Find value where max/min occurs

7

1. Express \(5 \cos \theta - 12 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation \(5 \cos \theta - 12 \sin \theta = 8\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
3. Find the greatest possible value of $$7 + 5 \cos \frac { 1 } { 2 } \phi - 12 \sin \frac { 1 } { 2 } \phi$$ as \(\phi\) varies, and determine the smallest positive value of \(\phi\) for which this greatest value occurs.
   [0pt] [4]

6

1. By first expanding \(\cos \left( x - 60 ^ { \circ } \right)\), show that the expression $$2 \cos \left( x - 60 ^ { \circ } \right) + \cos x$$ can be written in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places.
2. Hence find the value of \(x\) in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\) for which \(2 \cos \left( x - 60 ^ { \circ } \right) + \cos x\) takes its least possible value.

1. (a) Express \(12 \sin x - 5 \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\). Give the exact value of \(R\) and give the value of \(\alpha\) in radians, to 3 decimal places.

The function g is defined by $$g ( \theta ) = 10 + 12 \sin \left( 2 \theta - \frac { \pi } { 6 } \right) - 5 \cos \left( 2 \theta - \frac { \pi } { 6 } \right) \quad \theta > 0$$ Find
(b) (i) the minimum value of \(\mathrm { g } ( \theta )\) (ii) the smallest value of \(\theta\) at which the minimum value occurs. The function h is defined by $$\mathrm { h } ( \beta ) = 10 - ( 12 \sin \beta - 5 \cos \beta ) ^ { 2 }$$ (c) Find the range of h .

1. (a) Express \(35 \sin x - 12 \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\)

Give the exact value of \(R\), and give the value of \(\alpha\), in radians, to 4 significant figures.
(b) Hence solve, for \(0 \leqslant x < 2 \pi\), $$70 \sin x - 24 \cos x = 37$$ (Solutions based entirely on graphical or numerical methods are not acceptable.) $$y = \frac { 7000 } { 31 + ( 35 \sin x - 12 \cos x ) ^ { 2 } } , \quad x > 0$$ (c) Use your answer to part (a) to calculate

1. the minimum value of \(y\),
2. the smallest value of \(x , x > 0\), at which this minimum value occurs.

1. (a) Express \(6 \cos \theta + 8 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\).

Give the value of \(\alpha\) to 3 decimal places.
(b) $$\mathrm { p } ( \theta ) = \frac { 4 } { 12 + 6 \cos \theta + 8 \sin \theta } , \quad 0 \leqslant \theta \leqslant 2 \pi$$ Calculate

1. the maximum value of \(\mathrm { p } ( \theta )\),
2. the value of \(\theta\) at which the maximum occurs.

9. (a) Express \(2 \sin \theta - 4 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\) Give the value of \(\alpha\) to 3 decimal places. $$H ( \theta ) = 4 + 5 ( 2 \sin 3 \theta - 4 \cos 3 \theta ) ^ { 2 }$$ Find
(b) (i) the maximum value of \(\mathrm { H } ( \theta )\),
(ii) the smallest value of \(\theta\), for \(0 \leqslant \theta < \pi\), at which this maximum value occurs. Find
(c) (i) the minimum value of \(\mathrm { H } ( \theta )\),
(ii) the largest value of \(\theta\), for \(0 \leqslant \theta < \pi\), at which this minimum value occurs.

1. (a) Express \(\sin \theta - 2 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\)

Give the exact value of \(R\) and the value of \(\alpha\), in radians, to 3 decimal places. $$\mathrm { M } ( \theta ) = 40 + ( 3 \sin \theta - 6 \cos \theta ) ^ { 2 }$$ (b) Find

1. the maximum value of \(\mathrm { M } ( \theta )\),
2. the smallest value of \(\theta\), in the range \(0 < \theta \leqslant 2 \pi\), at which the maximum value of \(\mathrm { M } ( \theta )\) occurs. $$N ( \theta ) = \frac { 30 } { 5 + 2 ( \sin 2 \theta - 2 \cos 2 \theta ) ^ { 2 } }$$ (c) Find
3. the maximum value of \(\mathrm { N } ( \theta )\),
4. the largest value of \(\theta\), in the range \(0 < \theta \leqslant 2 \pi\), at which the maximum value of \(\mathrm { N } ( \theta )\) occurs.
   (Solutions based entirely on graphical or numerical methods are not acceptable.)

   END

5 Express \(\sqrt { 3 } \sin x - \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\). Express \(\alpha\) in the form \(k \pi\). Find the exact coordinates of the maximum point of the curve \(y = \sqrt { 3 } \sin x - \cos x\) for which \(0 < x < 2 \pi\).

7 Fig. 1 shows part of the graph of \(y = \sin x \quad \sqrt { 3 } \cos x\). \begin{figure}[h] \end{figure} Express \(\quad \sqrt { } \quad\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 \leqslant \alpha \leqslant \frac { 1 } { 2 } \pi\).
Hence write down the exact coordinates of the turning point P .

8

1. Express \(4 \cos \theta - 2 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
2. Hence
   1. solve the equation \(4 \cos \theta - 2 \sin \theta = 3\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\),
   2. determine the greatest and least values of $$25 - ( 4 \cos \theta - 2 \sin \theta ) ^ { 2 }$$ as \(\theta\) varies, and, in each case, find the smallest positive value of \(\theta\) for which that value occurs.

6. a. Express \(4 \sin x - 5 \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\). Give the exact value of \(R\), and give the value of \(\alpha\), in degrees, to 2 decimal places. $$T = \frac { 8400 } { 19 + ( 4 \sin x - 5 \cos x ) ^ { 2 } } , x > 0$$ b. Use your answer to part \(a\) to calculate
i. the minimum value of \(T\).
ii. the smallest value of \(x , x > 0\), at which this minimum value occurs.

1

1. Express \(2 \sin x + 5 \cos x\) in the form \(R \sin ( x + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
2. 1. Write down the maximum value of \(2 \sin x + 5 \cos x\).
   2. Find the value of \(x\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\) at which this maximum occurs, giving the value of \(x\) to the nearest \(0.1 ^ { \circ }\).

3

1. 1. Express \(3 \cos x + 2 \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
      (3 marks)
   2. Hence find the minimum value of \(3 \cos x + 2 \sin x\) and the value of \(x\) in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\) where the minimum occurs. Give your value of \(x\) to the nearest \(0.1 ^ { \circ }\).
2. 1. Show that \(\cot x - \sin 2 x = \cot x \cos 2 x\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\).
   2. Hence, or otherwise, solve the equation $$\cot x - \sin 2 x = 0$$ in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).

2

1. Express \(\sin x - 3 \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\). Give your value of \(\alpha\) in radians to two decimal places.
2. Hence:
   1. write down the minimum value of \(\sin x - 3 \cos x\);
   2. find the value of \(x\) in the interval \(0 < x < 2 \pi\) at which this minimum value occurs, giving your value of \(x\) in radians to two decimal places.

1 Fig. 1 shows part of the graph of \(y = \sin x - \sqrt { 3 } \cos x\). \begin{figure}[h] \end{figure} Express \(\sin x - \sqrt { 3 } \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 \leqslant \alpha \leqslant \frac { 1 } { 2 } \pi\).
Hence write down the exact coordinates of the turning point P .

7 Express \(\sqrt { 3 } \sin x - \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\). Express \(\alpha\) in the form \(k \pi\). Find the exact coordinates of the maximum point of the curve \(y = \sqrt { 3 } \sin x - \cos x\) for which \(0 < x < 2 \pi\).

8

1. Express \(\sin x - \sqrt { 8 } \cos x\) in the form \(R \sin ( x - \alpha )\) where \(R \geqslant 0\) and \(0 \leqslant \alpha \leqslant 90 ^ { \circ }\).
2. Hence write down the maximum value of \(\sin x - \sqrt { 8 } \cos x\) and find the smallest positive value of \(x\) for which it occurs.

Express \(3\sin x + 2\cos x\) in the form \(R\sin(x + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\). Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve \(y = f(x)\), where $$f(x) = 3\sin x + 2\cos x, \quad 0 < x < \pi.$$ [7]

The motion of a particle is modelled by the differential equation $$v \frac{dv}{dt} + 4x = 0,$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).

1. Solve the differential equation to show that \(v^2 = 20 - 4x^2\). [4]

Now consider motion for which \(x = \cos 2t + 2 \sin 2t\), where \(x\) is the displacement from a fixed point at time \(t\).

1. Verify that, when \(t = 0\), \(x = 1\). Use the fact that \(v = \frac{dx}{dt}\) to verify that when \(t = 0\), \(v = 4\). [4]
2. Express \(x\) in the form \(R \cos(2t - \alpha)\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v^2 = 20 - 4x^2\). [7]
3. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value. [3]

Express \(3\sin x + 2\cos x\) in the form \(R\sin(x + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\) Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve \(y = f(x)\), where $$f(x) = 3\sin x + 2\cos x, \quad 0 \leqslant x \leqslant \pi.$$ [7]

In Fig. 6, OAB is a thin bent rod, with OA = \(a\) metres, AB = \(b\) metres and angle OAB = 120°. 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. \includegraphics{figure_6}

1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a\sin\theta + b\sin(\theta - 60°).$$ [3]
2. Hence show that \(h = (a + \frac{1}{2}b)\sin\theta - \frac{\sqrt{3}}{2}b\cos\theta\). [3]

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.

1. Show that OB is horizontal when \(\tan\theta = \frac{\sqrt{3}b}{2a + b}\). [3]

In the case when \(a = 1\) and \(b = 2\), \(h = 2\sin\theta - \sqrt{3}\cos\theta\).

1. 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 \(\theta\). [7]

A new design for a company logo is to be made from two sectors of a circle, \(ORP\) and \(OQS\), and a rhombus \(OSTR\), as shown in the diagram below. \includegraphics{figure_7} The points \(P\), \(O\) and \(Q\) lie on a straight line and the angle \(ROS\) is \(\theta\) radians. A large copy of the logo, with \(PQ = 5\) metres, is to be put on a wall.

1. Show that the area of the logo, \(A\) square metres, is given by $$A = \frac{25}{8}(\pi - \theta + 2\sin\theta)$$ [4 marks]
2. 1. Show that the maximum value of \(A\) occurs when \(\theta = \frac{\pi}{3}\) Fully justify your answer. [6 marks]
   2. Find the exact maximum value of \(A\) [2 marks]
3. Without further calculation, state how your answers to parts (b)(i) and (b)(ii) would change if \(PQ\) were increased to 10 metres. [2 marks]