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I was solving a vector’s problem :

picture 1 The question was:

Find the length of $\vec{a} + \vec{b}$ vector in terms of $a,\,b,\,\alpha,\,\beta\,.$

Well, we know $\vec{a}$ equals to $\,(a\cos\beta,a\sin\beta)\,$ and $\vec{b}$ equals to $\,(b\cos\alpha,b\sin\alpha)\,.$

$\vec{a}+\vec{b}=(a\cos\beta+b\cos\alpha,a\sin\beta+b\sin \alpha)\,.$

So, the length of $\vec{a}+\vec{b}$ equals to :

$\sqrt{(a\cos\beta+b\cos\alpha)^2+(a\sin\beta+b\sin\alpha)^2}\,.$

Another way of solving this problem is drawing Cartesian coordinate system like this

picture 2 or using cosine rule which gives you:

$\left|\vec{a}+\vec{b}\right|=\sqrt{a^2+b^2+2\!\cdot\!a\!\cdot\!b\!\cdot\!\cos(\beta-\alpha)}$

As a result,

$\sqrt{(a\cos\beta+b\cos\alpha)^2+(a\sin\beta+b\sin\alpha)^2}=$

$=\sqrt{a^2+b^2+2\!\cdot\!a\!\cdot\!b\!\cdot\cos(\beta-\alpha)}\,.$

Now I'm trying to prove $\,\cos(\beta-\alpha)=\sin\alpha\sin\beta+\cos\alpha\cos\beta\,$ using this equality.

I solved this equation for $\,\cos(\beta-\alpha)\,$:

$\cos(\beta-\alpha)=\dfrac{a^2\cos^2\!\beta+2ab\cos\beta\cos\alpha+b^2\cos^2\!\alpha+a^2\sin^2\!\beta+2ab\sin\beta\sin\alpha+b^2\sin^2\!\alpha-b^2-a^2}{2ab}$

The problem is that $a$ and $b$ still exist in the equality. I don't get why this happens.
Does replacing $\,\sin^2\!\alpha\,$ and $\,\sin^2\!\beta\,$ with $\,1-\cos^2\!\alpha\,$ and $\,1-\cos^2\!\beta\,$ help to solve this problem ?

Cameron Williams
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forough m
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  • If I understand the question, you are to find the length of $\vec a + \vec b$ in terms of the lengths of $\vec a$ and $\vec b$ and the angles $\alpha$ and $\beta$. The answer should depend on the lengths of $\vec a$ and $\vec b$; after all, if you double both vectors the length of $\vec a + \vec b$ will double too. – Matthew Leingang Jul 26 '23 at 15:46
  • No. I'm trying to prove an equality with solving a question with two methods, as title says. – forough m Jul 26 '23 at 15:52
  • @foroughm, could read my answer? Tell me if you like it. – Angelo Jul 26 '23 at 16:42
  • Use \vec to make vector arrows above variables. – Cameron Williams Jul 27 '23 at 02:14

3 Answers3

2

You are on the right track. The strategy that you suggested will indeed help to solve the problem. Simplify the numerator: $$a^2\cos^2\beta + a^2\sin^2\beta=a^2(\cos^2+\sin^2\beta)=a^2.$$ Similarly note that $$b^2\cos^2\alpha + b^2\sin^2\alpha=b^2.$$ After simplification, the $a$ and $b$ will drop out and you'll have your identity.

grand_chat
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From the equality you wrote in your post

$\sqrt{(a\cos\beta+b\cos\alpha)^2+(a\sin\beta+b\sin\alpha)^2}=\\=\sqrt{a^2+b^2+2ab\cos(\beta-\alpha)}$

you can get

$\sqrt{\color{blue}{a^2\!\cos^2\!\beta}\!+\!\color{red}{b^2\!\cos^2\!\alpha}\!+\!2ab\cos\alpha\cos\beta\!+\!\color{blue}{a^2\!\sin^2\!\beta}\!+\!\color{red}{b^2\!\sin^2\!\alpha}\!+\!2ab\sin\alpha\sin\beta}=\sqrt{a^2+b^2+2ab\cos(\beta-\alpha)}\;\;,\\\\$

$\sqrt{\color{blue}{a^2}\!+\!\color{red}{b^2}\!+\!2ab(\cos\alpha\cos\beta\!+\!\sin\alpha\sin\beta)}=\sqrt{a^2\!+\!b^2\!+\!2ab\cos(\beta\!-\!\alpha)}\;,$

$\require{cancel}\color{#8888FF}{\cancel{\color{black}{a^2}}}\!\!\!+\!\!\color{#FF8888}{\cancel{\color{black}{b^2}}}\!\!\!+\!2ab(\cos\alpha\cos\beta\!+\!\sin\alpha\sin\beta)=\!\!\color{#8888FF}{\cancel{\color{black}{a^2}}}\!\!\!+\!\!\color{#FF8888}{\cancel{\color{black}{b^2}}}\!\!\!+\!2ab\cos(\beta\!-\!\alpha)\;,$

$\color{#8888FF}{\cancel{\color{black}{2ab}}}(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=\color{#8888FF}{\cancel{\color{black}{2ab}}}\cos(\beta-\alpha)\;,$

$\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\beta-\alpha)\;,$

that is

$\cos(\beta-\alpha)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\;.$

Angelo
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USING DOT PRODUCT

Using definition of dot product, we have $$ \begin{aligned} |\vec{a}||\vec{b}| \cos (\beta-\alpha)=&\vec{a} \cdot \vec{b}\\ = & \left[|\vec{a}|(\vec{i} \cos \beta+\vec{j} \sin \beta)\right]\cdot \left[ |\vec{b}|(\vec{i} \cos \alpha+\vec{j} \sin \alpha) \right] \\ = & |\vec{a}| |\vec{b}|(\vec{i} \cos \beta+\vec{j} \sin \beta)\cdot(\vec{i} \cos \alpha+\vec{j} \sin \alpha) \\ = & |\vec{a}||\vec{b}|(\cos \alpha \cos \beta+\sin \alpha \sin \beta )\\ \therefore \cos (\beta-\alpha)= & \cos \alpha \cos \beta+\sin \alpha \sin \beta \end{aligned} $$

Lai
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