Difference Identity for Cosine

We have to prove things in a bit of a funny order. First, we will prove the difference identity for cosine. Then we will detour to opposite angles, complementary angles, and supplementary angles. Finally, we will return to prove the remaining sum and difference identities.


The Euclidean Distance Formula

For this proof, we need to be able to find the straight-line distance between two points on the 2D plane. Doing this is just a simple application of the Pythagorean Theorem.


Thus, $d^2 = (y_2 - y_1)^2 + (x_2 - x_1)^2$ or $d = \sqrt{ (y_2 - y_1)^2 + (x_2 - x_1)^2 }$. This is called the Euclidean distance formula.


Derivation

Consider an isosceles triangle centered on the unit circle (left figure). Rotate this triangle so that one edge lies on the x-axis (right figure).

    

Since the triangles are congruent, the side length $d$ must be the same between the two. Thus, we will calculate $d$ in two different ways using the Euclidean distance formula.

Per the left figure,.

\[d_{\text{left}} = \sqrt{(\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2}\]

Per the right figure.

\[d_{\text{right}} = \sqrt{(\cos (\alpha -\beta) - 0)^2 + (\sin (\alpha - \beta) - 1)^2}\]

Equate the left and right figure.

\[\begin{align} d_{\text{left}} &= d_{\text{right}} \\[10pt] d_{\text{left}}^2 &= d_{\text{right}}^2 \\[10pt] (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 &= (\cos (\alpha -\beta) - 1)^2 + (\sin (\alpha - \beta) - 0)^2 \\[10pt] \cos^2 \alpha - 2 \cos \alpha \cos \beta + \cos^2 \beta + \sin^2 \alpha - 2 \sin \alpha \sin \beta + \sin^2 \beta &= \cos^2 (\alpha -\beta) - 2 \cos (\alpha - \beta) + 1 + \sin^2 (\alpha -\beta) \\[10pt] (\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta &= (\cos^2 (\alpha -\beta) + \sin^2 (\alpha -\beta)) + 1 - 2 \cos (\alpha - \beta) \\[10pt] 1 + 1 - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta &= 1 + 1 - 2 \cos (\alpha - \beta) \\[10pt] - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta &= - 2 \cos (\alpha - \beta) \\[10pt] \cos \alpha \cos \beta + \sin \alpha \sin \beta &= \cos (\alpha - \beta) \end{align}\]

Thus, we have shown

\[\boxed{\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta}\]

Trigonometry Series