16N.1.SL.TZ0.S_2

pestleMathematicsAASLPaper 116N· sl-3-6-pythagorean-identity-double-anglessource ↗

Let $\sin \theta = \frac{{\sqrt 5 }}{3}$ , where $\theta$ is acute.

Find $\cos \theta$ .

[3]

Find $\cos 2\theta$ .

[2]

Markscheme / solution

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

evidence of valid approach (M1)

eg $\,\,\,\,\,$ right triangle, ${\cos ^2}\theta = 1 - {\sin ^2}\theta$

correct working (A1)

eg $\,\,\,\,\,$ missing side is 2, $\sqrt {1 - {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}}$

$\cos \theta = \frac{2}{3}$ A1 N2

[3 marks]

correct substitution into formula for $\cos 2\theta$ (A1)

eg $\,\,\,\,\,$ $2 \times {\left( {\frac{2}{3}} \right)^2} - 1,{\text{ }}1 - 2{\left( {\frac{{\sqrt 5 }}{3}} \right)^2},{\text{ }}{\left( {\frac{2}{3}} \right)^2} - {\left( {\frac{{\sqrt 5 }}{3}} \right)^2}$

$\cos 2\theta = - \frac{1}{9}$ A1 N2

[2 marks]

Examiners’ report

[N/A]