Question

Chứng minh đẳng thức
a) \(\dfrac{cos\alpha+cos3\alpha+cos5\alpha}{sin\alpha+sin3\alpha+sin5\alpha}=cot3\alpha\)

b) \(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\dfrac{1-cos4b}{cos\left(a-b\right)}=4sin2b\)

Answer 1

\(\dfrac{cosa+cos5a+cos3a}{sina+sin5a+sin3a}=\dfrac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}\)

\(=\dfrac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\dfrac{cos3a}{sin3a}=cot3a\)

\(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\left(\dfrac{1-cos4b}{cos\left(a-b\right)}\right)=\dfrac{\left(cosa.cosb+sina.sinb\right)}{sinb.cosb}.\dfrac{2sin^22b}{cos\left(a-b\right)}\)

\(=\dfrac{cos\left(a-b\right)}{\dfrac{1}{2}sin2b}.\dfrac{2sin^22b}{cos\left(a-b\right)}=4sin2b\)