Construction Steps for the Difference of Cosine identity.

Use the applet on the right to follow the construction steps of the difference of cosine identitiy. Given the unit circle and A as the center of the unit circle construct a line segment AB, where B is on the unit circle, and forms an angle \alpha > 90 degrees (step 1). Next construct a line segment AC, where C is on the unit circle, that forms an angle \beta which is less then 90 degrees (step 2. Construct line segement BC and for the triangle ABC (step 3). Now we will construct a congruent triangle to ABC. Construct line segment AE, where E is a point on the unit circle, and form the angle \( \alpha - \beta \). Then construct the line segment D where D is (1,0) on the unit circle (step 4). Finally construct the line segment ED and create the triangle AED. We will show that BC = AE algebraically.

Proof:

First find the distance of AB

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

\[ \Rightarrow \sqrt{cos^2( \alpha ) - 2cos( \alpha )cos( \beta ) + cos^2( \beta ) + sin^2 ( \alpha ) -2sin( \alpha )sin( \beta ) + sin^2( \beta )} \]

\[ \Rightarrow \sqrt{1 + 1 - 2cos( \alpha )cos( \beta ) -2sin( \alpha )sin( \beta )} \]

\[ \Rightarrow \sqrt{ 2-2cos(\alpha)cos(\beta)-2sin(\alpha)(\beta)} \]

Now find the distance of ED

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

\[ \Rightarrow \sqrt{1 - 2cos(\alpha-\beta) + cos^2(\alpha-\beta) + sin^2(\alpha-\beta)} \]

\[ \Rightarrow \sqrt{2 - 2cos(\alpha - \beta)}

Now set AB = ED

\[ \sqrt{2 - 2cos(\alpha)cos(\beta) - 2sin(\alpha)sin(\beta)} = \sqrt{2 - 2cos(\alpha - \beta)} \]

\[ \Rightarrow 2-2cos(\alpha)cos(\beta)-2sin(\alpha)sin(\beta)= 2 - 2cos(\alpha - \beta) \]

\[ \Rightarrow -2cos(\alpha)cos(\beta) - 2sin(\alpha)sin(\beta) = -2cos(\alpha - \beta) \]

\[ \Rightarrow cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta) = cos(\alpha - \beta) Q.E.D \]

The proof and the construciton shows the geometric and algebraic relationship of how this identity is derived(Gutmann, 2003).

Construciton steps for the Sum of Cosine identity.

Use the applet on the right to follow the following construction steps.

Given the unit circle and A as the center of the unit circle construct a line segment AF where F is on the unit circle and forms an angle \alpha that is less then 90 degrees (step 1). Then the line segment AG is constructed where the angle \beta is less then 90 degrees in the clockwise direction (step 2). Then the line segment FG is formed and the triangle AFG is also created (step 3).Now a congruent triangle will be constructed to AFG. Construct a line segment AH where AH forms an angle \( \alpha + \beta \) and the line segment AI is constructed where I is (1,0) on the unit circle (step 4). Then the line segment HI is constructed and the triangle AHI is created (step 5). We will prove that HI = FG.

Proof:

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

\[ \Rightarrow \sqrt{cos^2(\alpha) - 2cos(\alpha)cos(\beta) + cos^2(\beta) + sin^2(\alpha) + 2sin(\alpha)sin(\beta) + sin^2(\alpha)} \]

\[ \Rightarrow \sqrt{-2cos(\alpha)cos(beta) + 2sin(\alpha)sin(\beta) + 2} \]

Now we will find the distance of HI

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

\[ \Rightarrow \sqrt{cos^2(\alpha + \beta) - 2cos(\alpha + \beta) + 1 + sin^2(\alpha + \beta)} \]

\[ \Rightarrow \sqrt{-2cos(\alpha +\beta) +2} \]

Now we will set HI = FG.

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

\{ -2cos(\alpha)cos(\beta) + 2sin(\alpha)sin(\beta) + 2 = -2cos(\alpha + \beta) + 2 \]

\[ -2cos(\alpha)cos(\beta) + 2sin(\alpha)sin(\beta) = -2cos(\alpha + \beta) \]

\[ cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta) = cos(\alpha + \beta) Q.E.D \]

The proof and the construction shows the geometric and algebraic relationship of the sum of cosine identity (Gutmann, 2003).