(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 \(\geq\)...

(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

Answer Details

a) Using the identity sin(p - q) = sin p cos q - cos p sin q, we can substitute the given values of sin p and cos q to get: sin(p - q) = sin p cos q - cos p sin q = (\(\frac{1}{2}\))(\(\frac{1}{3}\)) - cos p sin q To solve for cos p sin q, we can use the Pythagorean identity cos^2 p + sin^2 p = 1, which gives sin p = sqrt(1 - cos^2 p). Substituting the given value of sin p, we get: (\(\frac{1}{2}\))^2 + cos^2 p = 1 cos^2 p = \(\frac{3}{4}\) cos p = ±sqrt(\(\frac{3}{4}\)) Since p is in the range 0° ≤ p ≤ 90°, we take cos p = sqrt(\(\frac{3}{4}\)) = \(\frac{\sqrt{3}}{2}\) Similarly, we can use the Pythagorean identity to solve for sin q, which gives: sin q = sqrt(1 - cos^2 q) = sqrt(1 - (\(\frac{1}{3}\))^2) = \(\frac{2\sqrt{2}}{3}\) Substituting these values into the expression for sin(p - q), we get: sin(p - q) = (\(\frac{1}{2}\))(\(\frac{1}{3}\)) - (\(\frac{\sqrt{3}}{2}\))(\(\frac{2\sqrt{2}}{3}\)) = -\(\frac{\sqrt{2}}{3}\) Therefore, sin(p - q) = -\(\frac{\sqrt{2}}{3}\). b) The trapezoidal rule is a numerical method for approximating definite integrals by approximating the area under the curve with trapezoids. To apply the trapezoidal rule with seven ordinates to the integral \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx, we first divide the interval [1, 4] into seven subintervals of equal width: 1 = x0 < x1 < x2 < x3 < x4 < x5 < x6 < x7 = 4 where xi = 1 + ih for i = 0, 1, 2, ..., 7 and h = (4 - 1)/7 = \(\frac{1}{7}\). Next, we approximate the integral over each subinterval using the formula for the area of a trapezoid: \(\int^{x_{i+1}}_{x_i} f(x)dx \approx \frac{h}{2}(f(x_i) + f(x_{i+1}))\) The approximation of the integral over the whole interval [1, 4] is the sum of the approximations over each subinterval: \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx ≈ \(\frac{h}{2}\)(f(x0) + 2f(x1) + 2f(x2) + 2f(x3) + 2f(x4) + 2f(x5) + 2f(x6) + f(x7)) where f(x) = 2/(sqrt(x + 3)). Substituting the values for x0, x1, x2, ..., x7 and f(x0), f(x1), f(x