I am trying to calculate $\cos 11^\circ = \cos(10^\circ + 1^\circ)$, i.e., trying to derive the formula $\cos(x + y)$ using the following drawing:
Note that I was able to derive the formula $\sin (x + y)$ by using this drawing but I am stuck with $\cos 11^\circ$.
My steps:
$OD = 1^\circ$
$OK = OD\cdot \cos 11^\circ = 1 \cdot \cos 11^\circ = \cos 11^\circ$
But also:
$OK = OF \cdot \cos 10^\circ$
Now we have to find $OF$
$OF = OE – EF = (OD \cdot \cos 1^\circ) – EF = (1 \cdot \cos 1^\circ) – EF = \cos 1^\circ – EF$
So now we have to find $EF$ which is where I am stuck. If I do e.g.
$EF =DF\cdot \sin 10^\circ$ (I think the angle $FDE$ is $10$ degrees because the angles $EFD = KFO$ and both are right angled triangles) leads nowhere because now I have to find $DF$ and it seems impossible based on what I know.
Trying to switch to the triangle $FEN$ (I think that is also the similar to $OME$) also leads to nowhere.
What am I missing here? Please note I don't recall all the possible ways to figure out equal angles so if you mention an approach please be explicit on how that is found.
Also I am interested in the derivation based on this drawing, not a different approach
Best Answer
Another approach, since you already proved the sum identity for sines: $$\cos(a+b)=\sin(90^\circ- (a+b))\\ =\sin((90^\circ- a)+(-b))\\ =\sin(90^\circ- a)\cos(-b)+\cos(90^\circ- a)\sin(-b)\\ =\cos(a)\cos(b)-\sin(a)\sin(b) .$$