Determine the sum of angles $A,B,$ where $0^\circ \leq A,B, \leq 180^\circ$ and \[ \sin A + \sin B = \sqrt{\frac{3}{2}}, \cos A + \cos B = \sqrt{\frac{1}{2}} \]
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Tags: trigonometry, LaTeX
Silverfalcon
28.01.2006 01:29
My first attempt was to directly multiply this equation. Simplifying, I get $\sin(A+B) + \frac{\sin (2A) + \sin(2B)}{2} = \frac{\sqrt{3}{2}}$. But how can I go from here? I mean, if it's possible.
Okay.. Following the answer key, I squared the both equations and add them. This gave me $\cos(A-B) = 0$ so $A-B = \pm 90$. The answer then simply said that "consider possibility.." Well, how can I find which one is the right one? I'm guessing $A-B = 90$ but still, this doesn't help me to get $A+B$.
Thanks for the help! This is the last problem on COMC 1996 Part I by the way.
4everwise
28.01.2006 01:42
I think I have a method. Let me work it out on paper to see if it works. Edit: It dodn't really work. I'll think about this more later if nobody posts a solution.
Artzilla03
28.01.2006 02:32
ok, so, use sum-to-product. (sorry, i dont know LaTeX well enough yet)
2sin((A+B)/2)cos((A-B)/2) = sqrt(3/2)
and
2cos((A+B)/2)cos((A-B)/2)) = sqrt(1/2)
divide equations:
tan((A+B)/2) = sqrt(3)
at 60 degrees this is true.
(A+B)/2 = 60
A+B = 120
lowest_reynolds
28.01.2006 08:59
Having practised on this question myself, it's pretty hard to forget the solution
Let $A = x+y$ and $B = x-y$. It follows then that:
$\sin(x+y) + \sin(x-y)$
$= \sin x\cos y + \sin y\cos x + \sin x\cos y- \sin y\cos x$
$= 2\sin x\cos y$
$= 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$
And:
$\cos(x+y) + \cos(x-y)$
$= \cos x\cos y - \sin x\sin y + \cos x\cos y + \sin x\sin y$
$= 2\cos x\cos y$
$= 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$
Now looking at our two new equations, we see that we get something familiar if we divide the first equation by the second. We get:
$\tan(\frac{A+B}{2}) = \sqrt{3}$
Therefore:
$\frac{A+B}{2} = 60^{\circ}$
and
$A + B = 120^{\circ}$
Virgil Nicula
28.01.2006 09:54
The Artzilla's proof. $\sin A+\sin B=\sqrt{\frac 32},\ \cos A+\cos B=\sqrt{\frac 12}\ .$ $\frac{\sin A+\sin B}{\cos A+\cos B}=\frac{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}{2\cos \frac{A+B}{2}\cos \frac{A-B}{2}}=\tan \frac{A+B}{2}\ .$ $\Longrightarrow \tan \frac{A+B}{2}=\sqrt 3\Longrightarrow A+B=120^{\circ}\ .$
JesusFreak197
28.01.2006 21:13
$(\sin A+\sin B)^2=\sin ^2A+\sin ^2B+2\sin A\sin B=\frac{3}{2}$
$(\cos A+\cos B)^2=\cos ^2A+\cos ^2B+2\cos A\cos B=\frac{1}{2}$
Adding the two, we have $2+2(\cos A\cos B+\sin A\sin B)=2\Rightarrow \cos (A-B)=0$
Therefore, $A=B+90^\circ$, so $\sin (B+90^\circ)+\sin B=\sin B+\cos B=\sqrt{\frac{3}{2}}$ and $\cos (B+90^\circ)+\cos B=\cos B-\sin B=\sqrt{\frac{1}{2}}$. Adding the two, we have $2\cos B=\frac{\sqrt{6}+\sqrt{2}}{2}\Rightarrow \cos B=\frac{\sqrt{6}+\sqrt{2}}{4}$, and therefore $B=15^\circ$, $A=105^\circ$, $A+B=120^\circ$.
maokid7
28.01.2006 21:28
try squaring both equations and then adding them together. it should be pretty obvious from there.
1234567890
29.01.2006 00:14
Let $z_1=\cos{A}+i\sin{A}=\text{cis}A$ and $z_2=\cos{B}+i\sin{B}=\text{cis B}$
$\Rightarrow z_1+z_2=r\text{cis}(\frac{A+B}{2})=\text{cis}A+\text{cis}B$
$\Rightarrow \sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}=\sqrt{2}(\frac{1}{2}+i \frac{\sqrt{3}}{2})=\sqrt{2}\text{cis}60^\circ$
$\Rightarrow \frac{A+B}{2}=60^\circ \Rightarrow \boxed{A+B = 120^\circ}$
Silverfalcon
29.01.2006 03:29
Thanks guys! I liked all three solutions! P.S. 1234567890, you have typo in the first one. I assume you meant on the first line as $z_1 = \cos A + i \sin A = \text{cis} A$. You forgot $\sin A$ part.