1.05p - OCR Spec

Pre-U Pre-U 9794/1 2020 Specimen Q10

4 marks Standard +0.3

10

Prove that $\cot \theta + \frac { \sin \theta } { 1 + \cos \theta } = \operatorname { cosec } \theta$.
Hence solve the equation $\cot \left( \theta + \frac { \pi } { 4 } \right) + \frac { \sin \left( \theta + \frac { \pi } { 4 } \right) } { 1 + \cos \left( \theta + \frac { \pi } { 4 } \right) } = \frac { 5 } { 2 }$ for $0 \leqslant \theta \leqslant 2 \pi$.

CAIE P1 2023 June Q7

11 marks Standard +0.3

1. By first expanding $(\cos \theta + \sin \theta)^2$, find the three solutions of the equation $$(\cos \theta + \sin \theta)^2 = 1$$ for $0 \leqslant \theta \leqslant \pi$. [3]
2. Hence verify that the only solutions of the equation $\cos \theta + \sin \theta = 1$ for $0 < \theta < \pi$ are $0$ and $\frac{1}{2}\pi$. [2]
Prove the identity $\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 - \cos \theta}{\cos \theta - \sin \theta} \equiv \frac{\cos \theta + \sin \theta - 1}{1 - 2\sin^2 \theta}$. [3]
Using the results of (a)(ii) and (b), solve the equation $$\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 - \cos \theta}{\cos \theta - \sin \theta} = 2(\cos \theta + \sin \theta - 1)$$ for $0 \leqslant \theta \leqslant \pi$. [3]

CAIE P1 2012 June Q5

6 marks Moderate -0.3

Prove the identity $\tan x + \frac{1}{\tan x} = \frac{1}{\sin x \cos x}$. [2]
Solve the equation $\frac{2}{\sin x \cos x} = 1 + 3 \tan x$, for $0° \leqslant x \leqslant 180°$. [4]

CAIE P1 2012 June Q1

4 marks Moderate -0.8

Prove the identity $\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta$. [3]
Use this result to explain why $\tan \theta > \sin \theta$ for $0° < \theta < 90°$. [1]

CAIE P1 2015 June Q5

5 marks Moderate -0.3

Prove the identity $\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} \equiv \frac{\tan \theta - 1}{\tan \theta + 1}$. [1]
Hence solve the equation $\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{\tan \theta}{6}$, for $0° \leqslant \theta \leqslant 180°$. [4]

Edexcel C3 Q4

7 marks Moderate -0.3

Prove, by counter-example, that the statement "$\sec(A + B) = \sec A + \sec B$, for all $A$ and $B$" is false. [2]
Prove that $$\tan \theta + \cot \theta = 2 \cosec 2\theta, \quad \theta \neq \frac{n\pi}{2}, n \in \mathbb{Z}.$$ [5]

Edexcel C3 Q5

8 marks Moderate -0.3

Given that $\sin x = \frac{3}{5}$, use an appropriate double angle formula to find the exact value of $\sec 2x$. [4]
Prove that $$\cot 2x + \cosec 2x \equiv \cot x, \quad \left(x \neq \frac{n\pi}{2}, n \in \mathbb{Z}\right).$$ [4]

Edexcel C3 Q7

12 marks Standard +0.3

Use the identity $$\cos (A + B) = \cos A \cos B - \sin A \sin B$$ to prove that $$\cos x \equiv 1 - 2 \sin^2 \frac{x}{2}.$$ [3]
Prove that, for $\sin x \neq 0$, $$\frac{1 - \cos x}{\sin x} \equiv \tan \frac{x}{2}.$$ [3]
Find the values of $x$ in the interval $0 \leq x \leq 360^{\circ}$ for which $$\frac{1 - \cos x}{\sin x} = 2 \sec^2 \frac{x}{2} - 5,$$ giving your answers to 1 decimal place where appropriate. [6]

OCR MEI C4 2012 June Q4

4 marks Moderate -0.5

Prove that $\sec^2\theta + \cosec^2\theta = \sec^2\theta \cosec^2\theta$. [4]

OCR H240/03 2019 June Q4

14 marks Standard +0.3

\includegraphics{figure_4} The diagram shows the part of the curve $y = 3x \sin 2x$ for which $0 \leqslant x \leqslant \frac{1}{2}\pi$. The maximum point on the curve is denoted by $P$.

Show that the $x$-coordinate of $P$ satisfies the equation $\tan 2x + 2x = 0$. [3]
Use the Newton-Raphson method, with a suitable initial value, to find the $x$-coordinate of $P$, giving your answer correct to 4 decimal places. Show the result of each iteration. [4]
The trapezium rule, with four strips of equal width, is used to find an approximation to $$\int_0^{\frac{1}{2}\pi} 3x \sin 2x \, dx.$$ Show that the result can be expressed as $k\pi^2(\sqrt{2} + 1)$, where $k$ is a rational number to be determined. [4]
1. Evaluate $\int_0^{\frac{1}{2}\pi} 3x \sin 2x \, dx$. [1]
2. Hence determine whether using the trapezium rule with four strips of equal width gives an under- or over-estimate for the area of the region enclosed by the curve $y = 3x \sin 2x$ and the $x$-axis for $0 \leqslant x \leqslant \frac{1}{2}\pi$. [1]
3. Explain briefly why it is not easy to tell from the diagram alone whether the trapezium rule with four strips of equal width gives an under- or over-estimate for the area of the region in this case. [1]

SPS SPS SM Pure 2021 May Q4

3 marks Standard +0.3

Prove that $\sqrt{2}\cos(2\theta + 45°) \equiv \cos^2\theta - 2\sin\theta\cos\theta - \sin^2\theta$, where $\theta$ is measured in degrees. [3]

SPS SPS SM 2021 November Q6

7 marks Challenging +1.2

Prove the following trigonometric identities. You must show all of your algebraic steps clearly. $$(\cos x + \sin x)(\cos x - \sec x) \equiv 2 \cot 2x$$ [3]
Solve the following equation for $x$ in the interval $0 \leq x \leq \pi$. Giving your answers in terms of $\pi$. $$\sin\left(2x + \frac{\pi}{6}\right) = \frac{1}{2}\sin\left(2x - \frac{\pi}{6}\right)$$ [4]

OCR H240/03 2018 December Q11

16 marks Standard +0.3

A ball $B$ is projected with speed $V$ at an angle $\alpha$ above the horizontal from a point $O$ on horizontal ground. The greatest height of $B$ above $O$ is $H$ and the horizontal range of $B$ is $R$. The ball is modelled as a particle moving freely under gravity.

Show that
1. $H = \frac{V^2}{2g}\sin^2 \alpha$, [2]
2. $R = \frac{V^2}{g}\sin 2\alpha$. [3]
Hence show that $16H^2 - 8R_0 H + R^2 = 0$, where $R_0$ is the maximum range for the given speed of projection. [5]
Given that $R_0 = 200\text{m}$ and $R = 192\text{m}$, find
1. the two possible values of the greatest height of $B$, [2]
2. the corresponding values of the angle of projection. [3]
State one limitation of the model that could affect your answers to part (iii). [1]

1.05p Proof involving trig: functions and identities