#### Understand the concept of radian measure

• Convert the following angles into degree measure. (Give your answers correct to 2 decimal places.)
(a) $${0.84^c}$$                         (b) $${3.14^c}$$
• (a) $${0.84^c} = 0.84(\frac{{180^\circ }}{\pi })$$ $$= \underline{\underline {48.13^\circ }}$$ (corr. to 2 d.p.)
(b) $${3.14^c} = 3.14(\frac{{180^\circ }}{\pi })$$ $$= \underline{\underline {179.91^\circ }}$$ (corr. to 2 d.p.)

#### Find arc lengths and areas of sectors through radian measure

• In the figure, a machine is made of two rollers bounded by a belt. The centres of the two rollers are A and B, and AB = 50 cm. The radii of the two rollers are 15 cm and 10 cm. Let the belt touches the larger roller at C and D, and the smaller one at E and F.
[figure missing(Latex_QB_M2E04:Q21)]

(b) Find the length of the belt. (c) Find the area bounded by the belt. (Give your answers correct to 3 significant figures.)
• (a) Let G be a point on AD such that $$AD \bot BG$$.
[figure missing(Latex_QB_M2E04:Q21)]
$$AG = AD - GD = (15 - 10){\rm{ cm}} = 5{\rm{ cm}}$$
$$\cos \angle GAB = \frac{{AG}}{{AB}} = \frac{5}{{50}} = \frac{1}{{10}}$$
Since the figure is symmetric,
∴ $$\angle CAB = \angle GAB$$
$$\begin{array}{c}\angle CAB + \angle GAB + \alpha = 2\pi (\angle s at a pt) \\\alpha = 2\pi - 2\angle GAB\end{array}$$
$$= 3.34$$ rad. (corr. to 3 sig. fig.)
$$\begin{array}{c}\sin \angle ABG = \frac{{AG}}{{BG}}\\ = \frac{5}{{50}}\\ = \frac{1}{{10}}\end{array}$$
$$\angle ABF = \angle ABG + \frac{\pi }{2}$$

Since the figure is symmetric,

∴ $$\angle ABE = \angle ABF$$

$$\angle ABE + \angle ABF + \beta = 2\pi$$ (∠ s at a pt.)
$$\begin{array}{c}\beta = 2\pi - 2\angle ABF\\ = 2\pi - 2(\angle ABG + \frac{\pi }{2})\\ = \pi - 2\angle ABG\end{array}$$
$$= 2.94$$rad. (corr. to 3 sig. fig.)

(b)
$$\begin{array}{c}B{G^2} = A{B^2} - A{G^2}\\ BG = \sqrt {{{50}^2} - {5^2}} {\rm{ cm}} \\ = \sqrt {2{\rm{ }}475} {\rm{ cm}} \\DF = EC = BG = \sqrt {2{\rm{ }}475} {\rm{ cm}}\end{array}$$
Length of the belt$$\begin{array}{c} = C\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over D} + DF + F\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over E} + EC\\ = (15\alpha + \sqrt {2{\rm{ }}475} + 10\beta + \sqrt {2{\rm{ }}475} ){\rm{ cm}}\end{array}$$

$$= \underline{\underline {179{\rm{ cm}}}}$$ (corr. to 3 sig. fig.)

(c) Area of trapezium ABFD $$\begin{array}{l} = \frac{{(BF + AD)(DF)}}{2}\\ = \frac{{(10 + 15)(\sqrt {2{\rm{ }}475} )}}{2}{\rm{ c}}{{\rm{m}}^2}\\ = \frac{{25\sqrt {2{\rm{ }}475} }}{2}{\rm{ c}}{{\rm{m}}^2}\end{array}$$

Area bounded by the belt + Area of major sector ACD = Area of minor sector BEF + 2 x Area of trapezium ABFD

$$= [\frac{1}{2}({15^2})\alpha + \frac{1}{2}({10^2})\beta + 2 \times \frac{{25\sqrt {2{\rm{ }}475} }}{2}]{\rm{ c}}{{\rm{m}}^2}$$

$$= \underline{\underline {1{\rm{ }}770{\rm{ c}}{{\rm{m}}^2}}}$$ (corr. to 3 sig. fig.)

#### Understand the functions cosecant, secant and cotangent and their graphs

• If $${\cos ^2}\alpha + {\sec ^2}\beta = 3$$, prove that $${\cos ^2}\alpha (2 - {\sin ^2}\beta ) = 3{\cos ^2}\beta - {\sin ^2}\alpha$$.
• $$\begin{array}{c}{\cos ^2}\alpha (2 - {\sin ^2}\beta ) = (3 - {\sec ^2}\beta )(2 - {\sin ^2}\beta )\\ = 6 - 2{\sec ^2}\beta - 3{\sin ^2}\beta + {\sec ^2}\beta {\sin ^2}{\beta ^{^{^{}}}}\\ = 6 - 2{\sec ^2}\beta - 3{\sin ^2}\beta + {\tan ^2}{\beta ^{^{^{}}}}\\ = 6 - 2{\sec ^2}\beta - 3{\sin ^2}\beta + {\sec ^2}\beta - {1^{^{^{}}}}\\ = 5 - {\sec ^2}\beta - 3{\sin ^2}{\beta ^{^{^{}}}}\\ = 5 - (3 - {\cos ^2}\alpha ) - 3{(1 - {\cos ^2}\beta )^{^{^{}}}}\\ = - 1 + {\cos ^2}\alpha + 3{\cos ^2}{\beta ^{^{^{}}}}\\ = 3{\cos ^2}\beta - {\sin ^2}{\alpha ^{^{^{}}}}\end{array}$$