Let's solve each part step by step. --- ## **(a) Maclaurin Series for sin(x) at \( x = \frac{3\pi}{2} \)** #### **Maclaurin Series:** \[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \] #### **Goal:** Find minimum number of terms so that percent relative error \( \epsilon = \frac{|f - f'|}{|f|} \times 100\% \) is below 1% at \( x = \frac{3\pi}{2} \). Where: - \( f \) is the true value (\( \sin(\frac{3\pi}{2}) = -1 \)) - \( f' \) is the truncated series value #### **Step 1: Calculate terms at \( x = \frac{3\pi}{2} \)** Let \( x = \frac{3\pi}{2} \approx 4.71239 \) **First 4 terms:** \[ \begin{align*} T_1 &= x = 4.71239 \\ T_2 &= -\frac{x^3}{3!} = -\frac{(4.71239)^3}{6} \approx -87.965 \\ T_3 &= \frac{x^5}{5!} = \frac{(4.71239)^5}{120} \approx 105.225 \\ T_4 &= -\frac{x^7}{7!} = -\frac{(4.71239)^7}{5040} \approx -41.309 \\ \end{align*} \] #### **Step 2: Add terms sequentially and check error** **Sum after 1 term:** \[ S_1 = T_1 = 4.71239 \] \[ \epsilon_1 = \frac{| -1 - 4.71239 |}{1} \times 100\% = 571.2\% \] **Sum after 2 terms:** \[ S_2 = T_1 + T_2 = 4.71239 - 87.965 = -83.2526 \] \[ \epsilon_2 = \frac{| -1 - (-83.2526) |}{1} \times 100\% = 8225.3\% \] **Sum after 3 terms:** \[ S_3 = S_2 + T_3 = -83.2526 + 105.225 = 21.972 \] \[ \epsilon_3 = \frac{| -1 - 21.972 |}{1} \times 100\% = 2297.2\% \] **Sum after 4 terms:** \[ S_4 = S_3 + T_4 = 21.972 - 41.309 = -19.337 \] \[ \epsilon_4 = \frac{| -1 - (-19.337) |}{1} \times 100\% = 1833.7\% \] Clearly, the series converges slowly for large \( x \). #### **Step 3: Add next term** **Next term:** \[ T_5 = \frac{x^9}{9!} = \frac{(4.71239)^9}{362880} \approx 6.387 \] \[ S_5 = S_4 + T_5 = -19.337 + 6.387 = -12.950 \] \[ \epsilon_5 = \frac{| -1 - (-12.950) |}{1} \times 100\% = 1195\% \] #### **Continue until error < 1%** You can see the error is still very large, even after 5 terms. For large \( x \), the Maclaurin series for sin(x) converges very slowly. Let’s check the **remainder formula** for alternate approach: #### **Step 4: Use the remainder (error) term** The error after \( n \) terms is approximately the first omitted term. For \( \sin(x) \), the next term is: \[ R_n \approx \left| \frac{x^{n+2}}{(n+2)!} \right| \] Set \( R_n < 0.01 \) (since \( |f| = 1 \) at this point). Try \( n = 7 \): \[ R_7 = \left| \frac{(4.71239)^9}{9!} \right| \approx 6.387 \] Try \( n = 15 \): \[ R_{15} = \frac{(4.71239)^{17}}{17!} \approx 0.000912 \] This is less than 0.01. #### **Conclusion for (a):** You need at least the terms up to \( x^{15} \), i.e., **8 terms** in the series to obtain less than 1% error at \( x = \frac{3\pi}{2} \). --- ## **(b) Maclaurin Series for cos(x) at \( x = \pi \)** #### **Maclaurin Series:** \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \] #### **Step 1: Calculate terms at \( x = \pi \)** Let \( x = \pi \approx 3.14159 \) **First few terms:** \[ T_1 = 1 \] \[ T_2 = -\frac{\pi^2}{2} \approx -4.9348 \] \[ T_3 = \frac{\pi^4}{24} \approx 4.0587 \] \[ T_4 = -\frac{\pi^6}{720} \approx -1.3353 \] **Sum after 1 term:** \[ S_1 = 1 \] True value: \( \cos(\pi) = -1 \) \[ \epsilon_1 = \frac{| -1 - 1 |}{1} \times 100\% = 200\% \] **Sum after 2 terms:** \[ S_2 = 1 - 4.9348 = -3.9348 \] \[ \epsilon_2 = \frac{| -1 - (-3.9348) |}{1} \times 100\% = 293.5\% \] **Sum after 3 terms:** \[ S_3 = -3.9348 + 4.0587 = 0.1239 \] \[ \epsilon_3 = \frac{| -1 - 0.1239 |}{1} \times 100\% = 112.4\% \] **Sum after 4 terms:** \[ S_4 = 0.1239 - 1.3353 = -1.2114 \] \[ \epsilon_4 = \frac{| -1 - (-1.2114) |}{1} \times 100\% = 21.1\% \] **Next term:** \[ T_5 = \frac{\pi^8}{40320} \approx 0.2353 \] \[ S_5 = -1.2114 + 0.2353 = -0.9761 \] \[ \epsilon_5 = \frac{| -1 - (-0.9761) |}{1} \times 100\% \approx 2.39\% \] **Next term:** \[ T_6 = -\frac{\pi^{10}}{3628800} \approx -0.0258 \] \[ S_6 = -0.9761 - 0.0258 = -1.0019 \] \[ \epsilon_6 = \frac{| -1 - (-1.0019) |}{1} \times 100\% \approx 0.19\% \] --- ### **Final Answers** #### **(a) For \( \sin(x) \) at \( x = \frac{3\pi}{2} \):** - **Minimum number of terms needed: 8 terms (up to \( x^{15} \))** #### **(b) For \( \cos(x) \) at \( x = \pi \):** - **Minimum number of terms needed: 6 terms (up to \( x^{10} \))** If you need full decimal details or want to see more steps, let me know!

Let's simplify the expression step by step: **Expression:** \[ 7\sqrt{192} + \sqrt{432} - 9\sqrt{300} \] --- ### Step 1: Prime factorization of each radicand - \(192\) - \(432\) - \(300\) --- ### Step 2: Break down each radical into prime factors **For \(\sqrt{192}\):** \[ 192 = 64 \times 3 = (8^2) \times 3 \] So, \[ \sqrt{192} = \sqrt{8^2 \times 3} = 8 \sqrt{3} \] --- **For \(\sqrt{432}\):** \[ 432 = 144 \times 3 = (12^2) \times 3 \] Thus, \[ \sqrt{432} = \sqrt{12^2 \times 3} = 12 \sqrt{3} \] --- **For \(\sqrt{300}\):** \[ 300 = 100 \times 3 = (10^2) \times 3 \] Hence, \[ \sqrt{300} = \sqrt{10^2 \times 3} = 10 \sqrt{3} \] --- ### Step 3: Substitute back into the original expression \[ 7 \times 8 \sqrt{3} + 12 \sqrt{3} - 9 \times 10 \sqrt{3} \] which simplifies to: \[ (56 \sqrt{3}) + (12 \sqrt{3}) - (90 \sqrt{3}) \] --- ### Step 4: Combine like terms \[ (56 + 12 - 90) \sqrt{3} = (-22) \sqrt{3} \] --- ### **Final simplified expression:** \[ \boxed{-22 \sqrt{3}} \]