1.48: Past-paper question 11

4PM1/2/June/2025 — Question 11 · 15 marks

(a) Show that

\displaystyle \mathrm{cos}^4 \theta - \mathrm{sin}^4 \theta = \mathrm{cos} 2\theta

(4)

(b) Hence, or otherwise, solve the equation

8 \mathrm{cos}^2 \left( 2\theta + \frac{\pi}{4} \right) - 3 = 2 \mathrm{cos}^4 \left( \theta + \frac{\pi}{8} \right) - 2 \mathrm{sin}^4 \left( \theta + \frac{\pi}{8} \right) \quad \mathrm{for} 0 \leq \theta < \pi

Give your solutions to 2 decimal places.

(7)

\displaystyle \int_{\frac{\pi}{16}}^{\frac{\pi}{8}} (\mathrm{cos}^4 2x - \mathrm{sin}^4 2x - 8 \mathrm{sin} 4x) \mathrm{d}x

Give your answer in the form

\displaystyle a - b\sqrt{2}

where

\displaystyle a

and

\displaystyle b

are rational numbers.

(4)