1.05o Trigonometric equations: solve in given intervals

5

1. Write down the identity expressing \(\sin 2 \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).
2. Given that \(\sin \alpha = \frac { 1 } { 4 }\) and \(\alpha\) is acute, show that \(\sin 2 \alpha = \frac { 1 } { 8 } \sqrt { 15 }\).
3. Solve, for \(0 ^ { \circ } < \beta < 90 ^ { \circ }\), the equation \(5 \sin 2 \beta \sec \beta = 3\).

8

1. Express \(5 \cos x + 12 \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
2. Hence give details of a pair of transformations which transforms the curve \(y = \cos x\) to the curve \(y = 5 \cos x + 12 \sin x\).
3. Solve, for \(0 ^ { \circ } < x < 360 ^ { \circ }\), the equation \(5 \cos x + 12 \sin x = 2\), giving your answers correct to the nearest \(0.1 ^ { \circ }\).

7

1. Sketch the graph of \(y = \sec x\) for \(0 \leqslant x \leqslant 2 \pi\).
2. Solve the equation \(\sec x = 3\) for \(0 \leqslant x \leqslant 2 \pi\), giving the roots correct to 3 significant figures.
3. Solve the equation \(\sec \theta = 5 \operatorname { cosec } \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\), giving the roots correct to 3 significant figures.

9

1. Prove the identity $$\tan \left( \theta + 60 ^ { \circ } \right) \tan \left( \theta - 60 ^ { \circ } \right) \equiv \frac { \tan ^ { 2 } \theta - 3 } { 1 - 3 \tan ^ { 2 } \theta }$$
2. Solve, for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\), the equation $$\tan \left( \theta + 60 ^ { \circ } \right) \tan \left( \theta - 60 ^ { \circ } \right) = 4 \sec ^ { 2 } \theta - 3 ,$$ giving your answers correct to the nearest \(0.1 ^ { \circ }\).
3. Show that, for all values of the constant k , the equation $$\tan \left( \theta + 60 ^ { \circ } \right) \tan \left( \theta - 60 ^ { \circ } \right) = \mathrm { K } ^ { 2 }$$ has two roots in the interval \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

5

1. Express \(\tan 2 \alpha\) in terms of \(\tan \alpha\) and hence solve, for \(0 ^ { \circ } < \alpha < 180 ^ { \circ }\), the equation $$\tan 2 \alpha \tan \alpha = 8 .$$
2. Given that \(\beta\) is the acute angle such that \(\sin \beta = \frac { 6 } { 7 }\), find the exact value of
   1. \(\operatorname { cosec } \beta\),
   2. \(\cot ^ { 2 } \beta\).

7

1. Write down the formula for \(\tan 2 x\) in terms of \(\tan x\).
2. By letting \(\tan x = t\), show that the equation $$4 \tan 2 x + 3 \cot x \sec ^ { 2 } x = 0$$ becomes $$3 t ^ { 4 } - 8 t ^ { 2 } - 3 = 0$$
3. Hence find all the solutions of the equation $$4 \tan 2 x + 3 \cot x \sec ^ { 2 } x = 0$$ which lie in the interval \(0 \leqslant x \leqslant 2 \pi\).

6. One end of a light elastic string, of natural length 5 l and modulus of elasticity 20 mg , is attached to a fixed point \(A\). A particle \(P\) of mass \(2 m\) is attached to the free end of the string and \(P\) hangs freely in equilibrium at the point \(B\).

1. Find the distance \(A B\).
   (3) The particle is now pulled vertically downwards from \(B\) to the point \(C\) and released from rest. In the subsequent motion the string does not become slack.
2. Show that \(P\) moves with simple harmonic motion with centre \(B\).
3. Find the period of this motion. The greatest speed of \(P\) during this motion is \(\frac { 1 } { 5 } \sqrt { g l }\)
4. Find the amplitude of this motion. The point \(D\) is the midpoint of \(B C\) and the point \(E\) is the highest point reached by \(P\).
5. Find the time taken by \(P\) to move directly from \(D\) to \(E\).

1. A particle \(P\) of mass 0.5 kg moves on the \(x\)-axis under the action of a single force.

At time \(t\) seconds, \(t \geqslant 0\)

• \(O P = x\) metres, \(0 \leqslant x < \frac { \pi } { 2 }\)
• the force has magnitude \(\sin 2 x \mathrm {~N}\) and is directed towards the origin \(O\)
• \(P\) is moving in the positive \(x\) direction with speed \(v \mathrm {~ms} ^ { - 1 }\)

At time \(t = 0 , P\) passes through the origin with speed \(2 \mathrm {~ms} ^ { - 1 }\)

1. Show that \(v = 2 \cos x\)
2. Show that \(t = \frac { 1 } { 2 } \ln ( \sqrt { 2 } + 1 )\) when \(x = \frac { \pi } { 4 }\)

1. (i) Use the identity for \(\cos ( A + B )\) to prove that

$$\cos 2 x \equiv 2 \cos ^ { 2 } x - 1$$ (ii) Prove that, for \(\cos x \neq 0\), $$2 \cos x - \sec x \equiv \sec x \cos 2 x$$ (iii) Hence, or otherwise, find the values of \(x\) in the interval \(0 \leq x \leq 180 ^ { \circ }\) for which $$2 \cos x - \sec x \equiv 2 \cos 2 x$$

1. (i) Show that the equation

$$2 \sin x + \sec \left( x + \frac { \pi } { 6 } \right) = 0$$ can be written as $$\sqrt { 3 } \sin x \cos x + \cos ^ { 2 } x = 0$$ (ii) Hence, or otherwise, find in terms of \(\pi\) the solutions of the equation $$2 \sin x + \sec \left( x + \frac { \pi } { 6 } \right) = 0$$ for \(x\) in the interval \(0 \leq x \leq \pi\).

6 Fig. 8 shows part of the curve \(y = x \cos 3 x\). The curve crosses the \(x\)-axis at \(\mathrm { O } , \mathrm { P }\) and Q . \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P and Q .
2. Find the exact gradient of the curve at the point P . Show also that the turning points of the curve occur when \(x \tan 3 x = \frac { 1 } { 3 }\).
3. Find the area of the region enclosed by the curve and the \(x\)-axis between O and P , giving your answer in exact form.

1 Fig. 8 shows a sketch of part of the curve \(y = x \sin 2 x\), where \(x\) is in radians.
The curve crosses the \(x\)-axis at the point P . The tangent to the curve at P crosses the \(y\)-axis at Q . \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that the \(x\)-coordinates of the turning points of the curve satisfy the equation \(\tan 2 x + 2 x = 0\).
2. Find, in terms of \(\pi\), the \(x\)-coordinate of the point P . Show that the tangent PQ has equation \(2 \pi x + 2 y = \pi ^ { 2 }\).
   Find the exact coordinates of Q.
3. Show that the exact value of the area shaded in Fig. 8 is \(\frac { 1 } { 8 } \pi \left( \pi ^ { 2 } - 2 \right)\).

4 Show that \(\frac { 1 + \tan ^ { 2 } \theta } { 1 - \tan ^ { 2 } \theta } = \sec 2 \theta\).
Hence, or otherwise, solve the equation \(\frac { 1 + \tan ^ { 2 } \theta } { 1 - \tan ^ { 2 } \theta } = 2\), for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

1 Express \(3 \cos \theta + 4 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
Hence solve the equation \(3 \cos \theta + 4 \sin \theta = 2\) for \(- \pi \leqslant \theta \leqslant \pi\).

4 The angle \(\theta\) satisfies the equation \(\sin \left( \theta + 45 ^ { \circ } \right) = \cos \theta\).

1. Using the exact values of \(\sin 45 ^ { \circ }\) and \(\cos 45 ^ { \circ }\), show that \(\tan \theta = \sqrt { 2 } - 1\).
2. Find the values of \(\theta\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

6 Solve the equation \(\operatorname { cosec } \theta = 3\), for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

1 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 \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

6 Solve the equation \(\tan \left( \theta + 45 ^ { \circ } \right) = 1 - 2 \tan \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }\). Section B (36 marks)

1 Express \(\cos \theta - 3 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\).
Hence show that the equation \(\cos \theta - 3 \sin \theta = 4\) has no solution.

4 Solve the equation \(2 \sin 2 \theta = 1 + \cos 2 \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

3 Solve the equation \(\sec ^ { 2 } \theta = 2 \tan \theta + 4\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

1 Solve the equation \(2 \sin 2 \theta = \cos \theta\) for \(0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }\).

2 Solve the equation \(3 \operatorname { cosec } ^ { 2 } x = 2 \cot ^ { 2 } x + 3\) for values of \(x\) in the range \(0 ^ { \circ } < x < 360 ^ { \circ }\).

6 The function \(\mathrm { f } ( \theta ) = 3 \sin \theta + 4 \cos \theta\) is to be expressed in the form \(r \sin ( \theta + \alpha )\) where \(r > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).

1. Find the values of \(r\) and \(\alpha\).
2. Write down the maximum and minimum value of \(\mathrm { f } ( \theta )\).
3. Solve the equation \(\mathrm { f } ( \theta ) = 1\) for \(0 ^ { \circ } \leq \theta \leq 180 ^ { \circ }\).

1 Solve the equation for values of \(\theta\) in the range \(0 ^ { \circ } < \theta < 360 ^ { \circ }\). $$\cot 2 \theta = 5$$