Can not derive $\cos (x + y)$

Question

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

Answer 1

You nearly got it. You have to use that $DE=1·\sin(1°)$ from the ODE triangle, and also $DE=DF·\cos(10°)$ inside the DEF triangle. Thus $$ EF=DF·\sin(10°)=\frac{\sin(1°)\sin(10°)}{\cos(10°)} $$

Answer 2

$\overline{OE} = \cos(1^\circ)$.

Therefore
$\overline{OM} = \cos(10^\circ)\cos(1^\circ)$.

$\overline{ED} = \sin(1^\circ)$.

Also $\angle DEG = 10^\circ.$

Therefore,
$\overline{MK} = \overline{DG} = \sin(10^\circ)\sin(1^\circ)$.

Finally,
$\cos(11^\circ) = \overline{OK} = \overline{OM} - \overline{MK}.$

Answer 3

If you're trying to derive the formula $\ \cos(10+1)=$$\cos(10)\cos(1)-$$\sin(10)\sin(1)\ $, one way of approaching that problem is to write down $\ \cos(10),\cos(1),\sin(10), \sin(1)\ $ and $\ \cos(11)\ $ in terms of any and all ratios of line segments you can find, and then look for ones for which the numerators of one of $\ \cos(10)\ $ and $\ \cos(1)\ $ and one of $\ \sin(10)\ $ and $\ \sin(1)\ $ cancel out with the denominators of the other. Since the actual values of the angles are immaterial, I'll put $\ a=10\ $ and $\ b=1\ $. Then you have \begin{align} \cos(a+b)&=\frac{OK}{OD}\\ \cos(a)&=\frac{OK}{OF}=\frac{OM}{OE}=\frac{FN}{FE}\\ \cos(b)&=\frac{OE}{OD}\\ \sin(a)&=\frac{KF}{OF}=\frac{ME}{OE}=\frac{EN}{FE}=\frac{DG}{DE}\\ \sin(b)&=\frac{DE}{OD}\ . \end{align} Now notice that the numerator of $\ \cos(b)=\frac{OE}{OD}\ $ cancels out with the denominator of $\ \cos(a)=\frac{OM}{OE}\ $, so we get $\ \cos(a)\cos(b)=\frac{OM}{OD}\ $, and the numerator of $\ \sin(b)=\frac{DE}{OD}\ $ cancels out with the denominator of $\ \sin(a)=\frac{DG}{DE}\ $, giving $\ \sin(a)\sin(b)=\frac{DG}{OD}\ $. Therefore \begin{align} \cos(a)\cos(b)-\sin(a)\sin(b)&=\frac{OM-DG}{OD}\\ &=\frac{OM-KM}{OD}\\ &=\frac{OK}{OD}\\ &=\cos(a+b)\ . \end{align}

Answer 4

Another approach, since you already proved the sum identity for sines: $$\cos(a+b)=\sin(90^\circ- (a+b))\\ =\sin((90^\circ- a)+(-b))\\ =\sin(90^\circ- a)\cos(-b)+\cos(90^\circ- a)\sin(-b)\\ =\cos(a)\cos(b)-\sin(a)\sin(b) .$$