Step 1: Express $Z$ in the form $x+yi$. Given $Z = \frac{5+i}{2+3i}$. To express this in the form $x+yi$, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $2+3i$ is $2-3i$. $$Z = \frac{5+i}{2+3i} \times \frac{2-3i}{2-3i}$$ $$Z = \frac{(5+i)(2-3i)}{(2+3i)(2-3i)}$$ Expand the numerator: $$(5+i)(2-3i) = 5(2) + 5(-3i) + i(2) + i(-3i)$$ $$= 10 - 15i + 2i - 3i^2$$ Since $i^2 = -1$: $$= 10 - 13i - 3(-1)$$ $$= 10 - 13i + 3$$ $$= 13 - 13i$$ Expand the denominator using the difference of squares formula $(a+b)(a-b) = a^2-b^2$: $$(2+3i)(2-3i) = 2^2 - (3i)^2$$ $$= 4 - 9i^2$$ Since $i^2 = -1$: $$= 4 - 9(-1)$$ $$= 4 + 9$$ $$= 13$$ Now substitute the expanded numerator and denominator back into the expression for $Z$: $$Z = \frac{13 - 13i}{13}$$ $$Z = \frac{13}{13} - \frac{13i}{13}$$ $$Z = 1 - i$$ a) The complex number $Z$ in the form $x+yi$ is $\boxed{1-i}$. Step 2: Show that $Z = \sqrt{2}\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right)$. From part (a), we have $Z = 1 - i$. To express a complex number $x+yi$ in polar form $r(\cos\theta + i\sin\theta)$, we find the modulus $r = \sqrt{x^2+y^2}$ and the argument $\theta = \arctan\left(\frac{y}{x}\right)$ (adjusted for quadrant). For $Z = 1 - i$, we have $x=1$ and $y=-1$. Calculate the modulus $r$: $$r = \sqrt{1^2 + (-1)^2}$$ $$r = \sqrt{1+1}$$ $$r = \sqrt{2}$$ Calculate the argument $\theta$: Since $x=1$ (positive) and $y=-1$ (negative), $Z$ lies in the fourth quadrant. The reference angle $\alpha$ is given by $\tan\alpha = \left|\frac{y}{x}\right| = \left|\frac{-1}{1}\right| = 1$. So, $\alpha = \frac{\pi}{4}$. For a complex number in the fourth quadrant, $\theta = -\alpha$. $$\theta = -\frac{\pi}{4}$$ Now substitute $r$ and $\theta$ into the polar form $r(\cos\theta + i\sin\theta)$: $$Z = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)$$ Using the trigonometric identities $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$: $$Z = \sqrt{2}\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right)$$ b) We have shown that $Z = \sqrt{2}\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right)$. Step 3: By equating the form in (a) to that in (b), show that $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$. From part (a), $Z = 1 - i$. From part (b), $Z = \sqrt{2}\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right)$. Equate the two expressions for $Z$: $$1 - i = \sqrt{2}\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right)$$ $$1 - i = \sqrt{2}\cos\frac{\pi}{4} - i\sqrt{2}\sin\frac{\pi}{4}$$ For two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal. Equating the real parts: $$1 = \sqrt{2}\cos\frac{\pi}{4}$$ Divide by $\sqrt{2}$: $$\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$ c) We have shown that $\boxed{\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}}$.