I'm struggling with a math problem that involves finding the diagonal lengths and their sum in a rhombus with a known angle. Here's what I've got so far:
The rhombus $ABCD$ has sides of length $AB = BC = CD = DA = a$.
Angle $B$ in triangle $ABD$ is $\varphi$.
I need to find the lengths of the two diagonals and their sum. Can anyone please guide me through the steps to solve this problem? Any help would be greatly appreciated.
I was told:
"If you call the center $X$, then $BX = DX = a \cos(\frac{\varphi}{2})$ and $AX = CX = a \sin(\frac{\varphi}{2})$. So the sum of the diagonals is $2a (\sin(\frac{\varphi}{2}) + > \cos(\frac{\varphi}{2}))$. You might make some progress simplifying an expression of the form $\sin\theta + \cos\theta$ by writing it as $\sqrt{2}\sin(\theta+\frac{\pi}{4})$."
But I am not sure what it means visually, nor did they motivate why.
Here's the drawing they made, where t is instead the angle instead of $\varphi$:
Thank you in advance!
Step #0: Concise definition of $\varphi$
Define $$\angle ABC=\varphi$$
(avoiding confusion with diagonal angles).
Step #1: Express length of diagonal $e$
$$e=2a\times \cos(\frac{\varphi}{2})$$
(diagonal e measures twice the distance from vertex $B$ to center $X$, which itself is given by $a\times \cos(\frac{\varphi}{2})=BX$ and where the argument $(\frac{\varphi}{2})$ is the diagonal angle).
Step #2: Express length of diagonal $f$
$$f=2a\times \sin(\frac{\varphi}{2})$$
(diagonal f measures twice the distance from vertex $A$ to center $X$, which itself is given by $a\times \sin(\frac{\varphi}{2})=AX$).
Step #3: Express the sum $(e+f)$
$$(e+f)=2a\times \left( \cos(\frac{\varphi}{2})+\sin(\frac{\varphi}{2}) \right)$$
Step #4: 'Simplify' using trig identities
With the known and adviced identity
$\cos x+\sin x={\sqrt {2}}\cdot \sin \left(x+{\frac {\pi }{4}}\right)$
or with
$\cos x+\sin x={\sqrt {2}}\cdot \cos \left(x-{\frac {\pi }{4}}\right)\\$,
you can establish that
$$(e+f)=2a\times \sqrt2\times \sin \left(\frac{\varphi}{2}+\frac{\pi}{4}\right)$$
$$(e+f)=2a\times \sqrt2\times \cos \left(\frac{\varphi}{2}-\frac{\pi}{4}\right).$$