# What mistake am I making while deriving the expansion of $\cos(\alpha + \beta)$

I was trying to derive the formula for expansion of $$\cos (\alpha + \beta)$$ by equating the ratio of lengths of two specific chords to the ratio of angles opposite to them but I'm not getting the correct results. Here's how I'm doing it :

In the above diagram,$$\angle AOB = \alpha$$, $$\angle BOC = \beta$$, $$\angle AOC = (\alpha + \beta)$$, $$a = \cos{\alpha}$$, $$b = \sin{\alpha}$$, $$x = \cos{(\alpha + \beta)}$$ and $$y = \sin {(\alpha + \beta)}$$

And as $$a$$, $$b$$, $$x$$ and $$y$$ are sines and cosines of $$\alpha$$ and $$(\alpha+\beta)$$ respectively, so : $$a^2+b^2=x^2+y^2=1$$

Now, using the distance formula for coordinate geometry, which states that the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ on the Cartesian Plane is : $$\sqrt{(x_1-x_2)^2 - (y_1-y_2)^2}$$ units, we obtain : $$AB = \sqrt{(a-1)^2+(b-0)^2}=\sqrt{a^2+1-2a+b^2}=\sqrt{(a^2+b^2)+1-2a}=\sqrt{1+1-2a}$$ $$\therefore AB = \sqrt{2-2a}$$ $$AC = \sqrt{(x-1)^2+(y-0)^2}=\sqrt{a^2+1-2x+y^2}=\sqrt{(x^2+y^2)+1-2x}=\sqrt{1+1-2x}$$ $$\therefore AC = \sqrt{2-2x}$$

Now, the ratio of lengths of $$AB$$ and $$AC$$ would be equal to the ratio of angles opposite to them, that are $$\alpha$$ and $$(\alpha + \beta)$$ respectively (this is the part where I think I might be wrong but don't see how).
So, according to me, $$\dfrac{AB}{AC}=\dfrac{\alpha}{\alpha + \beta} \implies \dfrac{AC}{AB} = \dfrac{\alpha + \beta}{\alpha} = 1 + \dfrac{\beta}{\alpha}$$ $$\implies \dfrac{\sqrt{2-2x}}{\sqrt{2-2a}} = 1 + \dfrac{\beta}{\alpha} \implies \dfrac{2-2x}{2-2a} = \Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$ $$\implies \dfrac{1-x}{1-a} = \Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2 \implies 1-x = (1-a)\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$ This leads us to the conclusion that : $$\cos(\alpha + \beta) = x = 1-(1-a)\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2 = 1-(1-\cos{\alpha})\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$

which is not true...

So, where am I going wrong in this?

Thanks!

PS : I am really grateful to those people who are giving alternative methods of derivation but what I really want to know is the mistake in my derivation. Thanks!

• You have to use the property that chords that subtend the same angle at the center are equal.
– Boy
Jun 2, 2020 at 18:01
• @Boy Could you please elaborate and tell me where I have to use that property? Thanks. Jun 2, 2020 at 18:02
• You have to add an extra point D $[cos(2\alpha+ \beta), sin(2\alpha+ \beta)]$ Now join BD and AC and they must be equal.
– Boy
Jun 2, 2020 at 18:08
• Oh that’s because the angle is not directly proportional to the length of the chord. It’s proportional to the length of the arc.
– Boy
Jun 2, 2020 at 18:13
• Yes, you're wrong at the part where you think you're wrong. Ratio of length of chords is not equal to the angle they subtend at the centre. Jun 2, 2020 at 18:22

You make a mistake exactly where you suspected. Note the ratio of the lengths of the chords is not equal to the ratio of the angles subtended by them. You could see this by applying the law of cosines on $$\triangle AOB$$ and $$\triangle AOC$$: $$AB^2=OB^2 + OA^2 -2OA\cdot OB\cos\alpha \\ =1+1-2\cos\alpha\\=2(1-\cos\alpha)\\ \implies AB=2\sin\frac \alpha 2$$ Similarly, $$AC= 2\sin\frac{\alpha+\beta}{2}$$ and $$\frac{AB}{AC} \ne \frac{\alpha}{\alpha+\beta}$$ As for your question in the comments, recall that the arc length is really $$r\theta$$ and the chord length is $$2r\sin\frac{\theta}{2}$$. Two quantities will be directly proportional iff their ratio is a constant. But $$\frac{2r\sin\frac{\theta}{2}}{r\theta}=\frac{\sin\frac{\theta}{2}}{\frac{\theta}{2}}$$ which is clearly not constant, except maybe for the case when $$\theta\approx 0$$.
• @RajdeepSindhu No, the $2$ goes down. Jun 2, 2020 at 18:59
• @RajdeepSindhu In that case, $\sin \frac{\theta}{2} \approx \frac{\theta}{2}$ and the proportionality constant is $1$, meaning the arc length equals the chord length. This also makes sense intuitively as the chord gets closer and closer to being the arc! Jun 2, 2020 at 19:08