A curious trig identity

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Here is an identity that doesn’t look correct but it is. For real x and y,

|sin(x + iy)| = |sin x + sin iy|

I found the identity in [1]. The author’s proof is short. First of all,

begin{align*} sin(x + iy) &= sin x cos iy + cos x sin iy \ &= sin x cosh y + i cos x sinh y end{align*}

Then

begin{align*} |sin(x + iy)|^2 &= sin^2 x cosh^2 y + cos^2 x sinh^2 y \ &= sin^2 x (1 + sinh^2 y) + (1 -sin^2x) sinh^2 y \ &= sin^2 x + sinh^2 y \ &= |sin x + i sinh y|^2 \ &= |sin x + sin iy|^2 end{align*}

Taking square roots completes the proof.

Now note that the statement at the top assumed x and y are real. You can see that this assumption is necessary by, for example, setting x = 2 and y = i.

Where does the proof use the assumption that x and y are real? Are there weaker assumptions on x and y that are sufficient?

 

[1] R. M. Robinson. A curious trigonometric identity. American Mathematical Monthly. Vol 64, No 2. (Feb. 1957). pp 83–85

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