The double angle idnenities are useful when the angles are increased. These formulas are especially important when solving optimization problems involving angles. The following proofs are the double angle identity for sine and cosine (Gonzala).
Sine double angle identity proof:
We will use the sum of sine identity \( sin(\alpha + \beta) = sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \).
In the equation above let \( \beta = \alpha \).
\[ \Rightarrow sin(\alpha + \alpha) = sin(\alpha)cos(\alpha) + cos(\alpha)sin(\alpha) \]
\[ \Rightarrow sin(2\alpha) = 2sin(\alpha)cos(\alpha) Q.E.D \] (Gonzala).
Cosine double angle identity proof:
We will use the sum of cosine identity \( cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta) \).
In the equation above let \( \beta = \alpha \).
\[ \Rightarrow cos(\alpha + \alpha) = cos(\alpha)cos(\alpha) - sin(\alpha)sin(\alpha) \]
\[ \Rightarrow cos(2\alpha) = cos^2(\alpha) - sin^2(\alpha) Q.E.D \] (Gonzala).
We will discuss the construction and the proof of the sum to product for \( cos(\alpha)+cos(\beta) = 2cos(\frac{\alpha + \beta}{2})cos(\frac{\alpha - \beta}{2}) \) and \( sin(\alpha) + sin(\beta) = 2sin(\frac{\alpha + \beta}{2})cos(\frac{\alpha - \beta}{2}). The applet to the right shows the construction of the sum to product proofs. The unit circle is given where A is the center of the circle. First construct the line segment AB where B is on the unit circle and it forms an angle \alpha (Step 1). Then construct the line segment AC where C is on the unit circle and forms an angle \beta (Step 2). Since we want to add the two angles then we will construct the midpoint N of B and C. Then the line segment AN is constructed and is a perpendicular to BC (Step 3). The following below is the proof for the two identities mentioned above.
Proof:
By construction \( B = (cos(\alpha),sin(\alpha)), C = (cos(\beta), sin(\beta)), and N = (1/2(cos(\alpha) + cos(\beta)), 1/2(sin(\alpha) + sin(\beta))) \). Also by construction the triangle ANC is congruent to ANB, so the angle \( NAC = 1/2(\alpha - \beta) \). Then by the definiton of cosine \( AN = cos(1/2(\alpha - \beta)) \). Now since AN is the terminal segment then \( cos(1/2(\alpha + \beta)) = \frac{1/2(cos(\alpha) + cos(\beta))}{cos(1/2(\alpha - \beta))} \). Then \( cos(\alpha)+cos(\beta) = 2cos(\frac{\alpha + \beta}{2})cos(\frac{\alpha - \beta}{2}) \) and the the same process is used for sine Q.E.D (Jarrett, 1987).