Use an appropriate 'double angle' formula to simplify:
Problem 1 :
2 sin α cos α
Solution:
= 2 sin α cos α
= sin 2α
Problem 2 :
4 cos α sin α
Solution:
= 4 cos α sin α
= 2 × 2 cos α sin α
= 2 sin 2α
Problem 3 :
sin α cos α
Solution:
= sin α cos α
Multiplying and dividing by 2, we get
= (2/2) sin α cos α
= (1/2) (2sin α cos α)
= (sin 2α/2)
Problem 4 :
2 cos2 β - 1
Solution:
= 2 cos2 β - 1
= cos 2β
Problem 5 :
1 - 2 cos2 ϕ
Solution:
= 1 - 2 cos2 ϕ
= - (2 cos2 ϕ - 1)
= - cos 2ϕ
Problem 6 :
1 - 2 sin2 N
Solution:
= 1 - 2 sin2 N
= cos 2N
Problem 7 :
2 sin2 M - 1
Solution:
= 2 sin2 M - 1
= - (1 - 2 sin2 M)
= - cos 2M
Problem 8:
cos2 α - sin2 α
Solution:
= cos2 α - sin2 α
= cos 2α
Problem 9 :
sin2 α - cos2 α
Solution:
= sin2 α - cos2 α
= - (cos2 α - sin2 α)
= - cos 2α
Problem 10 :
2 sin 2A cos 2A
Solution:
= 2 sin 2A cos 2A
= sin (2 × 2A)
= sin 4A
Problem 11 :
2 cos 3α sin 3α
Solution:
= 2 cos 3α sin 3α
= sin (2 × 3α)
= sin 6α
Problem 12 :
2 cos2 4θ - 1
Solution:
= 2 cos2 4θ - 1
It is in the form of 2 cos2 θ - 1, instead of θ we have 4θ
= cos (2 × 4θ)
= cos 8θ
Problem 13 :
1 - 2 cos2 3β
Solution :
= 1 - 2 cos2 3β
= - (2 cos2 3β - 1)
= - cos (2 × 3β)
= - cos 6β
Problem 14 :
1 - 2 sin2 5α
Solution:
= 1 - 2 sin2 5α
= cos (2 × 5α)
= cos 10α
Problem 15 :
2 sin2 3D - 1
Solution:
= 2 sin2 3D - 1
= - (1 - 2 sin2 3D)
= - cos (2 × 3D)
= - cos 6D
Problem 16 :
cos2 2A - sin2 2A
Solution:
= cos2 2A - sin2 2A
The given question looks like cos2 A - sin2 A, Here A = 2A
cos2 A - sin2 A = cos 2A
cos2 2A - sin2 2A = cos 2(2A)
= cos 4A
Problem 17 :
cos2 (α/2) - sin2 (α/2)
Solution:
= cos2 (α/2) - sin2 (α/2)
The given question looks like cos2 A - sin2 A, Here A = α/2
cos2 A - sin2 A = cos 2A
= cos (2 × α/2)
= cos α
Problem 18 :
2 sin2 3P - 2 cos2 3P
Solution:
= 2 sin2 3P - 2 cos2 3P
Factoring 2
= 2(sin2 3P - cos2 3P)
= -2(cos2 3P - sin2 3P)
= 2(cos(2 × 3P))
= - 2 cos (2 × 3P)
= -2 cos 6P
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM