First slide

Trigonometric transformations

Question

Which of the following quantities are rational?

Difficult

A

$\sin (\frac{11 π}{12}) \sin (\frac{5 π}{12})$
B

$\cos e c (\frac{9 π}{10}) \sec (\frac{4 π}{5})$
C

$\sin^{4} (\frac{π}{8}) + \cos^{4} (\frac{π}{8})$
D

$(1 + \cos \frac{2 π}{9}) (1 + \cos \frac{4 π}{9}) (1 + \cos \frac{8 π}{9})$

Solution

a) $\sin (\frac{11 π}{12}) \sin (\frac{5 π}{12}) = \sin (\frac{π}{12}) \cos (\frac{π}{12})$
$= \frac{1}{2} \sin (\frac{π}{6}) = \frac{1}{4} \in Q$
b) $\cos e c (\frac{9 π}{10}) \sec (\frac{4 π}{5}) = - \cos e c (\frac{π}{10}) \sec (\frac{π}{5})$
$= - \frac{1}{\sin 18^{0} \cos 36^{0}}$
$= - \frac{16}{(\sqrt{5} - 1) (\sqrt{5} + 1)}$
$= - 4 \in Q$
c) $\sin^{4} (\frac{π}{8}) + \cos^{4} (\frac{π}{8})$
$= 1 - 2 \sin^{2} (\frac{π}{8}) \cos^{2} (\frac{π}{8})$
$= 1 - \frac{1}{2} \sin^{2} (\frac{π}{4}) = 1 - \frac{1}{4} = \frac{3}{4} \in Q$
d) $2 \cos^{2} \frac{π}{9} 2 \cos^{2} \frac{2 π}{9} 2 \cos^{2} \frac{4 π}{9}$
$= 8 {(\cos 20^{0} \cos 40^{0} \cos 80^{0})}^{2}$
$= 8 {[\frac{1}{2} (\cos 60 + \cos 20) \cos 80]}^{2}$
$= 8 {[\frac{1}{2} \cos 60 \cos 80 + \frac{1}{2} \cos 20 \cos 80]}^{2}$
$= 8 {[\frac{1}{4} \cos 80 + \frac{1}{4} (\cos 100 + \cos 60)]}^{2}$
$= 8 {[\frac{1}{4} \cos 80 - \frac{1}{4} \cos 80 + \frac{1}{4} \cos 60]}^{2}$
$= \frac{8}{64}$

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