Edexcel AS Paper 1 2019 June — Question 12 7 marks

Exam BoardEdexcel
ModuleAS Paper 1 (AS Paper 1)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrigonometric equations in context
TypeProve trig identity then solve
DifficultyStandard +0.8 Part (a) requires algebraic manipulation with the Pythagorean identity to prove a non-trivial trigonometric identity involving division. Part (b) then requires using this result to solve a complex equation involving both sin and cos terms, requiring rearrangement into a quadratic in cos x and finding all solutions in the given range. This is significantly harder than routine trig equation solving due to the multi-step algebraic manipulation and the need to strategically use the proven identity.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
  1. (a) Show that
$$\frac { 10 \sin ^ { 2 } \theta - 7 \cos \theta + 2 } { 3 + 2 \cos \theta } \equiv 4 - 5 \cos \theta$$ (b) Hence, or otherwise, solve, for \(0 \leqslant x < 360 ^ { \circ }\), the equation $$\frac { 10 \sin ^ { 2 } x - 7 \cos x + 2 } { 3 + 2 \cos x } = 4 + 3 \sin x$$
Question 12:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\frac{10\sin^2\theta - 7\cos\theta + 2}{3+2\cos\theta} \equiv \frac{10(1-\cos^2\theta)-7\cos\theta+2}{3+2\cos\theta}\)M1 Uses identity \(\sin^2\theta = 1-\cos^2\theta\) within the fraction
\(\equiv \frac{12-7\cos\theta - 10\cos^2\theta}{3+2\cos\theta}\)A1 Correct simplified expression in just \(\cos\theta\)
\(\equiv \frac{(3+2\cos\theta)(4-5\cos\theta)}{3+2\cos\theta}\)M1 Correct attempt to factorise numerator. Allow \(u=\cos\theta\)
\(\equiv 4-5\cos\theta\)A1* Fully correct proof with correct notation and no errors
Alternative proof: Multiplies across and forms 3TQ in \(\cos\theta\) on rhs:
\((4-5\cos\theta)(3+2\cos\theta) \Rightarrow 10\sin^2\theta-7\cos\theta+2 = A\cos^2\theta+B\cos\theta+C\)
AnswerMarks Guidance
\(10\sin^2\theta-7\cos\theta+2 = -10\cos^2\theta-7\cos\theta+12\)A1 Correct identity formed
Uses \(\cos^2\theta=1-\sin^2\theta\) on rhs or \(\sin^2\theta=1-\cos^2\theta\) on lhsdM1 Alternatively proceeds to \(10\sin^2\theta+10\cos^2\theta=10\) and makes statement about \(\sin^2\theta+\cos^2\theta=1\)
Shows \((4-5\cos\theta)(3+2\cos\theta)\equiv 10\sin^2\theta-7\cos\theta+2\) AND makes minimal statement "hence true"A1*
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(4+3\sin x = 4-5\cos x \Rightarrow \tan x = -\frac{5}{3}\)M1 Attempts to use part (a), proceeds to equation of form \(\tan x = k\), \(k\neq 0\). Condone \(\theta \leftrightarrow x\)
\(x = \text{awrt } 121°, 301°\)A1 A1 First A1 for either value. Second A1 for both and no other solutions 
# Question 12:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{10\sin^2\theta - 7\cos\theta + 2}{3+2\cos\theta} \equiv \frac{10(1-\cos^2\theta)-7\cos\theta+2}{3+2\cos\theta}$ | M1 | Uses identity $\sin^2\theta = 1-\cos^2\theta$ within the fraction |
| $\equiv \frac{12-7\cos\theta - 10\cos^2\theta}{3+2\cos\theta}$ | A1 | Correct simplified expression in just $\cos\theta$ |
| $\equiv \frac{(3+2\cos\theta)(4-5\cos\theta)}{3+2\cos\theta}$ | M1 | Correct attempt to factorise numerator. Allow $u=\cos\theta$ |
| $\equiv 4-5\cos\theta$ | A1* | Fully correct proof with correct notation and no errors |

**Alternative proof:** Multiplies across and forms 3TQ in $\cos\theta$ on rhs:
$(4-5\cos\theta)(3+2\cos\theta) \Rightarrow 10\sin^2\theta-7\cos\theta+2 = A\cos^2\theta+B\cos\theta+C$

| $10\sin^2\theta-7\cos\theta+2 = -10\cos^2\theta-7\cos\theta+12$ | A1 | Correct identity formed |
|---|---|---|
| Uses $\cos^2\theta=1-\sin^2\theta$ on rhs or $\sin^2\theta=1-\cos^2\theta$ on lhs | dM1 | Alternatively proceeds to $10\sin^2\theta+10\cos^2\theta=10$ and makes statement about $\sin^2\theta+\cos^2\theta=1$ |
| Shows $(4-5\cos\theta)(3+2\cos\theta)\equiv 10\sin^2\theta-7\cos\theta+2$ AND makes minimal statement "hence true" | A1* | |

## Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $4+3\sin x = 4-5\cos x \Rightarrow \tan x = -\frac{5}{3}$ | M1 | Attempts to use part (a), proceeds to equation of form $\tan x = k$, $k\neq 0$. Condone $\theta \leftrightarrow x$ |
| $x = \text{awrt } 121°, 301°$ | A1 A1 | First A1 for either value. Second A1 for both and no other solutions |

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\begin{enumerate}
  \item (a) Show that
\end{enumerate}

$$\frac { 10 \sin ^ { 2 } \theta - 7 \cos \theta + 2 } { 3 + 2 \cos \theta } \equiv 4 - 5 \cos \theta$$

(b) Hence, or otherwise, solve, for $0 \leqslant x < 360 ^ { \circ }$, the equation

$$\frac { 10 \sin ^ { 2 } x - 7 \cos x + 2 } { 3 + 2 \cos x } = 4 + 3 \sin x$$

\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q12 [7]}}
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