Q. 14

# Prove that: cos 5

Answer :

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

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