(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})