(a) Given that \(\sin(A + B) = \sin A \cos B + \cos A \sin B\). Without using mathematical tables or calculator, evaluate \(\sin 105°\), leaving your answer in the surd form.
(You may use 105° = 60° + 45°)
(b) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?
(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \(2\) and \(\frac{2}{9}\) respectively. Find :
(i) the common difference ; (ii) the 5th term.
Explanation
(a) \(\sin 105° = \sin(60 + 45)\)
= \(\sin 60° \cos 45° + \cos 60° \sin 45°\)
= \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} + \frac{1}{2} \times \frac{\sqrt{2}}{2}\)
= \(\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}\)
(b) a = 11, d = 2.
\(S_{n} = \frac{n}{2} (2a + (n - 1) d\)
\(551 = \frac{n}{2} (2(11) + (n - 1) 2\)
\(551 = \frac{n}{2} (22 + 2n - 2)\)
\(551 = \frac{n}{2} (2n + 20)\)
\(1102 = 2n^{2} + 20n \implies 2n^{2} + 20n - 1102 = 0\)
Divide through by 2.
\(n^{2} + 10n - 551 = 0\)
\(n^{2} - 19n + 29n - 551 = 0\)
\(n(n - 19) + 29(n - 19) = 0\)
\(\text{n = 19 or -29}\)
Since n cannot be negative, n = 19 houses.
(c)(i) \(T_{n} = ar^{n - 1}\) (terms of a G.P)
\(a = 2\)
\(T_{3} = 2 \times r^{3 - 1} = \frac{2}{9}\)
\(r^{2} = \frac{1}{9} \implies r = \frac{1}{3}\)
(ii) \(T_{5} = ar^{5 - 1}\)
= \(2 \times (\frac{1}{3})^{4}\)
= \(2 \times \frac{1}{81}\)
= \(\frac{2}{81}\)