Midpoints in a Square

the angle at MAN is 45 take sin 45 and cos 45 u will get 7/5

I wish I can take a pic of my solution and upload here - editor doesn't work for me. Step 1 - just draw a square and state the lengths ,place the midpoints and use simple surds to find the sides of the triangles formed. Step 2 - for the triangle within the square - MN = 4 ROOT 2 and AM is root 80 .The perpendicular bi sector of the triangle (height) is 6 root 2 . step 3 - use the sine rule a/sinA =b/sin B and find for sin x = 3/5 then just draw a right angle using sin x =3/5 and find cos x = 4/5 add and you will get 7/5 Guys can we upload pics as solutions ?

Right angle $DAB$ is expressible as $\frac{\pi}{2} = \theta + 2\arctan(4/8) \Rightarrow \theta = \frac{\pi}{2} - 2\arctan(1/2).$ Thus:

$\sin\theta + \cos\theta = \sin(\frac{\pi}{2} - 2\arctan(1/2)) + \cos(\frac{\pi}{2} - 2\arctan(1/2));$

or $\cos(2\arctan(1/2)) + \sin(2\arctan(1/2));$

or $2\cos^{2}(\arccos(2/\sqrt{5}) - 1 + 2\sin(\arcsin(1/\sqrt{5})\cos(\arccos(2/\sqrt{5});$

or $2(\frac{4}{5}) - 1 + 2(\frac{1}{\sqrt{5}})(\frac{2}{\sqrt{5}});$

or $\frac{12}{5} - 1;$

or $\boxed{\frac{7}{5}}.$