Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$.
The first couple of steps are the same: finding $\sin$ and $\cos$ values for both $x$ and $y$.
From $\cos 2x$ we have:
\begin{align*}
\cos 2x &=-\frac {63}{65} \\
\cos2x &= \cos^2x-\sin^2x = \cos^2x – (1-\cos^2x)=2\cos^2x -1 \\
2\cos^2x -1 &= -\frac {63}{65} \\
2\cos^2x &=\frac {-63+65}{65} \\
\cos^2x &=\frac {1}{65} \\
\cos x &=\frac {1} {\sqrt{65}}
\end{align*}
(taking only the positive value of $\cos x$ because $\cos x$ is always positive under the given domain)
\begin{align*}
\sin^2x &=1-\frac {1}{65} \\
\sin^2x &=\frac {64}{65} \\
\sin x &=\frac {8} {\sqrt{65}}
\end{align*}
(again, only positive value)
From $\cos y$ we have:
\begin{align*}
\cos y &=\frac {7} {\sqrt{130}} \\
\cos^2y &=\frac {49} {130} \\
\sin^2y &=1-\frac {49} {130} \\
\sin^2y &=\frac {81} {130} \\
\sin y &=\frac {9} {\sqrt{130}}
\end{align*}
Now that we've gathered necessary information, we proceed to calculate value of some trigonometric function of $x+y$, hoping we will get some basic angle:
sin(x+y):
\begin{align*}
\sin(x+y) &=\sin x \cos y + \sin y \cos x =\frac {8} {\sqrt{65}} \frac {7} {\sqrt{130}} + \frac {9} {\sqrt{130}}\frac {1} {\sqrt{65}} \\
\sin(x+y) &=\frac {65} {\sqrt{65}\sqrt{130}} \\
\sin(x+y) &=\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {\pi}{4}+2k{\pi}$ OR $x+y =\frac {3\pi}{4}+2k{\pi}$
Since $x$ and $y$ are in the first quadrant, their sum must lie in first or second quadrant.
Solutions are:
$x+y= \{\frac {\pi}{4}, \frac {3\pi}{4} \}$
cos(x+y):
\begin{align*}
\cos(x+y) &= \cos x \cos y – \sin x \sin y =\frac {1} {\sqrt{65}} \frac {7} {\sqrt{130}} – \frac {8} {\sqrt{65}}\frac {9} {\sqrt{130}} \\
\cos(x+y) &=-\frac {65} {\sqrt{65}\sqrt{130}} \\
\cos(x+y) &=-\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {3\pi}{4}+2k{\pi} $ OR $x+y =\frac {5\pi}{4}+2k{\pi}$
Now we can only have one solution: $x+y=\{\frac {3\pi}{4}\}$
Similar happens with $\cos(x-y)$.
My question is: why do these two formulas give two different solutions? General insight would be great, since I found a lot of examples with similar problems. Thank you in advance.
Best Answer
Andre already gave the answer, but let me explain the "generalities".
The main problem is the following:
You interpreted this to mean
This interpretation is false! The correct interpretation is that
There may be other constraints that rule out one (or both) of those values. In your case, knowing that $\cos x \approx 1/8$ should tell you already that $x$ is bigger than $\pi/4$.
Another way to look at this clearly is to look at the directions of implication. From the values of $\sin x$, $\sin y$ etc you can derive the value of $\sin(x+y)$. But just knowing the value of $\sin(x+y)$ you cannot derive the value of $\sin x$, $\sin y$, etc. So the implication going from the "necessary information" step to the "computing $\sin(x+y)$ and $\cos(x+y)$" step is one that loses information.
So that solutions to $\sin(x+y) = \sqrt{2}/2$ are not necessarily solutions to $\cos(x) = 1/\sqrt{65}$ etc.