1.05l Double angle formulae: and compound angle formulae

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

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

6 Prove that

1. \(\frac { \sin 2 \theta } { 2 \tan \theta } + \sin ^ { 2 } \theta = 1\),
2. \(\quad \sin \left( x + 45 ^ { \circ } \right) = \cos \left( x - 45 ^ { \circ } \right)\).

3 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]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-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]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_450_416_1562_940} \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.

4 Solve the equation \(\cos 2 \theta = \sin \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\), giving your answers in terms of \(\pi\).

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

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 \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

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 \(0 ^ { \circ } \quad \left[ \sqrt { 3 } 60 ^ { \circ } \right.\). $$\leqslant \leqslant$$

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

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

6 Solve the equation \(2 \cos 2 x = 1 + \cos x\), for \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).

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] \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 \(\theta\).

3 Using appropriate right-angled triangles, show that \(\tan 45 ^ { \circ } = 1\) and \(\tan 30 ^ { \circ } = \frac { 1 } { \sqrt { 3 } }\).
Hence show that \(\tan 75 ^ { \circ } = 2 + \sqrt { 3 }\).

2 Solve, correct to 2 decimal places, the equation \(\cot 2 \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

3 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at \(\alpha\) to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all \(\beta\). \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} In the following, all lengths are in metres.

1. Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
2. Show that \(\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )\). $$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$ Similarly, it can be shown that \(\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }\). [You are not required to derive this result.]
   You are now given that \(\alpha = 45 ^ { \circ }\) and that \(\tan \beta = t\).
3. Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
4. Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\). For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
5. Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\). Hence find this value of \(\beta\).

4 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ } .$$

2

1. Given that \(\mathrm { f } ( x ) = \sin \left( 2 x + \frac { 1 } { 4 } \pi \right)\), show that \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \sqrt { 2 } ( \sin 2 x + \cos 2 x )\).
2. Hence find the first four terms of the Maclaurin series for \(\mathrm { f } ( x )\). [You may use appropriate results given in the List of Formulae.]

2 In this question, \(\theta\) is a real number with \(0 < \theta < \frac { 1 } { 6 } \pi\), and \(w = \frac { 1 } { 2 } \mathrm { e } ^ { 3 \mathrm { j } \theta }\).

1. State the modulus and argument of each of the complex numbers $$w , \quad w ^ { * } \quad \text { and } \quad \mathrm { j } w .$$ Illustrate these three complex numbers on an Argand diagram.
2. Show that \(( 1 + w ) \left( 1 + w ^ { * } \right) = \frac { 5 } { 4 } + \cos 3 \theta\). Infinite series \(C\) and \(S\) are defined by $$\begin{aligned} & C = \cos 2 \theta - \frac { 1 } { 2 } \cos 5 \theta + \frac { 1 } { 4 } \cos 8 \theta - \frac { 1 } { 8 } \cos 11 \theta + \ldots \\ & S = \sin 2 \theta - \frac { 1 } { 2 } \sin 5 \theta + \frac { 1 } { 4 } \sin 8 \theta - \frac { 1 } { 8 } \sin 11 \theta + \ldots \end{aligned}$$
3. Show that \(C = \frac { 4 \cos 2 \theta + 2 \cos \theta } { 5 + 4 \cos 3 \theta }\), and find a similar expression for \(S\).

3.Solve for values of \(\theta\) ,in degrees,in the range \(0 \leq \theta \leq 360\) , $$\sqrt { } 2 \times ( \sin 2 \theta + \cos \theta ) + \cos 3 \theta = \sin 2 \theta + \cos \theta$$

7.The variable \(y\) is defined by $$y = \ln \left( \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x \right) \text { for } 0 < x < \frac { \pi } { 2 } .$$ A student was asked to prove that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - 4 \cot 2 x .$$ The attempted proof was as follows: $$\begin{aligned} y & = \ln \left( \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x \right) \\ & = \ln \left( \sec ^ { 2 } x \right) + \ln \left( \operatorname { cosec } ^ { 2 } x \right) \\ & = 2 \ln \sec x + 2 \ln \operatorname { cosec } x \\ \frac { \mathrm {~d} y } { \mathrm {~d} x } & = 2 \tan x - 2 \cot x \\ & = \frac { 2 \left( \sin ^ { 2 } x - \cos ^ { 2 } x \right) } { \sin x \cos x } \\ & = \frac { - 2 \cos 2 x } { \frac { 1 } { 2 } \sin 2 x } \\ & = - 4 \cot 2 x \end{aligned}$$

1. Identify the error in this attempt at a proof.
2. Give a correct version of the proof.
3. Find and simplify a general relationship between \(p\) and \(q\) ,where \(p\) and \(q\) are variables that depend on \(x\) ,such that the student would obtain the correct result when differentiating \(\ln ( p + q )\) with respect to \(x\) by the above incorrect method.
4. Given that \(p ( x ) = k \sec r x\) and \(q ( x ) = \operatorname { cosec } ^ { 2 } x\) ,where \(k\) and \(r\) are positive integers,find the values of \(k\) and \(r\) such that \(p\) and \(q\) satisfy the relationship found in part(c). \section*{END} Marks for presentation: 7
   TOTAL MARKS: 100

4.(a)Prove the identity $$( \sin x + \cos y ) \cos ( x - y ) \equiv ( 1 + \sin ( x - y ) ) ( \cos x + \sin y )$$ (b)Hence,or otherwise,show that $$\frac { \sin 5 \theta + \cos 3 \theta } { \cos 5 \theta + \sin 3 \theta } \equiv \frac { 1 + \tan \theta } { 1 - \tan \theta }$$ (c)Given that \(k > 1\) ,show that the equation \(\frac { \sin 5 \theta + \cos 3 \theta } { \cos 5 \theta + \sin 3 \theta } = k\) has a unique solution in the interval \(0 < \theta < \frac { \pi } { 4 }\)

5.(a)The box below shows a student's attempt to prove the following identity for \(a > b > 0\) $$\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }$$ Let \(x = \arctan a\) and \(y = \arctan b\) ,so that \(a = \tan x\) and \(b = \tan y\) $$\begin{aligned} \text { So } \tan ( \arctan a - \arctan b ) & \equiv \tan ( x - y ) \\ & \equiv \frac { \tan x - \tan y } { 1 - \tan ^ { 2 } ( x y ) } \\ & \equiv \frac { a - b } { 1 - ( a b ) ^ { 2 } } \\ & \equiv \frac { a - a b + a b - b } { ( 1 - a b ) ( 1 + a b ) } \\ & \equiv \frac { a ( 1 - a b ) - b ( 1 - a b ) } { ( 1 - a b ) ( 1 + a b ) } \\ & \equiv \frac { a - b } { 1 + a b } \end{aligned}$$ Taking arctan of both sides gives \(\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }\) as required. There are three errors in the proof where the working does not follow from the previous line.

1. Describe these three errors.
2. Write out a correct proof of the identity.
   (b)[In this question take \(g\) to be \(9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) ] \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4d5b914c-28b2-4485-a42e-627c95fa16e2-22_244_1267_1870_504} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} Balls are projected,one after another,from a point,\(A\) ,one metre above horizontal ground. Each ball travels in a vertical plane towards a 6 metre high vertical wall of negligible thickness,which is a horizontal distance of \(10 \sqrt { 2 }\) metres from \(A\) . The balls are modelled as particles and it is assumed that there is no air resistance.
   Each ball is projected with an initial speed of \(28 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and at a random angle \(\theta\) to the horizontal,where \(0 < \theta < 90 ^ { \circ }\) Given that a ball will pass over the wall precisely when \(\alpha \leqslant \theta \leqslant \beta\)
3. find, in degrees, the angle \(\beta - \alpha\)
4. Deduce that the probability that a particular ball will pass over the wall is \(\frac { 2 } { 3 }\)
5. Hence find the probability that exactly 2 of the first 10 balls projected pass over the wall. You should give your answer in the form \(\frac { P } { Q ^ { k } }\) where \(P , Q\) and \(k\) are integers and \(P\) is not a multiple of \(Q\).
6. Explain whether taking air resistance into account would increase or decrease the probability in (b)(iii).
7. find, in degrees, the angle \(\beta - \alpha\)

3.(a)Use the formulae for \(\sin ( A \pm B )\) and \(\cos ( A \pm B )\) to prove that \(\tan \left( 90 ^ { \circ } - \theta \right) \equiv \cot \theta\) (b)Solve for \(0 < \theta < 360 ^ { \circ }\) $$2 - \sec ^ { 2 } \left( \theta + 11 ^ { \circ } \right) = 2 \tan \left( \theta + 11 ^ { \circ } \right) \tan \left( \theta - 34 ^ { \circ } \right)$$ Give each answer as an integer in degrees.

1.(a)Write down the exact value of \(\cos 405 ^ { \circ }\) (b)Hence,using a double angle identity for cosine,or otherwise,determine the exact value of \(\cos 101.25 ^ { \circ }\) ,giving your answer in the form $$a \sqrt { b + c \sqrt { 2 + \sqrt { 2 } } }$$ where \(a\) ,\(b\) and \(c\) are rational numbers.