CAIE P1 2023 June — Question 7 11 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeProve identity then solve
DifficultyStandard +0.3 This is a structured multi-part question that guides students through standard techniques: expanding and using Pythagorean identity, verifying solutions, proving an algebraic identity with trigonometric manipulation, and combining previous results. While it requires careful algebra and multiple steps (11 marks total), each part uses routine A-level methods with clear scaffolding, making it slightly easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities
    1. By first expanding \((\cos \theta + \sin \theta)^2\), find the three solutions of the equation $$(\cos \theta + \sin \theta)^2 = 1$$ for \(0 \leqslant \theta \leqslant \pi\). [3]
    2. Hence verify that the only solutions of the equation \(\cos \theta + \sin \theta = 1\) for \(0 < \theta < \pi\) are \(0\) and \(\frac{1}{2}\pi\). [2]
  2. Prove the identity \(\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 - \cos \theta}{\cos \theta - \sin \theta} \equiv \frac{\cos \theta + \sin \theta - 1}{1 - 2\sin^2 \theta}\). [3]
  3. Using the results of (a)(ii) and (b), solve the equation $$\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 - \cos \theta}{\cos \theta - \sin \theta} = 2(\cos \theta + \sin \theta - 1)$$ for \(0 \leqslant \theta \leqslant \pi\). [3]
Question 7:
AnswerMarks Guidance
7(a)(i)cos22sincossin21 leading to 2sincos0 or sin20 *B1
expanding and www.
π
 0 , ,π
AnswerMarks
2DB
2,1,0B2 for three correct answers only.
B1 for two correct answers and one incorrect or 3 correct
answers plus other values in the range.
SC DB1 for correct 3 answers in degrees and no others.
Ignore extras outside of the range and allow decimal
equivalents.
AnswerMarks
3Verifying 3 answers rather than expanding and solving 0/3.
AnswerMarks
7(a)(ii)π π
cos0sin0101 and cos sin 011
AnswerMarks Guidance
2 2B1 Checking both correct values. Do not allow solving an
equation.
Condone use of 90 degrees.
AnswerMarks Guidance
cosπsinπ101 or ≠ 1B1 www
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(b)cossinsincossin1cos
cossincossinM1 Correct common denominator and correct products in the
numerator and no missing terms. Correct factors in the
denominator can be implied by cos2sin2. Condone
brackets missing if recovered.
cossin  sin2coscos2sin sincos
AnswerMarks
cos2sin2A1
sincoscos2sin2 cossin1
 
AnswerMarks Guidance
cos2sin2 12sin2A1 AG
Clear evidence of using sin2cos21 in either the
numerator or denominator. Condone c, s and/or omission
of .
Working from both sides of the identity and correctly
arriving at the same expression can score M1A1. A final
statement is then required for the A1.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(c)cossin1
2cossin1
12sin2
AnswerMarks Guidance
leading to 1 2(12sin2)*M1 Replacing LHS with the expression from (b) and
attempting to simplify i.e. condone omission of
cossin10 at this stage.
M0 for 0 = 2(12sin2)
1 or 3
ksin2=1 or 3 leading to sin
k
 1
4sin21 leading to sin
 
AnswerMarks Guidance
 2DM1 Dividing by k and taking the square root of a positive value
< 1.
1 5
This mark can be implied by the solutions π, π.
6 6
1 1 5
Solutions 0, π, π , π
AnswerMarks Guidance
6 2 6A1 Allow 0, 0.524, 1.57, 2.62 AWRT.
1
If M0 SCB1 for cossin10 ⇒ 0, π.
2
If M0 SCB1 for all four correct answers and no others.
Ignore answers outside of the range.
Answers in degrees A0.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(a)(i) ---
7(a)(i) | cos22sincossin21 leading to 2sincos0 or sin20 | *B1 | Or arriving at cos0 or sin0 or tan0 after first
expanding and www.
π
 0 , ,π
2 | DB
2,1,0 | B2 for three correct answers only.
B1 for two correct answers and one incorrect or 3 correct
answers plus other values in the range.
SC DB1 for correct 3 answers in degrees and no others.
Ignore extras outside of the range and allow decimal
equivalents.
3 | Verifying 3 answers rather than expanding and solving 0/3.
--- 7(a)(ii) ---
7(a)(ii) | π π
cos0sin0101 and cos sin 011
2 2 | B1 | Checking both correct values. Do not allow solving an
equation.
Condone use of 90 degrees.
cosπsinπ101 or ≠ 1 | B1 | www
2
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | cossinsincossin1cos
cossincossin | M1 | Correct common denominator and correct products in the
numerator and no missing terms. Correct factors in the
denominator can be implied by cos2sin2. Condone
brackets missing if recovered.
cossin  sin2coscos2sin sincos

cos2sin2 | A1
sincoscos2sin2 cossin1
 
cos2sin2 12sin2 | A1 | AG
Clear evidence of using sin2cos21 in either the
numerator or denominator. Condone c, s and/or omission
of .
Working from both sides of the identity and correctly
arriving at the same expression can score M1A1. A final
statement is then required for the A1.
3
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | cossin1
2cossin1
12sin2
leading to 1 2(12sin2) | *M1 | Replacing LHS with the expression from (b) and
attempting to simplify i.e. condone omission of
cossin10 at this stage.
M0 for 0 = 2(12sin2)
1 or 3
ksin2=1 or 3 leading to sin
k
 1
4sin21 leading to sin
 
 2 | DM1 | Dividing by k and taking the square root of a positive value
< 1.
1 5
This mark can be implied by the solutions π, π.
6 6
1 1 5
Solutions 0, π, π , π
6 2 6 | A1 | Allow 0, 0.524, 1.57, 2.62 AWRT.
1
If M0 SCB1 for cossin10 ⇒ 0, π.
2
If M0 SCB1 for all four correct answers and no others.
Ignore answers outside of the range.
Answers in degrees A0.
3
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item By first expanding $(\cos \theta + \sin \theta)^2$, find the three solutions of the equation
$$(\cos \theta + \sin \theta)^2 = 1$$
for $0 \leqslant \theta \leqslant \pi$. [3]

\item Hence verify that the only solutions of the equation $\cos \theta + \sin \theta = 1$ for $0 < \theta < \pi$ are $0$ and $\frac{1}{2}\pi$. [2]
\end{enumerate}

\item Prove the identity $\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 - \cos \theta}{\cos \theta - \sin \theta} \equiv \frac{\cos \theta + \sin \theta - 1}{1 - 2\sin^2 \theta}$. [3]

\item Using the results of (a)(ii) and (b), solve the equation
$$\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 - \cos \theta}{\cos \theta - \sin \theta} = 2(\cos \theta + \sin \theta - 1)$$
for $0 \leqslant \theta \leqslant \pi$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2023 Q7 [11]}}
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