Edexcel P2 2022 June — Question 5 6 marks

Exam BoardEdexcel
ModuleP2 (Pure Mathematics 2)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeRational trig expressions
DifficultyStandard +0.3 This is a standard trigonometric equation requiring the technique of dividing by cos to convert tan to sin/cos, then solving the resulting equation. It involves multiple steps (rewriting tan, rearranging to quadratic in sin, solving, finding angles in range) but uses routine A-level methods without requiring novel insight. The angle shift of 43° adds minor complexity but is standard bookwork. Slightly above average due to the multi-step nature and need to carefully handle the range, but well within typical P2 expectations.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Solve, for \(-180° < \theta \leq 180°\), the equation $$3\tan(\theta + 43°) = 2\cos(\theta + 43°)$$ [6]

Question 5:
AnswerMarks
5s i n (( 4 3 ))  + 
3 t a n ( 4 3 ) 2 c o s ( 4 3 ) 3 2 c o s ( 4 3 )    +  = +   = + 
c o s 4 3  + 
AnswerMarks
3 s i n ( 4 3 ) 2 c o s 2 ( 4 3 )    +  = + M1
3 s i n ( 4 3 ) 2 ( 1 s i n 2 ( 4 3 ) )    +  = − + M1
2 s i n 2 ( 4 3 ) 3 s i n ( 4 3 ) 2 0    +  + +  − =
AnswerMarks
( 2 s i n ( 4 3 ) 1 ) ( s i n ( 4 3 ) 2 ) 0 s i n ( 4 3 ) . ..     +  − +  + =  +  =M1
1
sin(+43)=
AnswerMarks
2A1
" 1 "
a r c s i n 4 3  = − 
AnswerMarks
2M1
1 3 , 1 0 7  = −  A1
(6)
(6 marks)
Notes:
sin..
M1: Uses tan..= and multiplies through to form an equation of the form A s i n ... = B c o s 2 ...
cos..
Condone poor notation e.g.:
sin
3tan(+43)=2cos(+43)3 (+43)=2cos(+43)
cos
3 s i n ( 4 3 ) 2 c o s 2 ( 4 3 )    +  = +  (with or without brackets)
M1: Applies Pythagorean identity to obtain a 3 term quadratic equation in sin.
Allow use of cos2...=1sin2...
M1: Solves a 3 term quadratic in sin(+43) by any valid means.
This may be implied by at least one correct root for their quadratic.
Allow if they have sin (+43)= x or another variable or e.g. sin α where 4 3   = + 
A1: Correct value of s i n ( 4 3 )  +  . If s i n ( 4 3 ) x  +  = is used, it must be clear they mean
s i n ( 4 3 )  +  but this may be implied if they have e.g. sin α = ½ where 4 3   = + 
1
If x = is left as a final answer it is A0.
2
AnswerMarks Guidance
M1: Correct method for solving s i n ( 4 3 ) k ,k 1  +  =  , look for use of inverse sine followed by
−1(their
subtraction of 43 from sin k). Implied by one correct solution for their k
Do not allow mixing of degrees and radians for this mark.
A1: Correct solutions and no others in the range.
AnswerMarks Guidance
QuestionScheme Marks
Question 5:
5 | s i n (( 4 3 ))  + 
3 t a n ( 4 3 ) 2 c o s ( 4 3 ) 3 2 c o s ( 4 3 )    +  = +   = + 
c o s 4 3  + 
3 s i n ( 4 3 ) 2 c o s 2 ( 4 3 )    +  = +  | M1
3 s i n ( 4 3 ) 2 ( 1 s i n 2 ( 4 3 ) )    +  = − +  | M1
2 s i n 2 ( 4 3 ) 3 s i n ( 4 3 ) 2 0    +  + +  − =
( 2 s i n ( 4 3 ) 1 ) ( s i n ( 4 3 ) 2 ) 0 s i n ( 4 3 ) . ..     +  − +  + =  +  = | M1
1
sin(+43)=
2 | A1
" 1 "
a r c s i n 4 3  = − 
2 | M1
1 3 , 1 0 7  = −   | A1
(6)
(6 marks)
Notes:
sin..
M1: Uses tan..= and multiplies through to form an equation of the form A s i n ... = B c o s 2 ...
cos..
Condone poor notation e.g.:
sin
3tan(+43)=2cos(+43)3 (+43)=2cos(+43)
cos
3 s i n ( 4 3 ) 2 c o s 2 ( 4 3 )    +  = +  (with or without brackets)
M1: Applies Pythagorean identity to obtain a 3 term quadratic equation in sin.
Allow use of cos2...=1sin2...
M1: Solves a 3 term quadratic in sin(+43) by any valid means.
This may be implied by at least one correct root for their quadratic.
Allow if they have sin (+43)= x or another variable or e.g. sin α where 4 3   = + 
A1: Correct value of s i n ( 4 3 )  +  . If s i n ( 4 3 ) x  +  = is used, it must be clear they mean
s i n ( 4 3 )  +  but this may be implied if they have e.g. sin α = ½ where 4 3   = + 
1
If x = is left as a final answer it is A0.
2
M1: Correct method for solving s i n ( 4 3 ) k , | k | 1  +  =  , look for use of inverse sine followed by
−1(their
subtraction of 43 from sin k). Implied by one correct solution for their k
Do not allow mixing of degrees and radians for this mark.
A1: Correct solutions and no others in the range.
Question | Scheme | Marks
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Solve, for $-180° < \theta \leq 180°$, the equation
$$3\tan(\theta + 43°) = 2\cos(\theta + 43°)$$
[6]

\hfill \mbox{\textit{Edexcel P2 2022 Q5 [6]}}