Question 5:
--- 5(a) ---
5(a) | sin3x=sin ( 2x+x)=sin2xcosx+cos2xsinx | M1
( )
=( 2sinxcosx) cosx+ 1−2sin2 x sinx | dM1
= 2 s i n x c o s 2 x + s i n x − 2 s i n 3 x | A1
( )
= 2 s i n x 1 − s i n 2 x + s i n x − 2 s i n 3 x = 3 s i n x − 4 s i n 3 x | A1
(4)
(b) | ( )
2 s i n 3 = 5 s i n 2  2 3 s i n − 4 s i n 3 = 1 0 s i n c o s       | M1
Divides or takes out s i n as a factor and uses s i n 2 = 1 − c o s 2   to set up
and solve a 3TQ in c o s 
E.g.  6 s i n − 8 s i n 3 = 1 0 s i n c o s  6 −  ( 1 − c o s 2 = 1 0 c o s     ) 
4cos2−5cos−1=0
5 − 4 1
 c o s = = ( − 0 . 1 7 5 . .. ) 
8 | dM1
Any two of the following four answers
s i n = 0  = 1 8 0  , 3 6 0   
5− 41
cos= =awrt100or awrt 260
8 | A1
All of 1 8 0  , 3 6 0  , a w r t 1 0 0 .1  , a w r t 2 5 9 .9  | A1
(4)
Total 8
(b) Allow if solved in terms of another variable, say x
The question states hence or otherwise solve 2 s i n 3 = 5 s i n 2  
In almost all cases they will use their part (a) and proceed by the ‘hence’ route.
‘Otherwise’ routes could start with earlier versions of 2 s i n 3
( )
such as 2 2sinxcos 2 x+sinx−2sin3 x
In almost all cases the main mark scheme provides a framework of how the marks are awarded
M1: Uses their part (a) and a double angle for sin2 of the form G s i n c o s  to form an equation in s i n 
and c o s . Look for R s i n + S s i n 3 = T s i n c o s     o.e. Condone slips such as forgetting to  2
dM1: The mark is scored for dividing or taking out a factor of s i n and using sin2 =1cos2 to set up
and solve a 3TQ in c o s . To solve you should expect to see a valid method used, leading to at least
one value forcos = k,−1 k 1,k  0
If a calculator is used to solve the 3TQ in c o s  the value(s) must be correct for their equation.
Candidates cannot go straight from an equation e.g. 4cos2−5cos−1= 0 to a solution e.g. 100.1
without sight of the value for cosin between.
A1: Finds two angles of 1 8 0  , 3 6 0  , a w r t 1 0 0  , a w r t 2 6 0  following at least one previous M1
The 1 8 0  a n d / o r 3 6 0  must be found via a correct method with s i n = 0  being seen or implied and also
produced from a factorised P s i n + Q s i n 3 = R s i n c o s    
Likewise, the 100and 260must have been achieved via solving a correct equation.
Allow this mark if the answers are given in radians. So, for two angles of , 2 , a w r t 1 . 7 5 , a w r t 4 . 5 4  
A1: All of 1 8 0  , 3 6 0  , a w r t 1 0 0 .1  , a w r t 2 5 9 .9 
PMT
with no extra values within the range following M1, dM1
There are no marks for answers appearing without valid method and working
There are other (usually more complicated) ways of attempting part (b) so you must look carefully through
each response. If you think a response deserves credit and you are unsure of how to mark, please send to
review.