General solution to sin ⁡ α + sin ⁡ β and cos ⁡ α + cos ⁡ β?

Question

General solution to   sin  ⁡  α  +  sin  ⁡  β and   cos  ⁡  α  +  cos  ⁡  β?

Jamiya Costa · Accepted Answer

Using the sum/difference identities of the sine function, we find that

\sin (α + β) = \sin (α) \cos (β) + \sin (β) \cos (α)

\sin (α - β) = \sin (α) \cos (β) - \sin (β) \cos (α)

Adding them together, we find that

\sin (α + β) + \sin (α - β) = 2 \sin (α) \cos (β)

Now suppose that we want to find

\sin (x) + \sin (y)

where x<y. Then we can rewrite x and y as

α \pm β

where

α

is the average of x and y and where

β

is the difference between y and the average. From there, we can plug it into the identity above. A similar computation can be arrived at using the sum/difference formulas for cosine.

α = \frac{x + y}{2}, β = y - \frac{x + y}{2}

x = α - β, y = α + β

\sin (x) + \sin (y) = \sin (α - β) + \sin (α + β) = 2 \sin (α) \cos (β)

General solution to sin ⁡

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