IB Mathematics AHL 1.13 Modulus–argument (polar) form AI HL Paper 1- Exam Style Questions- New Syllabus

(a)
Using the identity $\sin\theta = \operatorname{Im}(e^{i\theta})$:
$L_1(x) = \operatorname{Im}\left( 14.8 e^{i(0.35x + 1.27)} \right)$
$L_1(x) = \operatorname{Im}\left( 14.8 e^{0.35xi} e^{1.27i} \right)$

$L_2(x) = \operatorname{Im}\left( 18.5 e^{i(0.35x – 2.36)} \right)$
$L_2(x) = \operatorname{Im}\left( 18.5 e^{0.35xi} e^{-2.36i} \right)$

$L_1 = \operatorname{Im}(14.8 e^{0.35xi} e^{1.27i}), \quad L_2 = \operatorname{Im}(18.5 e^{0.35xi} e^{-2.36i})$

(b)
The sum can be factored using the common term $e^{0.35xi}$:
$L_1 + L_2 = \operatorname{Im}\left( e^{0.35xi} \left( 14.8 e^{1.27i} + 18.5 e^{-2.36 i} \right) \right)$

Calculate the sum of the complex amplitudes (phasors):
$Z = 14.8 e^{1.27i} + 18.5 e^{-2.36i}$

Using a GDC to convert the sum back to polar form:
Magnitude $q \approx 8.81559$
Argument $r \approx 3.01604$ radians

The resulting function is $L_3(x) = 8.816 \sin(0.35x + 3.016)$
$\boxed{q \approx 8.82}, \quad \boxed{r \approx 3.02}$ (to 3 s.f.)