User:Egm4313.s12.team11.Suh/R7.5

Report 7, Problem 7.5
Solved by Solved by Daniel Suh

Problem Statement
Consider the following, $$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\phi _{2j-1}(x) \cdot \phi _{2k-1}(x)dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\sin {j\omega x} \cdot \sin {k\omega x} \; dx\!$$

with $$j\neq k\; and\; j,k = 1,2,...\!$$, and $$p=2\pi, j=2, k=3\!$$

1. Find the integration with the given data.

2. Confirm the results with Matlab's trapz command for the trapezoidal rule.

Trigonometric Identities
Angle Sum and Difference Identities

$$(1) \cos{(a+b)} = \cos{a}\cos {b} - \sin {a}\sin{b}\!$$ $$(2) \cos{(a-b)} = \cos{a}\cos {b} + \sin {a}\sin{b}\!$$

Rearrange

$$(1) \sin {a}\sin{b} = \cos{a}\cos {b}-\cos{(a+b)}\!$$ $$(2) \cos{a}\cos {b} = \cos{(a-b)} - \sin{a}\sin {b}\!$$

Substitute and Combine

$$\sin {a}\sin{b} = \cos{(a-b)} - \sin{a}\sin {b} - \cos{(a+b)}\!$$ $$2\sin {a}\sin{b} = \cos{(a-b)} - \cos{(a+b)}\!$$ $$\sin {a}\sin{b} = \frac{1}{2}\cos{(a-b)} - \frac{1}{2}\cos{(a+b)}\!$$

Utilize Trig Identities
$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\sin {jwx} \cdot \sin {kwx} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{2\pi}\sin {2\omega x} \cdot \sin {3\omega x} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{2\pi}\frac{1}{2}\cos {(2\omega x - 3\omega x)} - \frac{1}{2}\cos{(2\omega x + 3\omega x)} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\frac{1}{2}\int_{0}^{2\pi}\cos {(-\omega x)} \; dx - \frac{1}{2}\int_{0}^{2\pi}\cos{(5\omega x)} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\frac{1}{2}\sin {(-\omega x)}|_{0}^{2\pi} - \frac{1}{2}\sin{(5\omega x)}|_{0}^{2\pi}\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle= (0-0) - (0-0) \!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle= 0 \!$$