Prove that: cos 5A = 16 cos5 A – 20 cos3 A + 5 cos A.
Formula: - (i) sin2A + cos2A = 1
(ii) cos2A = 2cos2 A – 1
(ii) cos3A = 4cos3 A – 3cosA
(iii) sin3A = 3sinA – 4sin3 A
taking L.H.S
Cos(5A) = cos(3A + 2A)
⇒ Cos5A = cos3Acos2A – sin3Asin2A
using formula (iii) and (ii)
⇒ Cos5A = (4cos3 A – 3cosA)(2cos2 A – 1) – (3sinA – 4sin2 A)(2sinAcosA)
⇒ Cos5A = (4cos3 A – 3cosA)(2cos2 A – 1) – (3 – 4 sin2 A) (2sin2 A cos A)
⇒ Cos5A = (4cos3 A – 3cosA)(2cos2 A – 1) – (3 – 4(1 – cos2 A)) 2 (1 – cos2 A) cosA
⇒ Cos5A = (8 cos5 A – 10 cos3 A + 3 cosA) – 2 cosA(1 – cos2 A)(4cos2 A – 1)
⇒ Cos5A = (8cos5 A – 10cos3 A + 3cosA) – 2cosA (5cos2 A – 4cos4 A – 1)
⇒ Cos5A = (8cos5 A – 10cos3 A + 3cosA) – 2cosA(1 – cos2 A)(4cos2 A – 1)
⇒ cos5A = 16cos5 A – 20cos3 A + 5cos A
L.H.S = R.H.S
Hence, PROVED