Prove the following trigonometric identities by expressing the left side in complex form: (a) \(\cos (x+y)=\cos x \cos y-\sin x \sin y\) (b) \(\sin (x+y)=\sin x \cos y+\cos x \sin y\) (c) \(\sum_{n=1}^{N} \cos n \theta=\frac{\cos \frac{1}{3}(N+1) \theta \sin \frac{1}{2} N \theta}{\sin \frac{1}{2} \theta}\) (d) \(\sum_{n=1}^{N} \sin n \theta=\frac{\sin \frac{1}{2}(N+1) \theta \sin \frac{1}{2} N \theta}{\sin \frac{1}{2} \theta}\) Many other trigonometric identities can be similarly proved. 1.3.3 Investigate the multiplication and division of two complex numbers. Show that in general the product (quotient) of their real parts does not equal the real part of the product (quotient) of the two numbers.

Recall the Euler's formula: \(e^{ix} = \cos x + i\sin x\). To compute the left side, \begin{align*} \cos(x+y) &= \text{Re}(e^{i(x+y)}) \\ &= \text{Re}(e^{ix}e^{iy}) \\ &= \text{Re}((\cos x + i\sin x)(\cos y + i\sin y)) \\ &= \cos x \cos y - \sin x \sin y. \end{align*} Hence, the identity holds.

To compute the left side, \begin{align*} \sin(x+y) &= \text{Im}(e^{i(x+y)}) \\ &= \text{Im}(e^{ix}e^{iy}) \\ &= \text{Im}((\cos x + i\sin x)(\cos y + i\sin y)) \\ &= \sin x \cos y + \cos x \sin y. \end{align*} Hence, the identity holds.

Using the geometric sequence formula, we can represent the sum of cosine terms as a sum of complex exponentials: \begin{align*} \sum_{n=1}^{N} \cos n \theta &= \text{Re}\left(\sum_{n=1}^{N}e^{in\theta}\right) \\ &= \text{Re}\left(\frac{1-e^{i(N+1)\theta}}{1-e^{i\theta}}\right) \\ &= \text{Re}\left(\frac{(1-e^{i(N+1)\theta})(1-e^{-i\theta})}{1-e^{i\theta}e^{-i\theta}}\right) \\ &= \text{Re}\left(\frac{1-\cos((N+1)\theta)+i\sin((N+1)\theta)-e^{-i\theta}}{1-\cos\theta}\right) \\ &= \frac{\cos \frac{1}{2}(N+1) \theta \sin \frac{1}{2} N \theta}{\sin \frac{1}{2} \theta}. \end{align*} Hence, the identity holds.

Similar to part (c), we can represent the sum of sine terms as a sum of complex exponentials: \begin{align*} \sum_{n=1}^{N} \sin n \theta &= \text{Im}\left(\sum_{n=1}^{N}e^{in\theta}\right) \\ &= \text{Im}\left(\frac{1-e^{i(N+1)\theta}}{1-e^{i\theta}}\right) \\ &= \text{Im}\left(\frac{(1-e^{i(N+1)\theta})(1-e^{-i\theta})}{1-e^{i\theta}e^{-i\theta}}\right) \\ &= \text{Im}\left(\frac{1-\cos((N+1)\theta)+i\sin((N+1)\theta)-e^{-i\theta}}{1-\cos\theta}\right) \\ &= \frac{\sin \frac{1}{2}(N+1) \theta \sin \frac{1}{2} N \theta}{\sin \frac{1}{2} \theta}. \end{align*} Hence, the identity holds.

Let \(z_1 = a + bi\) and \(z_2 = c + di\) be two complex numbers. Their product and quotient are given by: \begin{align*} z_1 z_2 &= (a + bi)(c + di) = (ac - bd) + (ad + bc)i, \\ \frac{z_1}{z_2} &= \frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}. \end{align*} For the product case, the real part is given by \(ac - bd\), while the product of their real parts is \(ac\). If \(ad \neq 0\) and \(b \neq 0\), then \(ac - bd \neq ac\). Similarly, for the quotient case, the real part is given by \(\frac{ac + bd}{c^2 + d^2}\), while the quotient of their real parts is \(\frac{a}{c}\). If \(b\neq0\) and \(d\neq0\), then \(\frac{ac + bd}{c^2 + d^2} \neq \frac{a}{c}\). Thus, in general, the product (quotient) of the real parts of two complex numbers does not equal the real part of the product (quotient) of the two numbers.