sinx\)
