**The sum and difference formulas** can be used to find exact values of trigonometric functions involving sums or differences of special angles. Here are the sum and difference formulas that I will be providing the proofs for:

**cos(u + v) = cos u cos v – sin u sin v **

**cos(u – v) = cos u cos v + sin u sin v**

**tan(u ± v) = (tan u – tan v)/(1 + tan u tan v) **

Proof: You can use the figures in the pictures below to better understand the proof of the formulas for cos(u ± v). In the left figure, let A be the point (1,0) and then use **u **and **v **to locate the points B=(x¹, y¹), C=(x², y²), and D=(x³, y³) on the unit circle. So, (x¹)*2 + (y¹)*2=1, (x²)*2 + (y²)*2=1, and (x³)*2 + (y³)*2=1. Now this notation might seem strange but allow me to explain the signification of particular characters and state that the choice of these strange characters can be attributed to the fact that I am limited to the characters in the Word Press Kitchen Sink. Let *n signify the operation of exponentiation where n is an integer, and let ¹, ², and ³ signify the coordinates corresponding to B, C, and D respectively. It is slightly counterintuitive but stays consistent with the logic.

Now assume that 0 < v < u < 2pi and note that in the figure to the right, arcs AC and BD have the same length. So, line segments AC and BD are also equal in length, which implies that **√[(x² – 1)*2 + (y² – 0)*2] = √[(x³ – x¹)*2 + (y³ – y¹)*2]. **

Now to do some algebra so as to isolate x².

After applying the proper foiling operations we obtain: (**x²)*2 – 2x² + 1 + (y²)*2 = (x³)*2 – 2x¹x³ + (x¹)*2 + (y³)*2 – 2y¹y³ + (y¹)*2. **Now we can group the squared coordinates of **C **on the left side of the equation and group the squared coordinates of **D** and **B **on the right side of the equation. After this operation the equation will look like this: **[****(x²)*2 + (y²)*2] +1 – 2x² = [(x³)*2 + (y³)*2 + [(x¹)*2 + (y¹)*2] – 2x¹x³ – 2y¹y³. **Now since we know that **x*2 + y*2 =1**, we know that by replacing the **x **or **y **coordinates with the respective coordinates of either **A, B, C, **or **D **the equation will still equal **1, **so we can replace **[(x²)*2 + (y²)*2], [(x³)*2 + (y³)*2], **and **[(x¹)*2 + (y¹)*2] **with **1. **After doing this we obtain **1 + 1 – 2x² = 1 + 1 – 2x¹x³ – 2y¹y³ **which simplifies to **x² = x³x¹ + y³y¹. **

Now we can substitute the values **x² = cox(u – v), x³ = cos u, x¹ = cos v, y³ = sin u, **and **y¹ = sin v **which changes the above equation to **cos(u – v) = cos u cos v + sin u sin v. **The formula for **cos(u + v) **can be obtained by considering **u + v = u – (-v) **and using the formula just derived above to obtain: **cos(u + v) = cos[u – (-v)] = cos u cos(-v) + sin u sin(-v) **which equals **cos u cos v – sin u sin v.**

Now to get the formula for **sin(u + v), **we are going to take **cos(u + v) = cos u cos v – sin u sin v **and replace **u ** by **pi/2 – u.** By doing this we the equation **cos(pi/2 – u + v) = cos(pi/2 – u) cos v – sin(pi/2 – u) sin v. **Since we know that **sin(pi/2 – u) = cos u**, **cos(pi/2 – u) = sin u, **and **cos[pi/2 -(u – v)] = sin (u – v) **we can conclude that **sin(u – v) = sin u cos v – cos u sin v.** If we want to obtain **sin(u + v) **all we have to do is replace **v **by **-v.**

** **Finally we can use the sum and difference formulas for sine and cosine to prove the formulas for **tan(u ± v).** The proof is as follows:

**tan(u ± v) = sin(u ± v)/cos(u ± v)**. Then if we plug in the appropriate sum and difference formulas for **sin** and **cos** then we obtain (**sin u cos v ± cos u sin v)/(cos u cos v ± sin u sin v).** Then if we divide the numerator and the denominator by **cos u cos v **we obtain the equation **[****(sin u cos v ± cos u sin v)/(cos u cos v)]/[(cos u cos v ± sin u sin v)/(cos u cos v)] **Now if we write the numerator of the fraction in the numerator of the expression and the numerator of the fraction in the denominator of the expression we obtain the equation: [(**sin u cos v)/(cos u cos v) ± (cos u sin v)/(cos u cos v)]/[(cos u cos v)/(cos u cos v) ± (sin u sin v)/(cos u cos v). **Now in the numerator of the expression **cos u /cos u **occurs twice and therefore cancels twice; once in the fraction containing **sin u **in its numerator and once in the fraction containing **sin v** in its numerator. In the denominator of the expression the fraction containing **cos u cos v** in its numerator cancels out completely because their is symmetry between the numerator and denominator leaving the value 1 remaining. After the operation of cancellation we are left with the expression: [(**sin u/cos u) ± (sin v/ cos v)]/[(1±sin u/cos u)·(sin v/cos v)]. **Using the tangent quotient identity we obtain the expression (**tan u ± tan v)/(1 ± tan u tan v) **which is the sum and difference formula for **tan(u ± v).**