CAIE P3 2018 November — Question 6 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2018
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeSolve reciprocal trig equation
DifficultyChallenging +1.2 This is a structured two-part question requiring conversion of reciprocal trig functions to harmonic form (a guided process) followed by routine solving. Part (i) involves algebraic manipulation with cosec and cot, then matching to R sin(x-α) form—moderately technical but scaffolded. Part (ii) is standard equation solving once the form is established. More challenging than basic trig equations due to reciprocal functions and harmonic form, but the 'show that' structure provides significant guidance, making it above average but not requiring deep insight.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals

6
  1. Show that the equation ( \(\sqrt { } 2\) ) \(\operatorname { cosec } x + \cot x = \sqrt { } 3\) can be expressed in the form \(R \sin ( x - \alpha ) = \sqrt { } 2\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
  2. Hence solve the equation \(( \sqrt { } 2 ) \operatorname { cosec } x + \cot x = \sqrt { } 3\), for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

Question 6(i):
AnswerMarks Guidance
AnswerMark Guidance
Rearrange in the form \(\sqrt{3}\sin x - \cos x = \sqrt{2}\)B1
State \(R = 2\)B1
Use trig formulae to obtain \(\alpha\)M1
Obtain \(\alpha = 30°\) with no errors seenA1
Question 6(ii):
AnswerMarks Guidance
AnswerMark Guidance
Evaluate \(\sin^{-1}\!\left(\frac{\sqrt{2}}{R}\right)\)B1ft
Carry out a correct method to find a value of \(x\) in the given intervalM1
Obtain answer \(x = 75°\)A1
Obtain a second answer e.g. \(x = 165°\) and no others [Treat answers in radians as a misread. Ignore answers outside the given interval.]A1ft
## Question 6(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Rearrange in the form $\sqrt{3}\sin x - \cos x = \sqrt{2}$ | B1 | |
| State $R = 2$ | B1 | |
| Use trig formulae to obtain $\alpha$ | M1 | |
| Obtain $\alpha = 30°$ with no errors seen | A1 | |

---

## Question 6(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Evaluate $\sin^{-1}\!\left(\frac{\sqrt{2}}{R}\right)$ | B1ft | |
| Carry out a correct method to find a value of $x$ in the given interval | M1 | |
| Obtain answer $x = 75°$ | A1 | |
| Obtain a second answer e.g. $x = 165°$ and no others [Treat answers in radians as a misread. Ignore answers outside the given interval.] | A1ft | |

---
6 (i) Show that the equation ( $\sqrt { } 2$ ) $\operatorname { cosec } x + \cot x = \sqrt { } 3$ can be expressed in the form $R \sin ( x - \alpha ) = \sqrt { } 2$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$.\\

(ii) Hence solve the equation $( \sqrt { } 2 ) \operatorname { cosec } x + \cot x = \sqrt { } 3$, for $0 ^ { \circ } < x < 180 ^ { \circ }$.\\

\hfill \mbox{\textit{CAIE P3 2018 Q6 [8]}}