(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 \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx
b) Using trapezum rule with seven ordinates, evaluate \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx
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Correct Answer: Option
Explanation:
(a)To find sin q =\(\frac{2\sqrt{2}}{3}\) and cos p = \(\frac{\sqrt{3}}{3}\). Then, substituting
sin(p - q) = sin p cos q - cos p sin q = (\(\frac{1}{2} \times -\frac{1}{3}\)) - (\(\frac{2\sqrt{2}}{3} \times \frac{\sqrt{3}}{2}\)
Which simplified to -\(\frac{1}{6} - \frac{2\sqrt{6}}{6}\) and can be written as \(\frac{-1 -2\sqrt{6}}{6}\)
(b), candidates were expected to obtain the following table
Then, using the table and applying the formula;
\(\int^4_1\frac{2}{\sqrt{x + 3}}\)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.
(a)To find sin q =\(\frac{2\sqrt{2}}{3}\) and cos p = \(\frac{\sqrt{3}}{3}\). Then, substituting
sin(p - q) = sin p cos q - cos p sin q = (\(\frac{1}{2} \times -\frac{1}{3}\)) - (\(\frac{2\sqrt{2}}{3} \times \frac{\sqrt{3}}{2}\)
Which simplified to -\(\frac{1}{6} - \frac{2\sqrt{6}}{6}\) and can be written as \(\frac{-1 -2\sqrt{6}}{6}\)
(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;
\(\int^4_1\frac{2}{\sqrt{x + 3}}\)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.