I was solving a vector’s problem :
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
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 ?
Best Answer
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.