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This question looks easy but I don't know why I can't get the answer.

If $x$ and $y$ are acute angles, such that $$\cos x + \cos y = \frac{3}{2} \qquad\text{and}\qquad \sin x + \sin y = \frac{3}{4}$$ then evaluate $\sin(x+y)$.

I think this question should be solved by using basic transformation formulas.

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  • $\begingroup$ What have you tried? Posting the question isn't the best way to ask a question. Check out this link: math.stackexchange.com/help/how-to-ask and welcome to Math SE! $\endgroup$ – Paras Khosla Feb 11 at 11:35
  • $\begingroup$ Think about the simplest trigonometric equation. $\endgroup$ – Claude Leibovici Feb 11 at 11:37
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Use sum to product formulas for sine and cosine functions stated as follows: $$\cos x+\cos y=2\cos\biggl(\dfrac{x+y}{2}\biggr)\sin \biggl(\dfrac{x-y}{2}\biggr) \tag1 $$ $$ \sin x+\sin y=2\sin \biggl(\dfrac{x+y}{2}\biggr)\cos\biggl(\dfrac{x-y}{2}\biggr) \tag2$$Notice that dividing $(2)$ by $(1)$ gets us rid of the expression of $\cos\bigl(\frac{x-y}{2}\bigr)$ and gives us the equation only in terms of $\tan\bigl(\frac{x+y}{2}\bigr)$.$$\dfrac{\sin x+\sin y}{\cos x+\cos y}=\frac{1}{2}=\tan\biggl(\dfrac{x+y}{2}\biggr)\implies \dfrac{1}{4}=\tan^2\biggl(\dfrac{x+y}{2}\biggr)$$Now using the half-angle formula for tangent gives us the expression in terms of $\cos(x+y)$.$$\dfrac{1}{4}=\dfrac{1-\cos(x+y)}{1+\cos(x+y)}\implies \dfrac{3}{5}=\cos(x+y)$$Now finding $\sin(x+y)$ is simply a matter of applying the Pythagorean identity. In fact doing so gives us $\sin(x+y)=4/5$.

Aliter: As suggested by @lab bhattacharjee, a quicker and better way would be to use the direct relation between $\sin 2\theta$ and $\tan\theta$. This makes sense because we're supposed to find $\sin (x+y)$ from $\tan\bigl(\frac{x+y}{2}\bigr)$ which we've extracted from the information provided. Also, in my opinion this is a better approach because it allows us to skip the step having to do with applying the Pythagorean identity. Also, I believe this is the thought process one should go through while attempting such questions because it minimizes the time taken to solve, which is key in exam situations and otherwise as well. $$\sin(x+y)=\dfrac{2\tan\bigl(\frac{x+y}{2}\bigr)}{1+\tan^2\bigl(\frac{x+y}{2}\bigr)}=\dfrac{1}{1+\frac{1}{4}}=\dfrac{4}{5}$$

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  • $\begingroup$ We don't need to find cosine. en.m.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee Feb 11 at 12:41
  • $\begingroup$ @labbhattacharjee I've found $\sin(x+y)$ ultimately by finding $\cos(x+y)$ first. $\endgroup$ – Paras Khosla Feb 11 at 12:43
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    $\begingroup$ We can directly use $$\sin2z=\dfrac{2\tan z}{1+\tan^2z}$$ $\endgroup$ – lab bhattacharjee Feb 11 at 13:29
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The hint:

From $$(\cos{x}+\cos{y})^2+(\sin{x}+\sin{y})^2=\frac{9}{4}+\frac{9}{16}$$ we can get a value of $\cos(x-y).$

From $$(\cos{x}+\cos{y})^2-(\sin{x}+\sin{y})^2=\frac{9}{4}-\frac{9}{16}$$ we can get a value of $\cos(x+y).$

Assume that $x\geq y$ and get a values of $x$ and of $y$ and end this problem.

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The problem is symmetric, in the sense that it is invariant under interchanging $x$ and $y$. So it makes sense to start from the arithmetic mean of $x$ and $y$.

Let $z=(x+y)/2$ and write $x=z-t$, $y=z+t$ (where $t=(y-x)/2$). Then your first equation can be rewritten as $$ \cos z\cos t+\sin z\sin t+\cos z\cos t-\sin z\sin t=3/2 $$ that is $$ \cos z\cos t=3/4 $$ Similarly, $$ \sin z\cos t=3/8 $$ This implies $\tan z=1/2$. Now you know that $$ \sin 2z=\frac{2\tan z}{1+\tan^2z} $$

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