IBDP Maths analysis and approaches Topic: AHL 3.9 :Pythagorean identities HL Paper 1

Question

In the triangle ABC, \({\rm{A\hat BC}} = 90^\circ\) , \({\text{AC}} = \sqrt {\text{2}}\) and AB = BC + 1.

a.Show that cos \(\hat A – \sin \hat A = \frac{1}{{\sqrt 2 }}\).[3]

b.By squaring both sides of the equation in part (a), solve the equation to find the angles in the triangle.[8]

c.Apply Pythagoras’ theorem in the triangle ABC to find BC, and hence show that \(\sin \hat A = \frac{{\sqrt 6 – \sqrt 2 }}{4}\).[6]

d.Hence, or otherwise, calculate the length of the perpendicular from B to [AC].[4]

▶️Answer/Explanation

Markscheme

\(\cos \hat A = \frac{{{\text{BA}}}}{{\sqrt 2 }}\)     A1

\(\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }}\)     A1

\(\cos \hat A – \sin \hat A = \frac{{{\text{BA}} – {\text{BC}}}}{{\sqrt 2 }}\)     R1

\( = \frac{1}{{\sqrt 2 }}\)     AG

[3 marks]

a.

\({\cos ^2}\hat A – 2\cos \hat A\sin \hat A + {\sin ^2}\hat A = \frac{1}{2}\)     M1A1

\(1 – 2\sin \hat A\cos \hat A = \frac{1}{2}\)     M1A1

\(\sin 2\hat A = \frac{1}{2}\)     M1

\(2\hat A = 30^\circ \)     A1

angles in the triangle are 15° and 75°     A1A1

Note: Accept answers in radians.

[8 marks]

b.

\({\text{B}}{{\text{C}}^2} + {({\text{BC}} + 1)^2} = 2\)     M1A1

\(2{\text{B}}{{\text{C}}^2} + 2{\text{BC}} – 1 = 0\)     A1

\({\text{BC}} = \frac{{ -2 + \sqrt {12} }}{4}\left( { = \frac{{\sqrt 3  – 1}}{2}} \right)\)     M1A1

\(\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }} = \frac{{\sqrt 3 – 1}}{{2\sqrt 2 }}\)     A1

\( = \frac{{\sqrt 6 – \sqrt 2 }}{4}\)     AG

[6 marks]

c.

EITHER

\(h = {\text{ABsin}}\hat A\)     M1

\( = ({\text{BC}} + 1)\sin \hat A\)     A1

\( = \frac{{\sqrt 3  + 1}}{2} \times \frac{{\sqrt 6 – \sqrt 2 }}{4} = \frac{{\sqrt 2 }}{4}\)     M1A1

OR

\(\tfrac{1}{2}AB.BC = \tfrac{1}{2}AC.h\)     M1

\(\frac{{\sqrt 3 – 1}}{2} \cdot \frac{{\sqrt {3 + 1} }}{2} = \sqrt {2h} \)     A1

\(\frac{2}{4} = \sqrt 2 h\)     M1

\(h = \frac{1}{{2\sqrt 2 }}\)     A1

[4 marks]

d.

Examiners report

Many good solutions to this question, although some students incorrectly stated the value of \(\arcsin \left( {\frac{1}{2}} \right)\). A surprising number of students had greater difficulties with part (d).

a.

Many good solutions to this question, although some students incorrectly stated the value of \(\arcsin \left( {\frac{1}{2}} \right)\). A surprising number of students had greater difficulties with part (d).

b.

Many good solutions to this question, although some students incorrectly stated the value of \(\arcsin \left( {\frac{1}{2}} \right)\). A surprising number of students had greater difficulties with part (d).

c.

Many good solutions to this question, although some students incorrectly stated the value of \(\arcsin \left( {\frac{1}{2}} \right)\). A surprising number of students had greater difficulties with part (d).

d.

Question

Show that \(\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \sec 2A + \tan 2A\) .

▶️Answer/Explanation

Markscheme

METHOD 1

\(\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \sec 2A + \tan 2A\)

consider right hand side

\(\sec 2A + \tan 2A = \frac{1}{{\cos 2A}} + \frac{{\sin 2A}}{{\cos 2A}}\)     M1A1

\( = \frac{{{{\cos }^2}A + 2\sin A\cos A + {{\sin }^2}A}}{{{{\cos }^2}A – {{\sin }^2}A}}\)     A1A1 

Note: Award A1 for recognizing the need for single angles and A1 for recognizing \({\cos ^2}A + {\sin ^2}A = 1\) .

\( = \frac{{{{(\cos A + \sin A)}^2}}}{{(\cos A + \sin A)(\cos A – \sin A)}}\)     M1A1

\( = \frac{{\cos A + \sin A}}{{\cos A – \sin A}}\)     AG

 

METHOD 2

\(\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \frac{{{{(\cos A + \sin A)}^2}}}{{(\cos A + \sin A)(\cos A – \sin A)}}\)     M1A1

\( = \frac{{{{\cos }^2}A + 2\sin A\cos A + {{\sin }^2}A}}{{{{\cos }^2}A – {{\sin }^2}A}}\)     A1A1 

Note: Award A1 for correct numerator and A1 for correct denominator.

 

\( = \frac{{1 + \sin 2A}}{{\cos 2A}}\)     M1A1

\( = \sec 2A + \tan 2A\)     AG

[6 marks]

Examiners report

Solutions to this question were good in general with many candidates realising that multiplying the numerator and denominator by \((\cos A + \sin A)\) might be helpful.

 

Question

a.Show that \(\cot \alpha  = \tan \left( {\frac{\pi }{2} – \alpha } \right)\) for \(0 < \alpha  < \frac{\pi }{2}\). [1]

b.Hence find \(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x,{\text{ }}0 < \alpha  < \frac{\pi }{2}} \).[4]

▶️Answer/Explanation

Markscheme

EITHER

use of a diagram and trig ratios

eg,

\(\tan \alpha  = \frac{O}{A} \Rightarrow \cot \alpha  = \frac{A}{O}\)

from diagram, \(\tan \left( {\frac{\pi }{2} – \alpha } \right) = \frac{A}{O}\)     R1

OR

use of \(\tan \left( {\frac{\pi }{2} – \alpha } \right) = \frac{{\sin \left( {\frac{\pi }{2} – \alpha } \right)}}{{\cos \left( {\frac{\pi }{2} – \alpha } \right)}} = \frac{{\cos \alpha }}{{\sin \alpha }}\)     R1

THEN

\(\cot \alpha  = \tan \left( {\frac{\pi }{2} – \alpha } \right)\)     AG

[1 mark]

a.

\(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x}  = [\arctan x]_{\tan \alpha }^{\cot \alpha }\)     (A1)

Note:     Limits (or absence of such) may be ignored at this stage.

\( = \arctan (\cot \alpha ) – \arctan (\tan \alpha )\)     (M1)

\( = \frac{\pi }{2} – \alpha  – \alpha \)     (A1)

\( = \frac{\pi }{2} – 2\alpha \)     A1

[4 marks]

b.

Examiners report

This was generally well done.

a.

This was generally well done. Some weaker candidates tried to solve part (b) through use of a substitution, though the standard result \(\arctan x\) was well known. A small number used \(\arctan x + c\) and went on to obtain an incorrect final answer.

b.

Question

a.Show that \(\cot \alpha  = \tan \left( {\frac{\pi }{2} – \alpha } \right)\) for \(0 < \alpha  < \frac{\pi }{2}\).[1]

b.Hence find \(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x,{\text{ }}0 < \alpha  < \frac{\pi }{2}} \).[4]

▶️Answer/Explanation

Markscheme

EITHER

use of a diagram and trig ratios

eg,

\(\tan \alpha  = \frac{O}{A} \Rightarrow \cot \alpha  = \frac{A}{O}\)

from diagram, \(\tan \left( {\frac{\pi }{2} – \alpha } \right) = \frac{A}{O}\)     R1

OR

use of \(\tan \left( {\frac{\pi }{2} – \alpha } \right) = \frac{{\sin \left( {\frac{\pi }{2} – \alpha } \right)}}{{\cos \left( {\frac{\pi }{2} – \alpha } \right)}} = \frac{{\cos \alpha }}{{\sin \alpha }}\)     R1

THEN

\(\cot \alpha  = \tan \left( {\frac{\pi }{2} – \alpha } \right)\)     AG

[1 mark]

a.

\(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x}  = [\arctan x]_{\tan \alpha }^{\cot \alpha }\)     (A1)

Note:     Limits (or absence of such) may be ignored at this stage.

\( = \arctan (\cot \alpha ) – \arctan (\tan \alpha )\)     (M1)

\( = \frac{\pi }{2} – \alpha  – \alpha \)     (A1)

\( = \frac{\pi }{2} – 2\alpha \)     A1

[4 marks]

b.

Examiners report

This was generally well done.

a.

This was generally well done. Some weaker candidates tried to solve part (b) through use of a substitution, though the standard result \(\arctan x\) was well known. A small number used \(\arctan x + c\) and went on to obtain an incorrect final answer.

b.

Question

a.Find the value of \(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}\).[2]

Show that \(\frac{{1 – \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi \) where \(k \in \mathbb{Z}\).[2]

b.Use the principle of mathematical induction to prove that

\(\sin x + \sin 3x +  \ldots  + \sin (2n – 1)x = \frac{{1 – \cos 2nx}}{{2\sin x}},{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}x \ne k\pi \) where \(k \in \mathbb{Z}\). [9]

c.Hence or otherwise solve the equation \(\sin x + \sin 3x = \cos x\) in the interval \(0 < x < \pi \). [6]

▶️Answer/Explanation

Markscheme

\(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} – \frac{{\sqrt 2 }}{2} – \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}\)    (M1)A1

Note: Award M1 for 5 equal terms with \) + \) or \( – \) signs.

[2 marks]

a.

\(\frac{{1 – \cos 2x}}{{2\sin x}} \equiv \frac{{1 – (1 – 2{{\sin }^2}x)}}{{2\sin x}}\)    M1

\( \equiv \frac{{2{{\sin }^2}x}}{{2\sin x}}\)    A1

\( \equiv \sin x\)    AG

[2 marks]

b.

let \({\text{P}}(n):\sin x + \sin 3x +  \ldots  + \sin (2n – 1)x \equiv \frac{{1 – \cos 2nx}}{{2\sin x}}\)

if \(n = 1\)

\({\text{P}}(1):\frac{{1 – \cos 2x}}{{2\sin x}} \equiv \sin x\) which is true (as proved in part (b))     R1

assume \({\text{P}}(k)\) true, \(\sin x + \sin 3x +  \ldots  + \sin (2k – 1)x \equiv \frac{{1 – \cos 2kx}}{{2\sin x}}\)     M1

Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let \(n = k\)only. Subsequent marks are independent of this M1.

consider \({\text{P}}(k + 1)\):

\({\text{P}}(k + 1):\sin x + \sin 3x +  \ldots  + \sin (2k – 1)x + \sin (2k + 1)x \equiv \frac{{1 – \cos 2(k + 1)x}}{{2\sin x}}\)

\(LHS = \sin x + \sin 3x +  \ldots  + \sin (2k – 1)x + \sin (2k + 1)x\)    M1

\( \equiv \frac{{1 – \cos 2kx}}{{2\sin x}} + \sin (2k + 1)x\)    A1

\( \equiv \frac{{1 – \cos 2kx + 2\sin x\sin (2k + 1)x}}{{2\sin x}}\)

\( \equiv \frac{{1 – \cos 2kx + 2\sin x\cos x\sin 2kx + 2{{\sin }^2}x\cos 2kx}}{{2\sin x}}\)    M1

\( \equiv \frac{{1 – \left( {(1 – 2{{\sin }^2}x)\cos 2kx – \sin 2x\sin 2kx} \right)}}{{2\sin x}}\)    M1

\( \equiv \frac{{1 – (\cos 2x\cos 2kx – \sin 2x\sin 2kx)}}{{2\sin x}}\)    A1

\( \equiv \frac{{1 – \cos (2kx + 2x)}}{{2\sin x}}\)    A1

\( \equiv \frac{{1 – \cos 2(k + 1)x}}{{2\sin x}}\)

so if true for \(n = k\) , then also true for \(n = k + 1\)

as true for \(n = 1\) then true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note: Accept answers using transformation formula for product of sines if steps are shown clearly.

Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.

[9 marks]

c.

EITHER

\(\sin x + \sin 3x = \cos x \Rightarrow \frac{{1 – \cos 4x}}{{2\sin x}} = \cos x\)    M1

\( \Rightarrow 1 – \cos 4x = 2\sin x\cos x,{\text{ }}(\sin x \ne 0)\)    A1

\( \Rightarrow 1 – (1 – 2{\sin ^2}2x) = \sin 2x\)    M1

\( \Rightarrow \sin 2x(2\sin 2x – 1) = 0\)    M1

\( \Rightarrow \sin 2x = 0\) or \(\sin 2x = \frac{1}{2}\)     A1

\(2x = \pi ,{\text{ }}2x = \frac{\pi }{6}\) and \(2x = \frac{{5\pi }}{6}\)

OR

\(\sin x + \sin 3x = \cos x \Rightarrow 2\sin 2x\cos x = \cos x\)    M1A1

\( \Rightarrow (2\sin 2x – 1)\cos x = 0,{\text{ }}(\sin x \ne 0)\)    M1A1

\( \Rightarrow \sin 2x = \frac{1}{2}\) of \(\cos x = 0\)    A1

\(2x = \frac{\pi }{6},{\text{ }}2x = \frac{{5\pi }}{6}\) and \(x = \frac{\pi }{2}\)

THEN

\(\therefore x = \frac{\pi }{2},{\text{ }}x = \frac{\pi }{{12}}\) and \(x = \frac{{5\pi }}{{12}}\)     A1

Note: Do not award the final A1 if extra solutions are seen.

[6 marks]

d.

Examiners report

[N/A]

a.

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b.

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c.

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d.
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