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]
\({\cos ^2}\hat A – 2\cos \hat A\sin \hat A + {\sin ^2}\hat A = \frac{1}{2}\) 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]
\({\text{B}}{{\text{C}}^2} + {({\text{BC}} + 1)^2} = 2\) 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]
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]
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).
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).
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).
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).
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]
\(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x} = [\arctan x]_{\tan \alpha }^{\cot \alpha }\) (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]
Examiners report
This was generally well done.
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.
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]
\(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x} = [\arctan x]_{\tan \alpha }^{\cot \alpha }\) (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]
Examiners report
This was generally well done.
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.
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]
\(\frac{{1 – \cos 2x}}{{2\sin x}} \equiv \frac{{1 – (1 – 2{{\sin }^2}x)}}{{2\sin x}}\) M1
\( \equiv \frac{{2{{\sin }^2}x}}{{2\sin x}}\) A1
\( \equiv \sin x\) AG
[2 marks]
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]
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]
Examiners report
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