| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Equation with half angles |
| Difficulty | Standard +0.3 This is a standard FP1 half-angle substitution question with straightforward application. Part (i) is a routine proof using the t-substitution formulas (which students memorize). Part (ii) involves direct substitution into given formulas with algebraic manipulation but no novel insight—the substitution is explicitly provided and the steps are mechanical. While it requires careful algebra across multiple parts, it's a typical textbook exercise testing technique rather than problem-solving. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{lhs}=\cot x+\tan\left(\frac{x}{2}\right)=\frac{1-t^2}{2t}+t\) | M1 | Selects correct expression for \(\cot x\) in terms of \(t\) and substitutes |
| \(\frac{1-t^2}{2t}+t=\frac{1+t^2}{2t}\left(=\frac{1}{\sin x}\right)=\text{cosec}\, x\) | A1* | Fully correct proof; allow correct work leading to \(\frac{1+t^2}{2t}=\text{cosec}\, x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x=0\Rightarrow H=90-30\cos(0)-40\sin(0)=90-30=60\) | B1 | Demonstrates that when \(x=0\), \(H=60\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(H=90-30\cos120x-40\sin120x=90-30\left(\frac{1-t^2}{1+t^2}\right)-40\left(\frac{2t}{1+t^2}\right)\) | M1 | Uses correct formulae to obtain \(H\) in terms of \(t\) |
| \(=\frac{90+90t^2-30+30t^2-80t}{1+t^2}\) | M1 | Correct method to obtain common denominator |
| \(=\frac{120t^2-80t+60}{1+t^2}\) | A1* | Collects terms and simplifies to printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{120t^2-80t+60}{1+t^2}=100\Rightarrow 120t^2-80t+60=100+100t^2\) | M1 | Uses \(H=100\) and multiplies up to obtain quadratic in \(t\) |
| \(20t^2-80t-40=0\) | A1 | Correct 3TQ |
| \(t=\frac{4\pm\sqrt{16+8}}{2}\Rightarrow 60x=\tan^{-1}(2+\sqrt{6})\) or \(60x=\tan^{-1}(2-\sqrt{6})\) | M1 | Solves 3TQ in \(t\), obtains values of \(60x\) as suggested by model |
| \(60x=\tan^{-1}(2+\sqrt{6})=77.33...\Rightarrow x=...\) | dM1 | Fully correct strategy to identify required value of \(x\) from positive root |
| \(x=1.29\) | A1 | awrt 1.29; attempts in radians score all but final mark |
## Question 3(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{lhs}=\cot x+\tan\left(\frac{x}{2}\right)=\frac{1-t^2}{2t}+t$ | M1 | Selects correct expression for $\cot x$ in terms of $t$ and substitutes |
| $\frac{1-t^2}{2t}+t=\frac{1+t^2}{2t}\left(=\frac{1}{\sin x}\right)=\text{cosec}\, x$ | A1* | Fully correct proof; allow correct work leading to $\frac{1+t^2}{2t}=\text{cosec}\, x$ |
---
## Question 3(ii)(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x=0\Rightarrow H=90-30\cos(0)-40\sin(0)=90-30=60$ | B1 | Demonstrates that when $x=0$, $H=60$ |
---
## Question 3(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $H=90-30\cos120x-40\sin120x=90-30\left(\frac{1-t^2}{1+t^2}\right)-40\left(\frac{2t}{1+t^2}\right)$ | M1 | Uses correct formulae to obtain $H$ in terms of $t$ |
| $=\frac{90+90t^2-30+30t^2-80t}{1+t^2}$ | M1 | Correct method to obtain common denominator |
| $=\frac{120t^2-80t+60}{1+t^2}$ | A1* | Collects terms and simplifies to printed answer with no errors |
---
## Question 3(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{120t^2-80t+60}{1+t^2}=100\Rightarrow 120t^2-80t+60=100+100t^2$ | M1 | Uses $H=100$ and multiplies up to obtain quadratic in $t$ |
| $20t^2-80t-40=0$ | A1 | Correct 3TQ |
| $t=\frac{4\pm\sqrt{16+8}}{2}\Rightarrow 60x=\tan^{-1}(2+\sqrt{6})$ or $60x=\tan^{-1}(2-\sqrt{6})$ | M1 | Solves 3TQ in $t$, obtains values of $60x$ as suggested by model |
| $60x=\tan^{-1}(2+\sqrt{6})=77.33...\Rightarrow x=...$ | dM1 | Fully correct strategy to identify required value of $x$ from positive root |
| $x=1.29$ | A1 | awrt 1.29; attempts in radians score all but final mark |
---\begin{enumerate}
\item (i) Use the substitution $t = \tan \left( \frac { x } { 2 } \right)$ to prove that
\end{enumerate}
$$\cot x + \tan \left( \frac { x } { 2 } \right) = \operatorname { cosec } x \quad x \neq n \pi , n \in \mathbb { Z }$$
(ii)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1e5324f5-a9bc-4041-bfbb-cb940417ea63-08_389_455_573_877}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
An engineer models the vertical height above the ground of the tip of one blade of a wind turbine, shown in Figure 1. The ground is assumed to be horizontal.
The vertical height of the tip of the blade above the ground, $H$ metres, at time $x$ seconds after the wind turbine has reached its constant operating speed, is modelled by the equation
$$H = 90 - 30 \cos ( 120 x ) ^ { \circ } - 40 \sin ( 120 x ) ^ { \circ }$$
(a) Show that $H = 60$ when $x = 0$
Using the substitution $t = \tan ( 60 x ) ^ { \circ }$\\
(b) show that equation (I) can be rewritten as
$$H = \frac { 120 t ^ { 2 } - 80 t + 60 } { 1 + t ^ { 2 } }$$
(c) Hence find, according to the model, the value of $x$ when the tip of the blade is 100 m above the ground for the first time after the wind turbine has reached its constant operating speed.
\hfill \mbox{\textit{Edexcel FP1 AS 2020 Q3 [11]}}