SPS SPS SM Mechanics 2022 February — Question 7 7 marks

Exam BoardSPS
ModuleSPS SM Mechanics (SPS SM Mechanics)
Year2022
SessionFebruary
Marks7
TopicStandard trigonometric equations
TypeSolve using given identity
DifficultyStandard +0.3 This is a straightforward trigonometric identity question requiring standard double-angle formulas and algebraic manipulation in part (a), followed by a routine equation solve in part (b) using the established identity. The techniques are well-practiced at A-level (using 1-cos2θ=2sin²θ and sin2θ=2sinθcosθ), and part (b) reduces to a simple quadratic in tanx. While it requires multiple steps, no novel insight or complex problem-solving is needed—slightly easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
  1. Show that $$\frac{1 - \cos 2\theta}{\sin^2 2\theta} = k \sec^2 \theta \quad \theta = \frac{n\pi}{2} \quad n \in \mathbb{Z}$$ where \(k\) is a constant to be found. [3]
  2. Hence solve, for \(-\frac{\pi}{2} < x < \frac{\pi}{2}\) $$\frac{1 - \cos 2x}{\sin^2 2x} = (1 + 2\tan x)^2$$ Give your answers to 3 significant figures where appropriate. [4]

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac{1 - \cos 2\theta}{\sin^2 2\theta} = k \sec^2 \theta \quad \theta = \frac{n\pi}{2} \quad n \in \mathbb{Z}$$
where $k$ is a constant to be found.
[3]

\item Hence solve, for $-\frac{\pi}{2} < x < \frac{\pi}{2}$
$$\frac{1 - \cos 2x}{\sin^2 2x} = (1 + 2\tan x)^2$$
Give your answers to 3 significant figures where appropriate.
[4]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM Mechanics 2022 Q7 [7]}}