More about trigonometric functions(4.1-4.3)

Understand the concept of radian measure

  • Theory
  • Examples
    Convert the following angles into degree measure. (Give your answers correct to 2 decimal places.)
    (a) \({0.84^c}\)                         (b) \({3.14^c}\)
  • Solutions
    (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

  • Theory
  • Examples
    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)]

    (a) Find α and β. (Express your answers in radian measure.)
    (b) Find the length of the belt. (c) Find the area bounded by the belt. (Give your answers correct to 3 significant figures.)
  • Solutions
    (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

  • Theory
  • Examples
    If \({\cos ^2}\alpha + {\sec ^2}\beta = 3\), prove that \({\cos ^2}\alpha (2 - {\sin ^2}\beta ) = 3{\cos ^2}\beta - {\sin ^2}\alpha \).
  • Solutions
    \(\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}\)