18N.1.SL.TZ0.S_7

pestleMathematicsAASLPaper 118N· sl-3-6-pythagorean-identity-double-anglessource ↗

Given that ${\text{sin}}\,x = \frac{1}{3}$ , where $0 < x < \frac{\pi }{2}$ , find the value of ${\text{cos}}\,4x$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

correct substitution into formula for ${\text{cos}}\,\left( {2x} \right)$ or ${\text{sin}}\,\left( {2x} \right)$ (A1)

eg $1 - 2{\left( {\frac{1}{3}} \right)^2}$ , $2{\left( {\frac{{\sqrt 8 }}{3}} \right)^2} - 1$ , $2\left( {\frac{1}{3}} \right)\left( {\frac{{\sqrt 8 }}{3}} \right)$ , ${\left( {\frac{{\sqrt 8 }}{3}} \right)^2} - {\left( {\frac{1}{3}} \right)^2}$

${\text{cos}}\,\left( {2x} \right) = \frac{7}{9}$ or ${\text{sin}}\,\left( {2x} \right) = \frac{{2\sqrt 8 }}{9}\,\,\,\,\left( { = \frac{{\sqrt {32} }}{9} = \frac{{4\sqrt 2 }}{9}} \right)$ (may be seen in substitution) A2

recognizing 4 $x$ is double angle of 2 $x$ (seen anywhere) (M1)

eg ${\text{cos}}\,\left( {2\left( {2x} \right)} \right)$ , $2\,{\text{co}}{{\text{s}}^2}\,\left( {2\theta } \right) - 1$ , $1 - 2\,{\text{si}}{{\text{n}}^2}\,\left( {2\theta } \right)$ , ${\text{co}}{{\text{s}}^2}\,\left( {2\theta } \right) - {\text{si}}{{\text{n}}^2}\,\left( {2\theta } \right)$

correct substitution of their value of ${\text{cos}}\,\left( {2x} \right)$ and/or ${\text{sin}}\,\left( {2x} \right)$ into formula for ${\text{cos}}\,\left( {4x} \right)$ (A1)

eg $2{\left( {\frac{7}{9}} \right)^2} - 1$ , $\frac{{98}}{{81}} - 1$ , $1 - 2{\left( {\frac{{2\sqrt 8 }}{9}} \right)^2}$ , $1 - \frac{{64}}{{81}}$ , ${\left( {\frac{7}{9}} \right)^2} - {\left( {\frac{{2\sqrt 8 }}{9}} \right)^2}$ , $\frac{{49}}{{81}} - \frac{{32}}{{81}}$

${\text{cos}}\,\left( {4x} \right) = \frac{{17}}{{81}}$ A1 N2

METHOD 2

recognizing 4 $x$ is double angle of 2 $x$ (seen anywhere) (M1)

eg ${\text{cos}}\,\left( {2\left( {2x} \right)} \right)$

double angle identity for 2 $x$ (M1)

eg $2\,{\text{co}}{{\text{s}}^2}\,\left( {2\theta } \right) - 1$ , $1 - 2\,{\text{si}}{{\text{n}}^2}\left( {2x} \right)$ , ${\text{co}}{{\text{s}}^2}\,\left( {2\theta } \right) - {\text{si}}{{\text{n}}^2}\,\left( {2\theta } \right)$

correct expression for ${\text{cos}}\,\left( {4x} \right)$ in terms of ${\text{sin}}\,x$ and/or ${\text{cos}}\,x$ (A1)

eg $2{\left( {1 - 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)^2} - 1$ , $1 - 2{\left( {2\,{\text{sin}}\,x\,{\text{cos}}\,x} \right)^2}$ , ${\left( {1 - 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)^2} - {\left( {2\,{\text{sin}}\,\theta \,{\text{cos}}\,\theta } \right)^2}$

correct substitution for ${\text{sin}}\,x$ and/or ${\text{cos}}\,x$ A1

eg $2{\left( {1 - 2{{\left( {\frac{1}{3}} \right)}^2}} \right)^2} - 1$ , $2\left( {1 - 4{{\left( {\frac{1}{3}} \right)}^2} + 4{{\left( {\frac{1}{3}} \right)}^4}} \right) - 1$ , $1 - 2{\left( {2 \times \frac{1}{3} \times \frac{{\sqrt 8 }}{3}} \right)^2}$

correct working (A1)

eg $2\left( {\frac{{49}}{{81}}} \right) - 1$ , $1 - 2\left( {\frac{{32}}{{81}}} \right)$ , $\frac{{49}}{{81}} - \frac{{32}}{{81}}$

${\text{cos}}\,\left( {4x} \right) = \frac{{17}}{{81}}$ A1 N2

[6 marks]

Examiners’ report

[N/A]