1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1

8 Show that the equation \(4 \cos ^ { 2 } \theta = 1 + \sin \theta\) can be expressed as $$4 \sin ^ { 2 } \theta + \sin \theta - 3 = 0 .$$ Hence solve the equation for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

7 Solve the equation \(\tan \theta = 2 \sin \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

7 Show that the equation \(\sin ^ { 2 } x = 3 \cos x - 2\) can be expressed as a quadratic equation in \(\cos x\) and hence solve the equation for values of \(x\) between 0 and \(2 \pi\).

3

1. Express \(2 \tan ^ { 2 } \theta - \frac { 1 } { \cos \theta }\) in terms of \(\sec \theta\).
2. Hence solve, for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), the equation $$2 \tan ^ { 2 } \theta - \frac { 1 } { \cos \theta } = 4$$

4 The acute angles \(\alpha\) and \(\beta\) are such that $$2 \cot \alpha = 1 \text { and } 24 + \sec ^ { 2 } \beta = 10 \tan \beta \text {. }$$

1. State the value of \(\tan \alpha\) and determine the value of \(\tan \beta\).
2. Hence find the exact value of \(\tan ( \alpha + \beta )\).

8

1. Express \(\cos 4 \theta\) in terms of \(\sin 2 \theta\) and hence show that \(\cos 4 \theta\) can be expressed in the form \(1 - k \sin ^ { 2 } \theta \cos ^ { 2 } \theta\), where \(k\) is a constant to be determined.
2. Hence find the exact value of \(\sin ^ { 2 } \left( \frac { 1 } { 24 } \pi \right) \cos ^ { 2 } \left( \frac { 1 } { 24 } \pi \right)\).
3. By expressing \(2 \cos ^ { 2 } 2 \theta - \frac { 8 } { 3 } \sin ^ { 2 } \theta \cos ^ { 2 } \theta\) in terms of \(\cos 4 \theta\), find the greatest and least possible values of $$2 \cos ^ { 2 } 2 \theta - \frac { 8 } { 3 } \sin ^ { 2 } \theta \cos ^ { 2 } \theta$$ as \(\theta\) varies. \includegraphics[max width=\textwidth, alt={}, center]{89e54367-bb83-483a-add5-0527b71a5cac-5_606_926_267_552} The function f is defined for all real values of \(x\) by $$\mathrm { f } ( x ) = k \left( x ^ { 2 } + 4 x \right) ,$$ where \(k\) is a positive constant. The diagram shows the curve with equation \(y = \mathrm { f } ( x )\).
4. The curve \(y = x ^ { 2 }\) can be transformed to the curve \(y = \mathrm { f } ( x )\) by the following sequence of transformations: a translation parallel to the \(x\)-axis,
   a translation parallel to the \(y\)-axis,
   a stretch. a translation parallel to the \(x\)-axis, a translation parallel to the \(y\)-axis, a stretch.
   Give details, in terms of \(k\) where appropriate, of these transformations.
5. Find the range of f in terms of \(k\).
6. It is given that there are three distinct values of \(x\) which satisfy the equation \(| \mathrm { f } ( x ) | = 20\). Find the value of \(k\) and determine exactly the three values of \(x\) which satisfy the equation in this case.

3

1. Given that \(7 \sin 2 \alpha = 3 \sin \alpha\), where \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), find the exact value of \(\cos \alpha\).
2. Given that \(3 \cos 2 \beta + 19 \cos \beta + 13 = 0\), where \(90 ^ { \circ } < \beta < 180 ^ { \circ }\), find the exact value of \(\sec \beta\).

3 It is given that \(\theta\) is the acute angle such that \(\sec \theta \sin \theta = 36 \cot \theta\).

1. Show that \(\tan \theta = 6\).
2. Hence, using an appropriate formula in each case, find the exact value of
   1. \(\tan \left( \theta - 45 ^ { \circ } \right)\),
   2. \(\quad \tan 2 \theta\).

2 Using an appropriate identity in each case, find the possible values of

1. \(\sin \alpha\) given that \(4 \cos 2 \alpha = \sin ^ { 2 } \alpha\),
2. \(\sec \beta\) given that \(2 \tan ^ { 2 } \beta = 3 + 9 \sec \beta\).

2 By first using appropriate identities, solve the equation $$5 \cos 2 \theta \operatorname { cosec } \theta = 2$$ for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

9 It is given that \(\mathrm { f } ( \theta ) = \sin \left( \theta + 30 ^ { \circ } \right) + \cos \left( \theta + 60 ^ { \circ } \right)\).

1. Show that \(\mathrm { f } ( \theta ) = \cos \theta\). Hence show that $$f ( 4 \theta ) + 4 f ( 2 \theta ) \equiv 8 \cos ^ { 4 } \theta - 3 .$$
2. Hence
   1. determine the greatest and least values of \(\frac { 1 } { \mathrm { f } ( 4 \theta ) + 4 \mathrm { f } ( 2 \theta ) + 7 }\) as \(\theta\) varies,
   2. solve the equation $$\sin \left( 12 \alpha + 30 ^ { \circ } \right) + \cos \left( 12 \alpha + 60 ^ { \circ } \right) + 4 \sin \left( 6 \alpha + 30 ^ { \circ } \right) + 4 \cos \left( 6 \alpha + 60 ^ { \circ } \right) = 1$$ for \(0 ^ { \circ } < \alpha < 60 ^ { \circ }\). \section*{END OF QUESTION PAPER}

8 The functions f and g are defined for all real values of \(x\) by $$\mathrm { f } ( x ) = | 2 x + a | + 3 a \quad \text { and } \quad \mathrm { g } ( x ) = 5 x - 4 a$$ where \(a\) is a positive constant.

1. State the range of f and the range of g .
2. State why f has no inverse, and find an expression for \(\mathrm { g } ^ { - 1 } ( x )\).
3. Solve for \(x\) the equation \(\operatorname { gf } ( x ) = 31 a\).
4. Show that \(\sin 2 \theta ( \tan \theta + \cot \theta ) \equiv 2\).
5. Hence
   1. find the exact value of \(\tan \frac { 1 } { 12 } \pi + \tan \frac { 1 } { 8 } \pi + \cot \frac { 1 } { 12 } \pi + \cot \frac { 1 } { 8 } \pi\),
   2. solve the equation \(\sin 4 \theta ( \tan \theta + \cot \theta ) = 1\) for \(0 < \theta < \frac { 1 } { 2 } \pi\),
   3. express \(( 1 - \cos 2 \theta ) ^ { 2 } \left( \tan \frac { 1 } { 2 } \theta + \cot \frac { 1 } { 2 } \theta \right) ^ { 3 }\) in terms of \(\sin \theta\).

5

1. Show that \(\frac { 1 } { 1 - \tan x } - \frac { 1 } { 1 + \tan x } \equiv \tan 2 x\).
2. Hence evaluate \(\int _ { \frac { 1 } { 12 } \pi } ^ { \frac { 1 } { 6 } \pi } \left( \frac { 1 } { 1 - \tan x } - \frac { 1 } { 1 + \tan x } \right) \mathrm { d } x\), giving your answer in the form \(a \ln b\).

4 Show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 - 2 \sin ^ { 2 } x } { 1 + 2 \sin x \cos x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

8 The points \(A\) and \(B\) have position vectors relative to the origin \(O\) given by $$\overrightarrow { O A } = \left( \begin{array} { c } 3 \sin \alpha \\ 2 \cos \alpha \\ - 1 \end{array} \right) \text { and } \overrightarrow { O B } = \left( \begin{array} { c } 2 \cos \alpha \\ 4 \sin \alpha \\ 3 \end{array} \right)$$ where \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). It is given that \(\overrightarrow { O A }\) and \(\overrightarrow { O B }\) are perpendicular.

1. Calculate the two possible values of \(\alpha\).
2. Calculate the area of triangle \(O A B\) for the smaller value of \(\alpha\) from part (i).

4 Prove that \(\cot \beta - \cot \alpha = \frac { \sin ( \alpha - \beta ) } { \sin \alpha \sin \beta }\).

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

5 Show that \(\frac { \sin 2 \theta } { 1 + \cos 2 \theta } = \tan \theta\).

8 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} 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 \(\beta\).

5 Solve the equation \(2 \sec ^ { 2 } \theta = 5 \tan \theta\), for \(0 \leqslant \theta \leqslant \pi\).

6 In Fig. 6, \(\mathrm { ABC } , \mathrm { ACD }\) and AED are right-angled triangles and \(\mathrm { BC } = 1\) unit. Angles CAB and CAD are \(\theta\) and \(\phi\) respectively. \begin{figure}[h] \end{figure}

1. Find AC and AD in terms of \(\theta\) and \(\phi\).
2. Hence show that \(\mathrm { DE } = 1 + \frac { \tan \phi } { \tan \theta }\). Section B (36 marks)

2 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

1 By using the substitution \(t = \tan \frac { 1 } { 2 } \theta\), find \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 1 + \cos \theta } \mathrm { d } \theta\).

By considering \(\sum _ { k = 0 } ^ { n - 1 } ( 1 + \mathrm { i } \tan \theta ) ^ { k }\), show that $$\sum _ { k = 0 } ^ { n - 1 } \cos k \theta \sec ^ { k } \theta = \cot \theta \sin n \theta \sec ^ { n } \theta$$ provided \(\theta\) is not an integer multiple of \(\frac { 1 } { 2 } \pi\). Hence or otherwise show that $$\sum _ { k = 0 } ^ { n - 1 } 2 ^ { k } \cos \left( \frac { 1 } { 3 } k \pi \right) = \frac { 2 ^ { n } } { \sqrt { 3 } } \sin \left( \frac { 1 } { 3 } n \pi \right)$$ Given that \(0 < x < 1\), show that $$\sum _ { k = 0 } ^ { n - 1 } \frac { \cos \left( k \cos ^ { - 1 } x \right) } { x ^ { k } } = \frac { \sin \left( n \cos ^ { - 1 } x \right) } { x ^ { n - 1 } \sqrt { } \left( 1 - x ^ { 2 } \right) }$$

8

1. The curve \(C _ { 1 }\) has equation \(y = - \ln ( \cos x )\). Show that the length of the arc of \(C _ { 1 }\) from the point where \(x = 0\) to the point where \(x = \frac { 1 } { 3 } \pi\) is \(\ln ( 2 + \sqrt { 3 } )\).
2. The curve \(C _ { 2 }\) has equation \(y = 2 \sqrt { } ( x + 3 )\). The arc of \(C _ { 2 }\) joining the point where \(x = 0\) to the point where \(x = 1\) is rotated through one complete revolution about the \(x\)-axis. Show that the area of the surface generated is $$\frac { 8 } { 3 } \pi ( 5 \sqrt { } 5 - 8 )$$