SPS SPS FM Pure 2025 June — Question 6 9 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2025
SessionJune
Marks9
TopicTrig Proofs
TypeSolve equation using proven identity
DifficultyStandard +0.3 Part (a) is a straightforward trigonometric identity proof using standard double angle formulas (cos 2θ = 1 - 2sin²θ and sin 2θ = 2sin θ cos θ). Part (b) applies this identity to solve an equation, requiring substitution of sec²x = 1 + tan²x and algebraic manipulation to reach a quadratic in tan²x. While it involves multiple steps, the techniques are standard A-level Further Maths material with no novel insight required, making it slightly easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

  1. Prove that $$1 - \cos 2\theta = \tan \theta \sin 2\theta, \quad \theta \neq \frac{(2n + 1)\pi}{2}, \quad n \in \mathbb{Z}$$ [3]
  2. Hence solve, for \(-\frac{\pi}{2} < x < \frac{\pi}{2}\), the equation $$(\sec^2 x - 5)(1 - \cos 2x) = 3\tan^2 x \sin 2x$$ Give any non-exact answer to 3 decimal places where appropriate. [6]

\begin{enumerate}[label=(\alph*)]
\item Prove that
$$1 - \cos 2\theta = \tan \theta \sin 2\theta, \quad \theta \neq \frac{(2n + 1)\pi}{2}, \quad n \in \mathbb{Z}$$
[3]

\item Hence solve, for $-\frac{\pi}{2} < x < \frac{\pi}{2}$, the equation
$$(\sec^2 x - 5)(1 - \cos 2x) = 3\tan^2 x \sin 2x$$

Give any non-exact answer to 3 decimal places where appropriate.
[6]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM Pure 2025 Q6 [9]}}