Question
Consider the functions \( f(x)=cos x\) and \(g(x)=sin 2x\), where \(0\leq x\leq \pi \).
The graph of f intersects the graph of g at the point A, the point B \(\left ( \frac{\pi }{2},0 \right )\) and the point C as shown on the following diagram.
(a) Find the x-coordinate of point A and the x-coordinate of point C.
The shaded region R is enclosed by the graph of f and the graph of g between the points B and C.
(b) Find the area of R.
▶️Answer/Explanation
Ans:
(a) recognising \(cos x=2sin x cos x\)
\(cos x\neq 0\) so \(sin x=\frac{1}{2}\) OR one correct value (accept degrees)
x-coordinates \(\frac{\pi }{6}\) and \(\frac{5\pi }{6}\)
(b) METHOD 1
attempt to integrate \(\pm (cos x-sin 2x)\)
\(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}(cos x – sin 2x)dx\) OR \(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}(cos x – 2 sin x cos x)dx\)
\(=\left [ sin x +\frac{1}{2}cos 2x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\) OR \(=\left [ sin x – sin^{2}x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\)
attempt to substitute their limits into their integral and subtract
\(=\left ( sin\left ( \frac{5\pi }{6} \right )+\frac{1}{2}cos\left ( \frac{5\pi }{3} \right ) \right )-\left ( sin\left ( \frac{\pi }{2} \right )+\frac{1}{2}cos(\pi ) \right )\) OR \(=\left ( sin\left ( \frac{5\pi }{6} \right )-sin^{2}\left ( \frac{5\pi }{6} \right ) \right )-\left ( sin\left ( \frac{\pi }{2} \right )-sin^{2}\left ( \frac{\pi }{2} \right )\right)\)
\(=\left ( \frac{1}{2}+\frac{1}{4} \right )-\left ( 1-\frac{1}{2} \right )\) OR \(=\left ( \frac{1}{2}-\frac{1}{4} \right )-\left ( 1-1 \right )\)
area = \(\frac{1}{4}\)
METHOD 2
\(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}cos xdx=\left [ sin x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\) and \(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}sin2xdx=\left [ -\frac{1}{2}cos2x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\)
attempt to substitute their limits into their integral and subtract (for both integrals)
\(sin\left ( \frac{5\pi }{6} \right )-sin\left ( \frac{\pi }{2} \right )\) and \(-\frac{1}{2}cos\left ( \frac{5\pi }{3} \right )+\frac{1}{2}cos\left ( \pi \right )\)
attempt to subtract the two integrals in either order (seen anywhere)
\(\left ( sin\left ( \frac{5\pi }{6} \right )-sin\left ( \frac{\pi }{2} \right ) \right )-\left ( -\frac{1}{2}cos\left ( \frac{5\pi }{3} \right )+\frac{1}{2}cos(\pi ) \right )\) OR \(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}cos xdx-\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}sin2xdx\)
\(=\left ( \frac{1}{2}-1 \right )-\left ( \frac{1}{4}-\frac{1}{2} \right )\left ( =-\frac{1}{4} \right )\)
area = \(\frac{1}{4}\)
Detailed Solution
(a) Given functions \( f(x)=cos x\) and \(g(x)=sin 2x\)
we know sin2x = 2sinxcosx
to find the intersection points between f(x) and (g(x), we equate f(x) = g(x)
therefore cos x = 2sinxcosx
cos x – 2sinxcosx = 0
cos x (1- 2sinx) = 0
cos x = 0 , therefore x =\( \frac{\Pi}{2} \) and
1- 2sinx = 0 , sinx = \( \frac{1}{2} \) therefore
x = \(\frac{\Pi}{6}\) , \(\frac{5\Pi }{6}\)
(b) The shaded region R is enclosed between the points B (x= \( \frac{\Pi}{2} \) ) and C (x=\(\frac{5\Pi }{6}\))
Therefore, area of the shaded region R = \(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}(f(x) – g(x))dx\)
= \(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}(cos x – sin 2x)dx\)
= \(=\left [ sin x +\frac{1}{2}cos 2x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\)
= \(=\left ( sin\left ( \frac{5\pi }{6} \right )+\frac{1}{2}cos\left ( \frac{5\pi }{3} \right ) \right )-\left ( sin\left ( \frac{\pi }{2} \right )+\frac{1}{2}cos(\pi ) \right )\)
= \(=\left ( sin\left ( \frac{5\pi }{6} \right )-sin^{2}\left ( \frac{5\pi }{6} \right ) \right )-\left ( sin\left ( \frac{\pi }{2} \right )-sin^{2}\left ( \frac{\pi }{2} \right )\right)\)
substituting the values we get
\(=\left ( \frac{1}{2}-1 \right )-\left ( \frac{1}{4}-\frac{1}{2} \right )\left ( =-\frac{1}{4} \right )\)
area = \(\frac{1}{4}\)
Question
In the diagram below, AD is perpendicular to BC.
CD = 4, BD = 2 and AD = 3. \({\rm{C}}\hat {\rm{A}}{\rm{D}} = \alpha \) and \({\rm{B}}\hat {\rm{A}}{\rm{D}} = \beta \) .
Find the exact value of \(\cos (\alpha – \beta )\) .
▶️Answer/Explanation
Markscheme
METHOD 1
\({\text{AC}} = 5\) and \(\text{AB} = \sqrt {13}\) (may be seen on diagram) (A1)
\(\cos \alpha = \frac{3}{5}\) and \(\sin \alpha = \frac{4}{5}\) (A1)
\(\cos \beta = \frac{3}{{\sqrt {13} }}\) and \(\sin \beta = \frac{2}{{\sqrt {13} }}\) (A1)
Note: If only the two cosines are correctly given award (A1)(A1)(A0).
Use of \(\cos (\alpha – \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \) (M1)
\( = \frac{3}{5} \times \frac{3}{{\sqrt {13} }} + \frac{4}{5} \times \frac{2}{{\sqrt {13} }}\) (substituting) M1
\( = \frac{{17}}{{5\sqrt {13} }}\) \(\left( { = \frac{{17\sqrt {13} }}{{65}}} \right)\) A1 N1
[6 marks]
METHOD 2
\({\text{AC}} = 5\) amd \({\text{AB}} = \sqrt {13} \) (may be seen on diagram) (A1)
Use of \(\cos (\alpha + \beta ) = \frac{{{\text{A}}{{\text{C}}^2} + {\text{A}}{{\text{B}}^2} – {\text{B}}{{\text{C}}^2}}}{{{\text{2(AC)(AB)}}}}\) (M1)
\( = \frac{{25 + 13 – 36}}{{2 \times 5 \times \sqrt {13} }}\,\,\,\,\,\left( { = \frac{1}{{5\sqrt {13} }}} \right)\) A1
Use of \(\cos (\alpha + \beta ) + \cos (\alpha – \beta ) = 2\cos \alpha \cos \beta \) (M1)
\(\cos \alpha = \frac{3}{5}\) and \(\cos \beta = \frac{3}{{\sqrt {13} }}\) (A1)
\(\cos (\alpha – \beta ) = \frac{{17}}{{5\sqrt {13} }}\,\,\,\,\,\left( { = 2 \times \frac{3}{5} \times \frac{3}{{\sqrt {13} }} – \frac{1}{{5\sqrt {13} }}} \right){\text{ }}\left( { = \frac{{17\sqrt {13} }}{{65}}} \right)\) A1 N1
[6 marks]
Question
(a) Show that \(\sin 2nx = \sin \left( {(2n + 1)x} \right)\cos x – \cos \left( {(2n + 1)x} \right)\sin x\).
(b) Hence prove, by induction, that
\[\cos x + \cos 3x + \cos 5x + \ldots + \cos \left( {(2n – 1)x} \right) = \frac{{\sin 2nx}}{{2\sin x}},\]
for all \(n \in {\mathbb{Z}^ + }{\text{, }}\sin x \ne 0\).
(c) Solve the equation \(\cos x + \cos 3x = \frac{1}{2},{\text{ }}0 < x < \pi \).
▶️Answer/Explanation
Markscheme
(a) \(\sin (2n + 1)x\cos x – \cos (2n + 1)x\sin x = \sin (2n + 1)x – x\) M1A1
\( = \sin 2nx\) AG
[2 marks]
(b) if n = 1 M1
\({\text{LHS}} = \cos x\)
\({\text{RHS}} = \frac{{\sin 2x}}{{2\sin x}} = \frac{{2\sin x\cos x}}{{2\sin x}} = \cos x\) M1
so LHS = RHS and the statement is true for n = 1 R1
assume true for n = k M1
Note: Only award M1 if the word true appears.
Do not award M1 for ‘let n = k’ only.
Subsequent marks are independent of this M1.
so \(\cos x + \cos 3x + \cos 5x + \ldots + \cos (2k – 1)x = \frac{{\sin 2kx}}{{2\sin x}}\)
if n = k + 1 then
\(\cos x + \cos 3x + \cos 5x + \ldots + \cos (2k – 1)x + \cos (2k + 1)x\) M1
\( = \frac{{\sin 2kx}}{{2\sin x}} + \cos (2k + 1)x\) A1
\( = \frac{{\sin 2kx + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\) M1
\( = \frac{{\sin (2k + 1)x\cos x – \cos (2k + 1)x\sin x + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\) M1
\( = \frac{{\sin (2k + 1)x\cos x + \cos (2k + 1)x\sin x}}{{2\sin x}}\) A1
\( = \frac{{\sin (2k + 2)x}}{{2\sin x}}\) M1
\( = \frac{{\sin 2(k + 1)x}}{{2\sin x}}\) A1
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: Final R1 is independent of previous work.
[12 marks]
(c) \(\frac{{\sin 4x}}{{2\sin x}} = \frac{1}{2}\) M1A1
\(\sin 4x = \sin x\)
\(4x = x \Rightarrow x = 0\) but this is impossible
\(4x = \pi – x \Rightarrow x = \frac{\pi }{5}\) A1
\(4x = 2\pi + x \Rightarrow x = \frac{{2\pi }}{3}\) A1
\(4x = 3\pi – x \Rightarrow x = \frac{{3\pi }}{5}\) A1
for not including any answers outside the domain R1
Note: Award the first M1A1 for correctly obtaining \(8{\cos ^3}x – 4\cos x – 1 = 0\) or equivalent and subsequent marks as appropriate including the answers \(\left( { – \frac{1}{2},\frac{{1 \pm \sqrt 5 }}{4}} \right)\).
[6 marks] Total [20 marks]
Question
If x satisfies the equation \(\sin \left( {x + \frac{\pi }{3}} \right) = 2\sin x\sin \left( {\frac{\pi }{3}} \right)\), show that \(11\tan x = a + b\sqrt 3 \), where a, b \( \in {\mathbb{Z}^ + }\).
▶️Answer/Explanation
Markscheme
\(\sin \left( {x + \frac{\pi }{3}} \right) = \sin x\cos \left( {\frac{\pi }{3}} \right) + \cos x\sin \left( {\frac{\pi }{3}} \right)\) (M1)
\(\sin x\cos \left( {\frac{\pi }{3}} \right) + \cos x\sin \left( {\frac{\pi }{3}} \right) = 2\sin x\sin \left( {\frac{\pi }{3}} \right)\)
\(\frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = 2 \times \frac{{\sqrt 3 }}{2}\sin x\) A1
dividing by \(\cos x\) and rearranging M1
\(\tan x = \frac{{\sqrt 3 }}{{2\sqrt 3 – 1}}\) A1
rationalizing the denominator M1
\(11\tan x = 6 + \sqrt 3 \) A1
[6 marks]