Questions C4 (1162 questions)

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OCR C4 Q3
3 Show that \(\frac { \sin 2 \theta } { 1 + \cos 2 \theta } = \tan \theta\).
OCR C4 Q5
5 Solve the equation \(2 \sin 2 \theta + \cos 2 \theta = 1\), for \(0 ^ { \circ } \leqslant \theta < 360 ^ { \circ }\).
OCR C4 Q6
6 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
OCR C4 Q7
7
  1. Show that \(\cos ( \alpha + \beta ) = \frac { 1 - \tan \alpha \tan \beta } { \sec \alpha \sec \beta }\).
  2. Hence show that \(\cos 2 \alpha = \frac { 1 - \tan ^ { 2 } \alpha } { 1 + \tan ^ { 2 } \alpha }\).
  3. Hence or otherwise solve the equation \(\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = \frac { 1 } { 2 }\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
OCR C4 Q8
8 In Fig. 6, OAB is a thin bent rod, with \(\mathrm { OA } = a\) metres, \(\mathrm { AB } = b\) metres and angle \(\mathrm { OAB } = 120 ^ { \circ }\). The bent rod lies in a vertical plane. OA makes an angle \(\theta\) above the horizontal. The vertical height BD of B above O is \(h\) metres. The horizontal through A meets BD at C and the vertical through A meets OD at E . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9a4946dc-ab5a-499c-922c-a9eb1cd0f756-3_426_889_507_665} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$
  2. Hence show that \(h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta\). The rod now rotates about O , so that \(\theta\) varies. You may assume that the formulae for \(h\) in parts (i) and (ii) remain valid.
  3. Show that OB is horizontal when \(\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }\). In the case when \(a = 1\) and \(b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta\).
  4. Express \(2 \sin \theta - \sqrt { 3 } \cos \theta\) in the form \(R \sin ( \theta - \alpha )\). Hence, for this case, write down the maximum value of \(h\) and the corresponding value of \(\theta\).
  5. Express \(\cos \theta + \sqrt { 3 } \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(\alpha\) is acute, expressing \(\alpha\) in terms of \(\pi\).
  6. Write down the derivative of \(\tan \theta\). Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { ( \cos \theta + \sqrt { 3 } \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = \frac { \sqrt { 3 } } { 4 }\).
OCR MEI C4 Q1
1 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
OCR MEI C4 Q2
2
  1. Show that \(\cos ( \alpha + \beta ) = \frac { 1 - \tan \alpha \tan \beta } { \sec \alpha \sec \beta }\).
  2. Hence show that \(\cos 2 \alpha = \frac { 1 - \tan ^ { 2 } \alpha } { 1 + \tan ^ { 2 } \alpha }\).
  3. Hence or otherwise solve the equation \(\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = \frac { 1 } { 2 }\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
OCR MEI C4 Q3
3 Given the equation \(\sin \left( + 45 ^ { \circ } \right) = 2 \cos [\), show that \(\sin + \cos = 22 \cos\). Hence solve, correct to 2 decimal places, the equation for \(0 ^ { \circ } \quad \left[ \sqrt { 3 } 60 ^ { \circ } \right.\). $$\leqslant \leqslant$$
OCR MEI C4 Q4
4 Solve the equation \(\tan \left( \theta + 45 ^ { \circ } \right) = 1 - 2 \tan \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }\).
OCR MEI C4 Q5
5 Given that \(\sin ( \theta + \alpha ) = 2 \sin \theta\), show that \(\tan \theta = \frac { \sin \alpha } { 2 - \cos \alpha }\). Hence solve the equation \(\sin \left( \theta + 40 ^ { \circ } \right) = 2 \sin \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
OCR MEI C4 Q6
6 Solve the equation \(2 \cos 2 x = 1 + \cos x\), for \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).
OCR MEI C4 Q1
1 In Fig. 6, OAB is a thin bent rod, with \(\mathrm { OA } = a\) metres, \(\mathrm { AB } = b\) metres and angle \(\mathrm { OAB } = 120 ^ { \circ }\). The bent rod lies in a vertical plane. OA makes an angle \(\theta\) above the horizontal. The vertical height BD of B above O is \(h\) metres. The horizontal through A meets BD at C and the vertical through A meets OD at E . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9ac55ae6-7a7f-47d0-a363-92da179be4ca-1_427_898_464_683} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$
  2. Hence show that \(h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta\). The rod now rotates about O , so that \(\theta\) varies. You may assume that the formulae for \(h\) in parts (i) and (ii) remain valid.
  3. Show that OB is horizontal when \(\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }\). In the case when \(a = 1\) and \(b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta\).
  4. Express \(2 \sin \theta - \sqrt { 3 } \cos \theta\) in the form \(R \sin ( \theta - \alpha )\). Hence, for this case, write down the maximum value of \(h\) and the corresponding value of \(\theta\).
OCR MEI C4 Q2
2 Express \(3 \cos \theta + 4 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { - \pi } { 2 }\).
Hence solve the equation \(3 \cos \theta + 4 \sin \theta = 2\) for \(\quad - \pi \leqslant \theta \leqslant \pi\).
OCR MEI C4 Q3
3 Show that the equation \(\operatorname { cosec } x + 5 \cot x = 3 \sin x\) may be rearranged as $$3 \cos ^ { 2 } x + 5 \cos x - 2 = 0$$ Hence solve the equation for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\), giving your answers to 1 decimal place.
OCR MEI C4 Q4
4 Part of the track of a roller-coaster is modelled by a curve with the parametric equations $$x = 2 \theta - \sin \theta , \quad y = 4 \cos \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$ This is shown in Fig. 8. B is a minimum point, and BC is vertical. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9ac55ae6-7a7f-47d0-a363-92da179be4ca-3_591_1437_433_391} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the values of the parameter at A and B . Hence show that the ratio of the lengths OA and AC is \(( \pi - 1 ) : ( \pi + 1 )\).
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Find the gradient of the track at A .
  3. Show that, when the gradient of the track is \(1 , \theta\) satisfies the equation $$\cos \theta - 4 \sin \theta = 2$$
  4. Express \(\cos \theta - 4 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\). Hence solve the equation \(\cos \theta - 4 \sin \theta = 2\) for \(0 \leqslant \theta \leqslant 2 \pi\).
OCR MEI C4 Q5
5 Express \(4 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
Hence solve the equation \(4 \cos \theta - \sin \theta = 3\), for \(0 \leqslant \theta \leqslant 2 \pi\).
OCR MEI C4 Q1
1 Given that \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\), show that \(\cot ^ { 2 } \theta - \cot \theta - 2 = 0\).
Hence solve the equation \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
OCR MEI C4 Q2
2 Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for \(\pi\).
  1. Fig. 8.1 shows a regular 12-sided polygon inscribed in a circle of radius 1 unit, centre \(\mathrm { O } . \mathrm { AB }\) is one of the sides of the polygon. C is the midpoint of AB . Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{54a69773-651f-4e2f-9a3c-06ea7c07098b-2_457_422_457_936} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
    \end{figure} (A) Show that \(\mathrm { AB } = 2 \sin 15 ^ { \circ }\).
    (B) Use a double angle formula to express \(\cos 30 ^ { \circ }\) in terms of \(\sin 15 ^ { \circ }\). Using the exact value of \(\cos 30 ^ { \circ }\), show that \(\sin 15 ^ { \circ } = \frac { 1 } { 2 } \sqrt { 2 - \sqrt { 3 } }\).
    (C) Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6 \sqrt { 2 - \sqrt { 3 } }\).
  2. In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{54a69773-651f-4e2f-9a3c-06ea7c07098b-2_450_420_1562_938} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
    \end{figure} (A) Show that \(\mathrm { DE } = 2 \tan 15 ^ { \circ }\).
    (B) Let \(t = \tan 15 ^ { \circ }\). Use a double angle formula to express \(\tan 30 ^ { \circ }\) in terms of \(t\). Hence show that \(t ^ { 2 } + 2 \sqrt { 3 } t - 1 = 0\).
    (C) Solve this equation, and hence show that \(\pi < 12 ( 2 - \sqrt { 3 } )\).
  3. Use the results in parts (i)( \(C\) ) and (ii)( \(C\) ) to establish upper and lower bounds for the value of \(\pi\), giving your answers in decimal form.
OCR MEI C4 Q3
3 Express \(\sin \theta - 3 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Hence solve the equation \(\sin \theta - 3 \cos \theta = 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{54a69773-651f-4e2f-9a3c-06ea7c07098b-4_606_624_236_754} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2 \theta , \quad y = 10 \sin \theta + 5 \sin 2 \theta , \quad ( 0 \leqslant \theta < 2 \pi )$$ where \(x\) and \(y\) are in metres.
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { \cos \theta + \cos 2 \theta } { \sin \theta + \sin 2 \theta }\). Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 3 } \pi\). Hence find the exact coordinates of the highest point A on the path of C .
  2. Express \(x ^ { 2 } + y ^ { 2 }\) in terms of \(\theta\). Hence show that $$x ^ { 2 } + y ^ { 2 } = 125 + 100 \cos \theta$$
  3. Using this result, or otherwise, find the greatest and least distances of C from O . You are given that, at the point B on the path vertically above O , $$2 \cos ^ { 2 } \theta + 2 \cos \theta - 1 = 0$$
  4. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures.
OCR MEI C4 Q5
5 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ } .$$
OCR MEI C4 Q1
1 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).
  1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
  2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1e788cb0-36b0-42a9-9e0c-077022d410ae-1_556_867_580_588} \captionsetup{labelformat=empty} \caption{Fig. 5}
    \end{figure}
  3. Find the volume of revolution produced, giving your answer in exact form.
OCR MEI C4 Q2
2 Show that the equation \(\operatorname { cosec } x + 5 \cot x = 3 \sin x\) may be rearranged as $$3 \cos ^ { 2 } x + 5 \cos x - 2 = 0$$ Hence solve the equation for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\), giving your answers to 1 decimal place.
OCR MEI C4 Q3
3 Using appropriate right-angled triangles, show that \(\tan 45 ^ { \circ } = 1\) and \(\tan 30 ^ { \circ } = \frac { 1 } { \sqrt { 3 } }\).
Hence show that \(\tan 75 ^ { \circ } = 2 + \sqrt { 3 }\).
OCR MEI C4 Q4
4 Prove that \(\sec ^ { 2 } \theta + \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta \operatorname { cosec } ^ { 2 } \theta\).
OCR MEI C4 Q5
5 Solve the equation \(\operatorname { cosec } ^ { 2 } \theta = 1 + 2 \cot \theta\), for \(- 180 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).