Expert answers to "Prove trigonometric equivalencesin⁡x+sin⁡y=2sin⁡(x+y2)cos⁡(x−y2)"

svrstanojpkqx

Answered question

2022-03-30

Prove trigonometric equivalence

\sin x + \sin y = 2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2})

Madilyn Shah

Beginner2022-03-31Added 11 answers

As requested, we will prove the identity of interest through direct use Euler's Identity,

e^{i z} = \cos z + i \sin z

From this we have

\sin z = \frac{e^{i z} - e^{- i z}}{2 i}

and

\cos z = \frac{e^{i z} + e^{- i z}}{2}

. Then, we can write

2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2}) = 2 (\frac{e^{i \frac{x + y}{2}} - e^{- i \frac{x + y}{2}}}{2 i}) (\frac{e^{i \frac{x - y}{2}} + e^{- i \frac{x - y}{2}}}{2})

= \frac{e^{i x} + e^{i y} - e^{- y} - e^{- i x}}{2 i}

= \frac{e^{i x} - e^{- i x}}{2 i} + \frac{e^{i y} - e^{- i y}}{2 i}

= \sin x + \sin y

as was to be shown!

undodaonePvopxl24

Beginner2022-04-01Added 13 answers

From

\sin (A + B) = \sin A \cos B + \cos A \sin B

and

\sin (A - B) = \sin A \cos B - \cos A \sin B

, get:

\sin x + \sin y = \sin (\frac{x + y}{2} + \frac{x - y}{2}) + \sin (\frac{x + y}{2} - \frac{x - y}{2})

= \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2}) + \cos (\frac{x + y}{2}) \sin (\frac{x - y}{2}) + \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2}) - \cos (\frac{x + y}{2}) \sin (\frac{x - y}{2})

= 2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2})

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