# Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers.

The question:

Determine $$A + B$$ if $$A$$ and $$B$$ are acute angles such that:

$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$

$$\cos A - \cos B = \sqrt{\frac{1}{2}}$$

Here are the two solutions that I found:

I think the problem with the second solution may have to do with the assumption that:

$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$

is equivalent to

$$A + \frac{\pi}{6} = -B + \frac{\pi}{6}$$

But can't you say that $$\cos N = \cos M$$ is equivalent to $$\pm N = \pm M$$ for all values of $$N$$ and $$M$$ (because cosine is an even function)?

• And no, it's not an algebraic mistake somewhere that I'm too lazy to find myself. I have checked both of these solutions multiple times (the top is the correct answer) and there aren't any obvious mistakes in the bottom. Feb 16, 2018 at 2:14
• NOTE: A lot of people are saying that the solution A+B=0 (or $0+\pi*k$ where $k \in \mathbb{Z}$ to be exact) is extraneous/impossible because the terms of the question specify that A and B are acute angles. But that is arbitrary. If the terms of the question were asking for all possible values for A+B, 0 would still be incorrect because we know the set of values is $\frac{\pi}{2} + k$ where $k \in \mathbb{Z}$ . So why does this incorrect solution emerge? Feb 16, 2018 at 11:51

Using Complex Addition $$e^{ia}+e^{i(\pi-b)}=\sqrt{\frac12}+i\sqrt{\frac32}=\color{#C00}{\sqrt2}e^{i\color{#090}{\pi/3}}$$

1. $b+a=\frac\pi2$ since $\left|e^{ia}+e^{i(\pi-b)}\right|^2=\color{#C00}{2}\implies\overbrace{\color{#C00}{2}=2+2\cos((\pi-b)-a)}^{\text{Law of Cosines}}$

2. $b-a=\frac\pi3$ since $\frac{a+(\pi-b)}2=\color{#090}{\frac\pi3}$

Therefore, $$a=\frac\pi{12}\text{ and }b=\frac{5\pi}{12}$$

• Since the question was to find $A+B$, I could have stopped at 1.
– robjohn
Feb 16, 2018 at 19:08

In the second solution $$A+\frac{\pi}{6}=B-\frac{\pi}{6},$$ which is very well or $$A+B=0,$$ which is impossible.

Another way: $$2\sin\frac{A+B}{2}\cos\frac{A-B}{2}=\sqrt{\frac{3}{2}}$$ and $$2\sin\frac{A+B}{2}\sin\frac{B-A}{2}=\sqrt{\frac{1}{2}}.$$ Thus, $$\tan\frac{B-A}{2}=\frac{1}{\sqrt3}$$ and since $A$ and $B$ they are acute angles, we obtain $$B-A=60^{\circ},$$ $$\sin\frac{A+B}{2}=\frac{1}{\sqrt2}$$ and $$A+B=90^{\circ}.$$

• Thank you — I was more looking for an answer explaining why the second solution doesn’t work, however. I can’t find anything wrong with it but it yields an incorrect answer. Feb 16, 2018 at 2:52
• I think your second solution works. Just you need to end it. I added something. See now. Feb 16, 2018 at 2:54
• Yes, I got to A + B = 0 myself... but why does it not work? Does $cos(N) = cos(M)$ not equate to $\pm N = \pm M$? Feb 16, 2018 at 3:07
• It works. You got $B-A=\frac{\pi}{3}$ and just continue to solve. Feb 16, 2018 at 3:19
• I guess I'm just confused as to how you got from $$B-A=60^{\circ},$$ to $$\sin\frac{A+B}{2}=\frac{1}{\sqrt2}$$ Also why is the A+B=0 solution impossible? Does taking the inverse cosine function of both sides of the equation not always work? Feb 16, 2018 at 3:26

$$\sqrt3\cos A-\sin A=\sqrt3\cos B+\sin B$$

$$\implies\cos(A+30^\circ)=\cos(B-30^\circ)$$

$$A+30^\circ=360^\circ n\pm(B-30^\circ)$$

$$-\implies A+B=360^\circ n$$ which is impossible as $0<A+B<180^\circ$

$+\implies A-B=360^\circ n-60^\circ$

As $-90<A-B<90,n=0$

Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html in anyone of the given relation

• See my comment on my question -- even if the terms of the question specify only to sample from the set of the values of A+B when A and B are acute, why does this solution that A+B=0 emerge at all? Shouldn't the set of all possible solutions be $\frac{\pi}{2} + k$ where $k \in \mathbb{Z}$ ? Feb 16, 2018 at 12:02
• @MureyTasroc, $$A=360^\circ n-B\implies$$ $$\cos(A)=\cos(360^\circ n-B)=?$$ $$\sin(A)=\sin(360^\circ n-B)=?$$ Feb 17, 2018 at 12:55

My question was more about why my second approach yields an incorrect solution, ignoring the arbitrary assertion that $A$ and $B$ must be acute in the final answer selected from the solution set for $A+B$. I have answered my own question, however, with the help/tips from everyone who kindly provided comments/answers.

Given the values for $A$ and $B$ that robjohn kindly found using Complex Addition (+1) the $B-A=\frac{\pi}{3}$ solution is correct using my second strategy where this is the antepenultimate step:

$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$

and then:

$$B-A=\frac{\pi}{3}$$

And $\frac{\pi}{3}$ is indeed an element of the correct solution set for $B - A$



The $A+B = 0$  solution that you can get by multiplying the inputs of either of the cosine functions (but not the other) by $-1$ is not correct, however; it does not match any valid solutions in the solution set for $A + B$: $\frac{\pi}{2}+\pi*k ,$  $k \in \mathbb{z}$.  Also, as one or two answers pointed out, $A$ and $B$ must be acute so $A + B$ cannot equal $0$ (although I was less concerned with this than I was with the fact that the answer didn't match any of the valid solutions in the solution set for $A+B$). I now realize that the reason for this is because multiplying the inputs of the cosine functions in such an equation does not change the validity of the equation (because cosine is even) but it does change the value of the variables in that equation.

In other words:

If $\pm N = \pm M$ then $\cos N = \cos M$. However, if $\cos N = \cos M$ then $\pm N$ does not necessarily equal $\pm M$.  Another example:

Let $M = \frac{\pi}{4}$  ,  Let $N = \frac{7*\pi}{4}$

$cosM =cosN$

But $M \ne N$

This seems obvious when put this way but I did not realize or think of it when doing out this solution.