(a) If sin p = (frac{1}{2}) and cos q = (frac{1}{3}), evaluate sin(p – q), where 0(^o) (geq) p (geq) 90(^o) and 90(^o) (geq) q (geq) 180(^o)
b) Using trapezum rule with seven ordinates, evaluate (int^4_1frac{2}{sqrt{x + 3}})dx
Explanation
(a)To find sin q =(frac{2sqrt{2}}{3}) and cos p = (frac{sqrt{3}}{3}). Then, substituting
sin(p – q) = sin p cos q – cos p sin q = ((frac{1}{2} times -frac{1}{3})) – ((frac{2sqrt{2}}{3} times frac{sqrt{3}}{2})
Which simplified to -(frac{1}{6} – frac{2sqrt{6}}{6}) and can be written as (frac{-1 -2sqrt{6}}{6})
(b), candidates were expected to obtain the following table
| x | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
| y | 1 | 0.9428 | 0.8944 | 0.8528 | 0.8165 | 0.7845 | 0.7559 |
Then, using the table and applying the formula;
(int^4_1frac{2}{sqrt{x + 3}})dx = 0.5{(frac{1}{2}) + 0.9428 + 0.8944 + 0.8528 + 0.8165 + 0.7845 + (frac{0.7559}{2})}
= 0.5 (5.16895)
= 2.584475
= 2.58 correct to two decimal places.