Sketch the graph $y=3\text{sin}x+4\text{cos}x$, for $-2\pi ⩽x⩽2\pi$

Write down the amplitude of this graph

Write down the period of this graph

Use your answers from part (a) to write down the value of $A$, $B$ and $D$.

Find the value of $C$.

Find $\text{arctan}\frac{3}{4}$, giving the answer to 3 significant figures.

Comment on your answer to part (c)(i).

By considering the graph of $y=5\text{sin}x+12\text{cos}x$, find the value of $A$, $B$, $C$ and $D$.

$A$.

$B$.

$C$.

$D$.

Show that $\frac{b}{\sqrt{a^2+b^2}}=\text{cos}\theta$.

Show that $\frac{a}{b}=\text{tan}\theta$.

Hence prove your conjectures in part (e).

      A1

[1 mark]

5       A1

[1 mark]

$2\pi$      A1

[1 mark]

$A=5$, $B=1$, $D=0$      A1

[1 mark]

maximum at $x=0.644$      M1

So $C=-0.644$      A1

[2 marks]

0.644      A1

[1 mark]

it appears that $C=-\text{arctan}\frac{3}{4}$      A1

[1 mark]

        M1

$A=13$        A1

$B=1$ and $D=0$        A1

maximum at $x=0.395$        M1

So C = −0.395  $\left(=-\text{arctan}\frac{5}{12}\right)$      A1

[5 marks]

$A=\sqrt{a^2+b^2}$       A1

[1 mark]

$B=1$       A1

[1 mark]

$C=-\text{arctan}\frac{a}{b}$      A1

[1 mark]

$D=0$      A1

[1 mark]

EITHER

use of a right triangle and Pythgoras’ to show the missing side length is $b$         M1A1

OR

Use of $\text{si}{\text{n}}^2\theta +\text{co}{\text{s}}^2\theta =1$, leading to the required result         M1A1

[2 marks]

EITHER

use of a right triangle, leading to the required result.         M1

OR

Use of $\text{tan}\theta =\frac{\text{sin}\theta }{\text{cos}\theta }$, leading to the required result.       M1

[1 mark]

$a\text{sin}x+b\text{cos}x=\sqrt{a^2+b^2}\left(\text{sin}\theta \text{sin}x+\text{cos}\theta \text{cos}x\right)$        M1

$a\text{sin}x+b\text{cos}x=\sqrt{a^2+b^2}\left(\text{cos}\left(x-\theta \right)\right)$        M1A1

So $A=\sqrt{a^2+b^2}$, $B=1$ and $D=0$       A1

And $C=-\theta$        M1

So $C=-\text{arctan}\frac{a}{b}$       A1

[6 marks]

On the same set of axes, sketch the graphs of $y=f_1(x)$ and $y=f_3(x)$ for −1 ≤ $x$ ≤ 1.

local maximum points;

local minimum points;

On a new set of axes, sketch the graphs of $y=f_2(x)$ and $y=f_4(x)$ for −1 ≤ $x$ ≤ 1.

local maximum points;

local minimum points.

Solve the equation ${f_n}^{\mathrm{\prime }}(x)=0$ and hence show that the stationary points on the graph of $y=f_n(x)$ occur at $x=\text{cos}\frac{k\pi }{n}$ where $k\in {\mathbb{Z}}^{+}$ and 0 < $k$ < $n$.

Use an appropriate trigonometric identity to show that $f_2(x)=2x^2-1$.

Use an appropriate trigonometric identity to show that $f_{n+1}(x)=\text{cos}\left(n\text{arccos}x\right)\text{cos}\left(\text{arccos}x\right)-\text{sin}\left(n\text{arccos}x\right)\text{sin}\left(\text{arccos}x\right)$.

Hence show that $f_{n+1}(x)+f_{n-1}(x)=2x{f}_n\left(x\right)$, $n\in {\mathbb{Z}}^{+}$.

Hence express $f_3(x)$ as a cubic polynomial.

correct graph of $y=f_1(x)$      A1

correct graph of $y=f_3(x)$      A1

[2 marks]

graphical or tabular evidence that $n$ has been systematically varied        M1

eg $n$ = 3, 1 local maximum point and 1 local minimum point

$n$ = 5, 2 local maximum points and 2 local minimum points

$n$ = 7, 3 local maximum points and 3 local minimum points        (A1)

$\frac{n-1}{2}$ local maximum points      A1

[3 marks]

$\frac{n-1}{2}$ local minimum points      A1

Note: Allow follow through from an incorrect local maximum formula expression.

[1 mark]

correct graph of $y=f_2(x)$       A1

correct graph of $y=f_4(x)$       A1

[2 marks]

graphical or tabular evidence that $n$ has been systematically varied       M1

eg $n$ = 2, 0 local maximum point and 1 local minimum point

$n$ = 4, 1 local maximum points and 2 local minimum points

$n$ = 6, 2 local maximum points and 3 local minimum points       (A1)

$\frac{n-2}{2}$ local maximum points     A1

[3 marks]

$\frac{n}{2}$ local minimum points     A1

[1 mark]

$f_n(x)=\text{cos}\left(n\text{arccos}\left(x\right)\right)$

${f_n}^{\mathrm{\prime }}(x)=\frac{n\text{sin}\left(n\text{arccos}\left(x\right)\right)}{\sqrt{1-x^2}}$      M1A1

Note: Award M1 for attempting to use the chain rule.

${f_n}^{\mathrm{\prime }}(x)=0\Rightarrow n\text{sin}\left(n\text{arccos}\left(x\right)\right)=0$     M1

$n\text{arccos}\left(x\right)=k\pi \left(k\in {\mathbb{Z}}^{+}\right)$     A1

leading to

$x=\text{cos}\frac{k\pi }{n}$  ($k\in {\mathbb{Z}}^{+}$ and 0 < $k$ < $n$)     AG

[4 marks]

$f_2(x)=\text{cos}\left(2\text{arccos}x\right)$

$=2{\left(\text{cos}\left(\text{arccos}x\right)\right)}^2-1$     M1

stating that $\left(\text{cos}\left(\text{arccos}x\right)\right)=x$     A1

so $f_2(x)=2x^2-1$     AG

[2 marks]

$f_{n+1}(x)=\text{cos}\left(\left(n+1\right)\text{arccos}x\right)$

$=\text{cos}\left(n\text{arccos}x+\text{arccos}x\right)$     A1

use of cos(A + B) = cos A cos B − sin A sin B leading to      M1

$=\text{cos}\left(n\text{arccos}x\right)\text{cos}\left(\text{arccos}x\right)-\text{sin}\left(n\text{arccos}x\right)\text{sin}\left(\text{arccos}x\right)$      AG

[2 marks]

$f_{n-1}(x)=\text{cos}\left(\left(n-1\right)\text{arccos}x\right)$     A1

$=\text{cos}\left(n\text{arccos}x\right)\text{cos}\left(\text{arccos}x\right)+\text{sin}\left(n\text{arccos}x\right)\text{sin}\left(\text{arccos}x\right)$    M1

$f_{n+1}(x)+f_{n-1}(x)=2\text{cos}\left(n\text{arccos}x\right)\text{cos}\left(\text{arccos}x\right)$     A1

$=2x{f}_n\left(x\right)$     AG

[3 marks]

$f_3(x)=2x{f}_2\left(x\right)-f_1(x)$      (M1)

$=2x\left(2x^2-1\right)-x$

$=4x^3-3x$   A1

[2 marks]

Find the side length, $s$, where $s>0$, of a square such that $A=P$.

Write down, in terms of $x$ and $n$, an expression for the area, $A_T$, of one of these isosceles triangles.

Show that $y=2x \sin \frac{\mathrm{\pi }}{n}$.

Use the results from parts (b) and (c) to show that $A=P=4n \tan \frac{\mathrm{\pi }}{n}$.

Use the Maclaurin series for $\tan x$ to find $\underset{n\to \infty }{\lim }\left(4n \tan \frac{\mathrm{\pi }}{n}\right)$.

Interpret your answer to part (e)(i) geometrically.

Show that $a=\frac{8}{b-4}+4$.

By using the result of part (f) or otherwise, determine the three side lengths of the only two right-angled triangles for which $a, b, A, P\in \mathrm{\mathbb{Z}}$.

Determine the area and perimeter of these two right-angled triangles.

$A=s^2$ and $P=4s$              (A1)

$A=P\Rightarrow s^2=4s$              (M1)

$s\left(s-4\right)=0$

$\Rightarrow s=4\left(s>0\right)$        A1

Note: Award A1M1A0 if both $s=4$ and $s=0$ are stated as final answers.

[3 marks]

$A_T=\frac{1}{2}{x}^{2 }\sin \frac{2\mathrm{\pi }}{n}$        A1

Note: Award A1 for a correct alternative form expressed in terms of $x$ and $n$ only.

          For example, using Pythagoras’ theorem, $A_T=x \sin \frac{\mathrm{\pi }}{n}\sqrt{x^2-x^{2 }{\sin }^2\frac{\mathrm{\pi }}{n}}$  or  $A_T=2\left(\frac{1}{2}\left(x \sin \frac{\mathrm{\pi }}{n}\right)\left(x \cos \frac{\mathrm{\pi }}{n}\right)\right)$  or  $A_T=x^2 \sin \frac{\mathrm{\pi }}{n}\cos \frac{\mathrm{\pi }}{n}$.

[1 mark]

METHOD 1

uses $\sin \theta =\frac{\text{opp}}{\text{hyp}}$         (M1)

$\frac{{\displaystyle \frac{y}{2}}}{x}=\sin \frac{\mathrm{\pi }}{n}$        A1

$y=2x \sin \frac{\mathrm{\pi }}{n}$       AG

METHOD 2

uses Pythagoras’ theorem ${\left(\frac{y}{2}\right)}^2+h^2=x^2$  and  $h=x \cos \frac{\mathrm{\pi }}{n}$         (M1)

${\left(\frac{y}{2}\right)}^2+{\left(x \cos \frac{\mathrm{\pi }}{n}\right)}^2=x^2 \left(y^2=4x^2\left(1-{\cos }^2\frac{\mathrm{\pi }}{n}\right)\right)$

$=4x^2 {\sin }^2\frac{\mathrm{\pi }}{n}$        A1

$y=2x \sin \frac{\mathrm{\pi }}{n}$       AG

METHOD 3

uses the cosine rule         (M1)

$y^2=2x^2-2x^2 \cos \frac{2\mathrm{\pi }}{n} \left(=2x^2\left(1-\cos \frac{2\mathrm{\pi }}{n}\right)\right)$

$=4x^2 {\sin }^2\frac{\mathrm{\pi }}{n}$        A1

$y=2x \sin \frac{\mathrm{\pi }}{n}$       AG

METHOD 4

uses the sine rule         (M1)

$\frac{y}{\sin {\displaystyle \frac{2\mathrm{\pi }}{n}}}=\frac{x}{\sin {\displaystyle \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{n}\right)}}$

$y \cos \frac{\mathrm{\pi }}{n}=2x \sin \frac{\mathrm{\pi }}{n}\cos \frac{\mathrm{\pi }}{n}$        A1

$y=2x \sin \frac{\mathrm{\pi }}{n}$       AG

[2 marks]

$A=P\Rightarrow n{A}_T=ny$         (M1)

Note: Award M1 for equating correct expressions for $A$ and $P$.

$\frac{1}{2}n{x}^2 \sin \frac{2\mathrm{\pi }}{n}=2nx \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}} \left(n{x}^2 \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}=2nx \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\right)$

$\frac{1}{2}{x}^2 \sin \frac{2\mathrm{\pi }}{n}=2x \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}} \left(x^2 \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}=2x \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\right)$        A1

uses $\sin \frac{2\mathrm{\pi }}{n}=2 \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}$ (seen anywhere in part (d) or in part (b))         (M1)

$x^2 \sin \frac{\mathrm{\pi }}{n}\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}=2x \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}$

attempts to either factorise or divide their expression         (M1)

$x \sin \frac{\mathrm{\pi }}{n}\left(x \cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}-2\right)=0$

$x=\frac{2}{\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}}, \left(x \sin \frac{\mathrm{\pi }}{n}\ne 0\right)$ (or equivalent)        A1

EITHER

substitutes $x=\frac{2}{\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}}$ (or equivalent) into $P=ny$         (M1)

$P=2n\left(\frac{2}{\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}}\right)\left(\sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\right)$        A1

Note: Other approaches are possible. For example, award A1 for $P=2nx \cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\tan \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}$ and M1 for substituting $x=\frac{2}{\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}}$ into $P$.

OR

substitutes $x=\frac{2}{\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}}$ (or equivalent) into $A=n{A}_T$          (M1)

$A=\frac{1}{2}n{\left(\frac{2}{\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}}\right)}^2\left(\sin \frac{{\displaystyle 2\mathrm{\pi }}}{{\displaystyle n}}\right)$

$A=\frac{1}{2}n{\left(\frac{2}{\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}}\right)}^2\left(2 \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\cos \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}\right)$        A1

THEN

$A=P=4n \tan \frac{\mathrm{\pi }}{n}$       AG

[7 marks]

attempts to use the Maclaurin series for $\tan x$ with $x=\frac{\mathrm{\pi }}{n}$         (M1)

$\tan \frac{\mathrm{\pi }}{n}=\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle n}}+\frac{{\displaystyle {\left(\frac{\mathrm{\pi }}{n}\right)}^3}}{3}+\frac{{\displaystyle 2{\left(\frac{\mathrm{\pi }}{n}\right)}^5}}{15}\left(+\dots \right)$

$4n \tan \frac{\mathrm{\pi }}{n}=4n\left(\frac{\mathrm{\pi }}{n}+\frac{{\mathrm{\pi }}^3}{3n^3}+\frac{2{\mathrm{\pi }}^5}{15n^5}\left(+\dots \right)\right)$ (or equivalent)        A1

$=4\left(\mathrm{\pi }+\frac{{\mathrm{\pi }}^3}{3n^2}+\frac{2{\mathrm{\pi }}^5}{15n^4}+\dots \right)$

$\Rightarrow \underset{n\to \infty }{\lim }\left(4n \tan \frac{\mathrm{\pi }}{n}\right)=4\mathrm{\pi }$        A1

Note: Award a maximum of M1A1A0 if $\underset{n\to \infty }{\lim }$ is not stated anywhere.

[3 marks]

(as $n\to \infty , P\to 4\mathrm{\pi }$ and $A\to 4\mathrm{\pi }$)

the polygon becomes a circle of radius $2$                   R1

Note: Award R1 for alternative responses such as:
the polygon becomes a circle of area $4\mathrm{\pi }$ OR
the polygon becomes a circle of perimeter $4\mathrm{\pi }$ OR
the polygon becomes a circle with $A=P=4\mathrm{\pi }$.
Award R0 for polygon becomes a circle.

[1 mark]

$A=\frac{1}{2}ab$ and $P=a+b+\sqrt{a^2+b^2}$                   (A1)(A1)

equates their expressions for $A$ and $P$                 M1

$A=P\Rightarrow a+b+\sqrt{a^2+b^2}=\frac{1}{2}ab$

$\sqrt{a^2+b^2}=\frac{1}{2}ab-\left(a+b\right)$                M1

Note: Award M1 for isolating $\sqrt{a^2+b^2}$ or $\pm 2\sqrt{a^2+b^2}$. This step may be seen later.

$a^2+b^2={\left(\frac{1}{2}ab-\left(a+b\right)\right)}^2$

$a^2+b^2=\frac{1}{4}{a}^2{b}^2-2\left(\frac{1}{2}ab\right)\left(a+b\right)+{\left(a+b\right)}^2$                M1

$\left(=\frac{1}{4}{a}^2{b}^2-a^{2}b-a{b}^2+a^2+2ab+b^2\right)$

Note: Award M1 for attempting to expand their RHS of either $a^2+b^2=\dots$ or $4\left(a^2+b^2\right)=\dots$.

EITHER

$ab\left(\frac{1}{4}ab-a-b+2\right)=0 \left(ab\ne 0\right)$               A1

$\frac{1}{4}ab-a-b+2=0$

$ab-4a=4b-8$

OR

$\frac{1}{4}{a}^2{b}^2-a^{2}b-a{b}^2+2ab=0$

$a\left(\frac{1}{4}{b}^2-b\right)+\left(2b-b^2\right)=0 \left(a\left(b^2-4b\right)+\left(8b-4b^2\right)=0\right)$               A1

$a=\frac{4b^2-8b}{b^2-4b}$

THEN

$\Rightarrow a=\frac{4b-8}{b-4}$               A1

$a=\frac{4b-16+8}{b-4}$

$a=\frac{8}{b-4}+4$               AG

Note: Award a maximum of A1A1M1M1M0A0A0 for attempting to verify.
For example, verifying that $A=P=\frac{16}{b-4}+2b+4$ gains $4$ of the $7$ marks.

[7 marks]

using an appropriate method          (M1)

eg substituting values for $b$ or using divisibility properties

$\left(5, 12, 13\right)$ and $\left(6, 8, 10\right)$             A1A1

Note: Award A1A0 for either one set of three correct side lengths or two sets of two correct side lengths.

[3 marks]

$A=P=30$  and  $A=P=24$            A1

Note: Do not award A1FT.

[1 mark]