Question 13:
--- 13(a) ---
13(a) | Expands
(cos θ + isin θ)3
PI correct real part | 1.1a | M1 | cos3 θ + isin3 θ = (cos θ + isin θ)3
= cos3 θ + 3icos2 θsin θ – 3cos θsin2 θ – isin3 θ
Equating real parts
cos3 θ = cos3 θ – 3cos θsin2 θ
= cos3 θ – 3cos θ(1 – cos2 θ)
= 4cos3 θ – 3cos θ
Equates real parts
and uses
sin2 θ = 1 – cos2 θ | 1.1a | M1
Completes a
reasoned
argument using de
Moivre’s theorem
to show
cos3 θ
= 4cos3 θ – 3cos θ
AG | 2.1 | R1
Subtotal | 3
Q | Marking
instructions | AO | Marks | Typical solution
--- 13(b) ---
13(b) | Equates imaginary
parts and uses
cos2 θ = 1 – sin2 θ | 1.1a | M1 | Equating imaginary parts
sin3 θ = 3cos2 θsin θ – sin3 θ
= 3(1 – sin2 θ)sin θ – sin3 θ
= 3sin θ – 4sin3 θ
Obtains
3sin θ – 4sin3 θ | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 13(c) ---
13(c) | Substitutes expressions
for cos3θ and
their sin3θ into
c o s 3 θ
c o t 3 θ = or
s in 3 θ
s in 3 θ
t a n 3 θ =
c o s 3 θ | 1.1b | B1F | θ c o s 3 3 θ – 4 c o s 3 c o s θ
θ c o t 3 = =
θ s in 3 3 θ – θ 3 s in 4 s in
( c 4 o s 3 θ )  θ c o s – 3  
3 θ s in 3 θ s in 
=
 s in  θ 3  θ s in 
3   – 4 
3 θ s in  3 θ s in 
4 3 θ – c o t 3 2 θ θ c o t c o s e c
=
3 c o s 2 θ – e c 4
( )
4 c 3 θ – o t 3 2 θ + θ c o t 1 c o t
=
( )
+ 3 1 2 θ – c o t 4
c 3 θ – θ o t 3 c o t
c θ = o t 3
2 θ – 3 c o t 1
Manipulates their rational
function of s in θ and
c o s θ to obtain at least
one instance of
c o t θ or t a n θ | 3.1a | M1
Manipulates their rational
function of s in θ and
cosθ to obtain only cotθ
(and c o s e c θ ) terms | 3.1a | M1
Completes a reasoned
argument from the final
or intermediate results in
parts (a) and (b) to show
c o t 3 θ – 3 c o θ t
c o t 3 θ =
3 c o t 2 θ – 1
AG | 2.1 | R1
Subtotal | 4
Question total | 9
Q | Marking instructions | AO | Marks | Typical solution