IB DP Maths Topic 3.3 :Double angle identities HL Paper 1

(i)     Show that \({(1 + {\text{i}}\tan \theta )^n} + {(1 – {\text{i}}\tan \theta )^n} = \frac{{2\cos n\theta }}{{{{\cos }^n}\theta }},\;\;\;\cos \theta  \ne 0\).

(ii)     Hence verify that \({\text{i}}\tan \frac{{3\pi }}{8}\) is a root of the equation \({(1 + z)^4} + {(1 – z)^4} = 0,\;\;\;z \in \mathbb{C}\).

(iii)     State another root of the equation \({(1 + z)^4} + {(1 – z)^4} = 0,\;\;\;z \in \mathbb{C}\).

(i)     Use the double angle identity \(\tan 2\theta  = \frac{{2\tan \theta }}{{1 – {{\tan }^2}\theta }}\) to show that \(\tan \frac{\pi }{8} = \sqrt 2  – 1\).

(ii)     Show that \(\cos 4x = 8{\cos ^4}x – 8{\cos ^2}x + 1\).

(iii)     Hence find the value of \(\int_0^{\frac{\pi }{8}} {\frac{{2\cos 4x}}{{{{\cos }^2}x}}{\text{d}}x} \).

(i)     METHOD 1

\({(1 + {\text{i}}\tan \theta )^n} + {(1 – {\text{i}}\tan \theta )^n} = {\left( {1 + {\text{i}}\frac{{\sin \theta }}{{\cos \theta }}} \right)^n} + {\left( {1 – {\text{i}}\frac{{\sin \theta }}{{\cos \theta }}} \right)^n}\)     M1

\( = {\left( {\frac{{\cos \theta  + i\sin \theta }}{{\cos \theta }}} \right)^n} + {\left( {\frac{{\cos \theta  – i\sin \theta }}{{\cos \theta }}} \right)^n}\)     A1

by de Moivre’s theorem     (M1)

\({\left( {\frac{{\cos \theta  + i\sin \theta }}{{\cos \theta }}} \right)^n} = \frac{{\cos n\theta  + i\sin n\theta }}{{{{\cos }^n}\theta }}\)     A1

recognition that \(\cos \theta  – i\sin \theta \) is the complex conjugate of \(\cos \theta  + i\sin \theta \)     (R1)

use of the fact that the operation of complex conjugation commutes with the operation of raising to an integer power:

\({\left( {\frac{{\cos \theta  – i\sin \theta }}{{\cos \theta }}} \right)^n} = \frac{{\cos n\theta  – i\sin n\theta }}{{{{\cos }^n}\theta }}\)     A1

\({(1 + {\text{i}}\tan \theta )^n} + {(1 – {\text{i}}\tan \theta )^n} = \frac{{2\cos n\theta }}{{{{\cos }^n}\theta }}\)     AG

METHOD 2

\({(1 + {\text{i}}\tan \theta )^n} + {(1 – {\text{i}}\tan \theta )^n} = {(1 + {\text{i}}\tan \theta )^n} + {\left( {1 + {\text{i}}\tan ( – \theta )} \right)^n}\)     (M1)

\( = \frac{{{{(\cos \theta  + i\sin \theta )}^n}}}{{{{\cos }^n}\theta }} + \frac{{{{\left( {\cos ( – \theta ) + i\sin ( – \theta )} \right)}^n}}}{{{{\cos }^n}\theta }}\)     M1A1

Note:     Award M1 for converting to cosine and sine terms.

use of de Moivre’s theorem     (M1)

\( = \frac{1}{{{{\cos }^n}\theta }}\left( {\cos n\theta  + {\text{i}}\sin n\theta  + \cos ( – n\theta ) + {\text{i}}\sin ( – n\theta )} \right)\)     A1

\( = \frac{{2\cos n\theta }}{{{{\cos }^2}\theta }}\;\;\;{\text{as}}\;\;\;\cos ( – n\theta ) = \cos n\theta \;\;\;{\text{and}}\;\;\;\sin ( – n\theta ) =  – \sin n\theta \)     R1AG

(ii)     \({\left( {1 + {\text{i}}\tan \frac{{3\pi }}{8}} \right)^4} + {\left( {1 – {\text{i}}\tan \frac{{3\pi }}{8}} \right)^4} = \frac{{2\cos \left( {4 \times \frac{{3\pi }}{8}} \right)}}{{{{\cos }^4}\frac{{3\pi }}{8}}}\)     (A1)

\( = \frac{{2\cos \frac{{3\pi }}{2}}}{{{{\cos }^4}\frac{{3\pi }}{8}}}\)     A1

\( = 0\;\;\;{\text{as}}\;\;\;\cos \frac{{3\pi }}{2} = 0\)     R1

Note:     The above working could involve theta and the solution of \(\cos (4\theta ) = 0\).

so \({\text{i}}\tan \frac{{3\pi }}{8}\) is a root of the equation     AG

(iii)     either \( – {\text{i}}\tan \frac{{3\pi }}{8}\;\;\;{\text{or}}\;\;\; – {\text{i}}\tan \frac{\pi }{8}\;\;\;{\text{or}}\;\;\;{\text{i}}\tan \frac{\pi }{8}\)     A1

Note:     Accept \({\text{i}}\tan \frac{{5\pi }}{8}\;\;\;{\text{or}}\;\;\;{\text{i}}\tan \frac{{7\pi }}{8}\).

Accept \( – \left( {1 + \sqrt 2 } \right){\text{i}}\;\;\;{\text{or}}\;\;\;\left( {1 – \sqrt 2 } \right){\text{i}}\;\;\;{\text{or}}\;\;\;\left( { – 1 + \sqrt 2 } \right){\text{i}}\).

[10 marks]

(i)     \(\tan \frac{\pi }{4} = \frac{{2\tan \frac{\pi }{8}}}{{1 – {{\tan }^2}\frac{\pi }{8}}}\)     (M1)

\({\tan ^2}\frac{\pi }{8} + 2\tan \frac{\pi }{8} – 1 = 0\)     A1

let \(t = \tan \frac{\pi }{8}\)

attempting to solve \({t^2} + 2t – 1 = 0\;\;\;{\text{for}}\;\;\;t\)     M1

\(t =  – 1 \pm \sqrt 2 \)     A1

\(\frac{\pi }{8}\) is a first quadrant angle and tan is positive in this quadrant, so

\(\tan \frac{\pi }{8} > 0\)     R1

\(\tan \frac{\pi }{8} = \sqrt 2  – 1\)     AG

(ii)     \(\cos 4x = 2{\cos ^2}2x – 1\)     A1

\( = 2{\left( {2{{\cos }^2}x – 1} \right)^2} – 1\)     M1

\( = 2\left( {4{{\cos }^4}x – 4{{\cos }^2}x + 1} \right) – 1\)     A1

\( = 8{\cos ^4}x – 8{\cos ^2}x + 1\)     AG

Note:     Accept equivalent complex number derivation.

(iii)     \(\int_0^{\frac{\pi }{8}} {\frac{{2\cos 4x}}{{{{\cos }^2}x}}{\text{d}}x = 2} \int_0^{\frac{\pi }{8}} {\frac{{8{{\cos }^4}x – 8{{\cos }^2}x + 1}}{{{{\cos }^2}x}}{\text{d}}x} \)

\( = 2\int_0^{\frac{\pi }{8}} {8{{\cos }^2}x – 8 + {{\sec }^2}x{\text{d}}x} \)     M1

Note:     The M1 is for an integrand involving no fractions.

use of \({\cos ^2}x = \frac{1}{2}(\cos 2x + 1)\)     M1

\( = 2\int_0^{\frac{\pi }{8}} {4\cos 2x – 4 + {{\sec }^2}x{\text{d}}x} \)     A1

\( = [4\sin 2x – 8x + 2\tan x]_0^{\frac{\pi }{8}}\)     A1

\( = 4\sqrt 2  – \pi  – 2\;\;\;\)(or equivalent)     A1

[13 marks]

Total [23 marks]

Fairly successful.

(i)     Most candidates attempted to use the hint. Those who doubled the angle were usually successful – but many lost the final mark by not giving a convincing reason to reject the negative solution to the intermediate quadratic equation. Those who halved the angle got nowhere.

(ii)     The majority of candidates obtained full marks.

(iii)     This was poorly answered, few candidates realising that part of the integrand could be re-expressed using \(\frac{1}{{{{\cos }^2}x}} = {\sec ^2}x\), which can be immediately integrated.