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Question: For all $x,y\in\mathbb{R}, \quad \sin{(x)}+\sin{(y)}$ is equal to

a) $\quad 2\sin{\left(\frac{x+y}{2}\right)}\cos{\left(\frac{x-y}{2}\right)};$

b) $\quad 2\sin{\left(\frac{x-y}{2}\right)}\cos{\left(\frac{x+y}{2}\right)};$

c) $\quad 2\sin{\left(\frac{x+y}{2}\right)}\sin{\left(\frac{x-y}{2}\right)};$

c) $\quad 2\cos{\left(\frac{x+y}{2}\right)}\cos{\left(\frac{x-y}{2}\right)}.$


I know that one can use the product formula for the trig functions to find the correct answer. But If one forgets this formula and doesn't have time to deduce it during a test, can one just assume that $x=y=\pi/4 \Longrightarrow \sin{x}+\sin{y}=2\sin{x}=2\sin{(\pi/4)}=\sqrt{2}$, because there are no given restrictions on $x$ nor $y$ in $\mathbb{R}$?

So, logically the following should work: substituting $x=y=\pi/4$ in every answer and simplifying, the expression that equals $\sqrt{2}$ is the correct answer. However when I do this, I always get zero in each.

EDIT: I'm sorry guys, It was supposed to be $\sin{(x)}+\sin{(y)}.$ I edited it now.

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    $\begingroup$ do you mean $$\sin(x)+\cos(y)?$$ $\endgroup$ Sep 1 '17 at 15:56
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    $\begingroup$ (1) look again at $\cos{0}$, you should not get $0$ in every case. (2) you should get the same (non-zero) answers for a and the second c (perhaps d?) $\endgroup$
    – Jim H
    Sep 1 '17 at 16:05
  • $\begingroup$ it's logic that sin (x) -sin (y) gives zero when x=y $\endgroup$
    – MtGlasser
    Sep 1 '17 at 16:13
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    $\begingroup$ Actually, none of the formulas work. I'm guessing it should be $\sin x+\sin y$ or similar. To be more precise, a) corresponds to $\sin x+\sin y$, b) corresponds to $\sin x-\sin y$, c) corresponds to $\cos y - \cos x$ and d) corresponds to $\cos x+\cos y$ $\endgroup$
    – Ennar
    Sep 1 '17 at 16:16
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    $\begingroup$ Edited the question now. Sorry about that guys. $\endgroup$
    – Parseval
    Sep 1 '17 at 17:45
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Yes, it is true that this will work, with $\frac{\pi}{4}$ or any other chosen values. Though note that the logic only works one way: any option which does not give the correct answer is incorrect; however, options which give the correct answer may still be incorrect.

With $x = y = \frac{\pi}{4}$:

  • Correct value [of $\sin(x)+\cos(y)$] is $\sqrt{2}$ as you say.
  • (a) yields $2\sin(\frac{\pi}{4})\cos(0) = 2\times\frac{1}{\sqrt{2}}\times1 = \sqrt{2}$.
  • (b) yields $2\sin(0)\cos(\frac{\pi}{4}) = 0 $
  • (c) yields $2\sin(\frac{\pi}{4})\sin(0) = 0 $
  • (d) yields $2\cos(\frac{\pi}{4})\cos(0) = 2\times\frac{1}{\sqrt{2}}\times1 = \sqrt{2} $

So we have eliminated (b) and (c).

With $x = y = \pi$, the correct value is -1. (a) yields $2\sin(\pi)\cos(0) = 0$ while (d) yields $2\cos(\pi)\cos(0) = -2$, so neither appear to be correct.

So in fact, none of the options are equivalent to $\sin(x) + \cos(y)$.

EDIT: So the question is supposed to read $\sin(x) + \sin(y)$. In that case, the $\frac{\pi}{4}$ example is still exactly the same, as $\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, so we can eliminate (b) and (c). To decide between (a) and (d), perhaps the simplest example is $x = y = 0$: the correct value is 0; (a) evaluates to $2\times 0 \times 1 = 0$ while (d) is $2\times1\times1 = 2$, so (d) is clearly incorrect. This leaves only (a).

Just to reiterate the unidirectional logic point: this means we can conclude that (a) is an identity if and only if it is safe to assume that exactly one of the answers is correct (e.g. if it is implicit in the structure of a multiple choice test). If the question required proof, no number of examples would be sufficient, and one would have to resort to such methods as derivation from the product formula.

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  • $\begingroup$ Edited the question, sorry about it man. $\endgroup$
    – Parseval
    Sep 1 '17 at 17:45
  • $\begingroup$ Yes, it is implicit in these tests that only one question can be true. Otherwise it would have said "none of the above" in d). Wonderful, thank you man! $\endgroup$
    – Parseval
    Sep 2 '17 at 8:43
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$y = \sin(x) + \sin(y)$

Let's try to exclude options.

1) $x \rightarrow y$, and $y \rightarrow x$.

$y= \sin(y) + \sin(x)$ is unchanged.

b) and c) change sign, exclude.

Only a) and d) are left;

2) $x \rightarrow -x$, and $y \rightarrow -y$.

$y = \sin(-x) + \sin(-y) =$

$ - \sin(x) - \sin(y)$,

the expression changes sign .

Exclude d) ( no sign change), remains

option a).

Last check: $ x= y$:

$y = 2\sin(x)$.

Option a) : $2\sin(x)$.

Assuming one of the given options is correct for $x,y \in \mathbb{R}$:

It is a).

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No, because for $x=y=\frac{\pi}{4}$, the first and last formulas produce the same result of $\sqrt{2}$. In this situation you could try $x=y=\frac{\pi}{2}$, which uniquely points to the first formula.

Advice: You should remember the addition formulas to derive the multiplication formulas: $$\sin{x}+\sin{y}=\sin{(a+b)}+\sin{(a-b)}=2\sin{a}\cos{b}=2\sin{\frac{x+y}{2}}\cos{\frac{x-y}{2}}.$$ Similarly for others.

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