I proved the cosine subtraction formula, but I don't know why it is correct

[Four Muffins]

Hello again. I got stuck for a while on proving $cos(\alpha-\beta) = cos \alpha cos\beta+sin\alpha sin\beta$. I did solve it, but only by doing something that seems obviously wrong.

With the Cosine Law, $c^2=a^2+b^2-2ab*cos(\theta)$, I normally set $c$ to be the hypotenuse. The triangle in the question has the shortest side of an isosceles triangle as $c$, so I set one of the equal sides to be the pseudo-hypotenuse $b$ and solved for $c$, getting $c^2=b^2-a^2+2ab*cos(\alpha -\beta)$, which did not let me prove the formula.

Does the Cosine Law allow for solving any one unknown side without algebraic manipulation, or am I missing something about what $c$ is in this triangle?



Correct, then incorrect working.

Spoiler

 

[skeeter]

The cosine law can be used to solve for the length of an unknown side of any triangle as long as one has the angle opposite to that side and the lengths of the two sides adjacent to that unknown side.

From the diagram’s hint …

$c^2 = 1^2 + 1^2 - 2(1)(1) \cdot \cos(\alpha - \beta)$

$c^2 = (\cos{\alpha}-\cos{\beta})^2 + (\sin{\alpha}-\sin{\beta})^2$

 

[blamocur]

I did solve it, but only by doing something that seems obviously wrong.

Why do you think your solution is wrong?

 

[Four Muffins]

The cosine law can be used to solve for the length of an unknown side of any triangle as long as one has the angle opposite to that side and the lengths of the two sides adjacent to that unknown side.

From the diagram’s hint …

$c^2 = 1^2 + 1^2 - 2(1)(1) \cdot \cos(\alpha - \beta)$

$c^2 = (\cos{\alpha}-\cos{\beta})^2 + (\sin{\alpha}-\sin{\beta})^2$

Thank you Skeeter, that makes sense. I was under the impression that the Cosine Law had to follow the same rules as Pythagoras' Theorum.

Why do you think your solution is wrong?

I didn't think it was wrong, I just didn't understand why it was right. I did a thing I thought was incorrect to get the right answer, and I wasn't sure what I was misunderstanding.