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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 :

Figure - 1

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 : $$AB = \sqrt{(a-1)^2+(b-0)^2}=\sqrt{a^2+1-2a+b^2}=\sqrt{(a^2+b^2)+1-2a}=\sqrt{1+1-2a}$$ $$\therefore AB = \sqrt{2-2a}$$ $$AC = \sqrt{(x-1)^2+(y-0)^2}=\sqrt{a^2+1-2x+y^2}=\sqrt{(x^2+y^2)+1-2x}=\sqrt{1+1-2x}$$ $$\therefore AC = \sqrt{2-2x}$$

Figure - 2

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!
Rajdeep Sindhu
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  • You have to use the property that chords that subtend the same angle at the center are equal. – Boy Jun 02 '20 at 18:01
  • @Boy Could you please elaborate and tell me where I have to use that property? Thanks. – Rajdeep Sindhu Jun 02 '20 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 02 '20 at 18:08
  • I'm really grateful to you for telling me the alternative method to do this but what I wanted to know was the mistake in my derivation, and that's why I put that title. – Rajdeep Sindhu Jun 02 '20 at 18:10
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    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 02 '20 at 18:13
  • I think that you might just have opened my mind, I totally forgot that the length of the chord is $2r \sin {\dfrac{\alpha}{2}}$ where $\alpha$ is the angle between the initial and terminal sides of the angle an $r$ is the radius. I will see if I can come up with something. Thanks! Also, isn't the length of the chord directly proportional to the length of the arc too? – Rajdeep Sindhu Jun 02 '20 at 18:17
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    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. – UmbQbify Jun 02 '20 at 18:22

1 Answers1

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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 $\theta\approx 0$.

Vishu
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  • Thanks! I think that I know a way to find the length of the chord without the law of cosines, which is a topic that I have not yet covered. It can be done by splitting the angle into two parts and then finding the length of half of the chord which will be $r \sin {\theta}$, so length of the chord will be : $2r \sin {\theta}$. This is very simple compared to the law of cosines, I think. Shall I elaborate? – Rajdeep Sindhu Jun 02 '20 at 18:43
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    @RajdeepSindhu Yes, that’s a good method as well. – Vishu Jun 02 '20 at 18:46
  • Shouldn't it be $\dfrac{2 \sin{\dfrac{\theta}{2}}}{\dfrac{\theta}{2}}$ at the end? – Rajdeep Sindhu Jun 02 '20 at 18:58
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    @RajdeepSindhu No, the $2$ goes down. – Vishu Jun 02 '20 at 18:59
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    @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! – Vishu Jun 02 '20 at 19:08
  • Got it, that was silly. And also, you say that when $ \theta \approx 0$, then the ratio may be a constant because $\sin{ \alpha} \approx \alpha$ for $\alpha \approx 0$ and in that case, $k \approx 1$, right? – Rajdeep Sindhu Jun 02 '20 at 19:09
  • Thanks a lot! I just have one more question, a general doubt though. You see, my trigonometry curriculum is like this : first, they introduce trigonometric ratios and identities in triangles and then it's applications, then, they extend the definition of trigonometric ratios to functions using the unit circle and remove the $0\leq x \leq \frac{\pi}{2}$ constraint and derive formulae for values of trigonometric functions at allied angles and give some introduction about the periodic nature of trigonometric functions. Then, they start with graphs of trigonometric functions. – Rajdeep Sindhu Jun 02 '20 at 19:15
  • Now, the graphs chapter basically teaches us how to make graphs of $f(x) = m.f_x(ax+b)$, where $f_x$ is some trigonometric function. Now, what I want to ask is : What are the applications of this chapter (the graphs one)? Is it necessary to do it? Or is it okay to skip it? I'm sorry if I'm being annoying. – Rajdeep Sindhu Jun 02 '20 at 19:17
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    I would say learning about graphs is more important than learning about the identities and formulas, it’s the reason they say ‘a picture is worth a thousand words’. You should definitely give your time to it. – Vishu Jun 02 '20 at 19:20
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    Thanks a lot for answering my 'annoying' questions :) – Rajdeep Sindhu Jun 02 '20 at 19:20
  • I did a complete (almost) derivation of the expansion of $\cos (\alpha + \beta)$. I'd be glad if you can check it out. Thanks! Here's the link : https://math.stackexchange.com/questions/3704308/is-this-derivation-i-did-for-the-expansion-of-cos-alpha-beta-correct-and – Rajdeep Sindhu Jun 03 '20 at 18:40