The Exercises for The Feynman Lectures on Physics (New Millenium Edition 2014) asks the reader to prove the following formula in exercise 2.4 (a) pg. 2-1: $$\Delta L = \frac{d_1 d_2}{L} \sin\alpha \Delta\alpha \quad(1)$$ where $d_1$ & $d_2$ correspond to the fixed lengths of a triangle, $\alpha$ is the angle between $d_1$ & $d_2$, and L is the length of the side opposite to $\alpha$ whose length varies by $\Delta L$ when $\alpha$ is varied by a small amount, $\Delta\alpha$.
I attempted to find a justification by applying the law of cosines & varying a starting angle, $\alpha_1$, by $\Delta \alpha$:
Let $\ L_1$ correspond to the length of L associated with angle $\alpha_1$ and $\ L_2$ correspond to the length of L associated with angle $\alpha_2$. Then $$ \ {L_2}^2 - {L_1}^2 = d_1^2 + d_2^2 -2d_1d_2\cos\alpha_2-(d_1^2 + d_2^2 -2d_1d_2\cos\alpha_1) \\ (L_2-L_1)(L_2+L_1) = -2d_1d_2\cos\alpha_2+2d_1d_2\cos\alpha_1 \\ L_2-L_1 = \frac{2 d_1 d_2}{L_2+L_1} (\cos\alpha_1-\cos\alpha_2) \quad (2). $$ Since $\Delta\alpha$ is small, $L_2 \approx L_1$ and thus $L_2+L_1=2L_1$, or simply $2L$. Also, $L_2-L_1$ in (2) can be rewritten as $\Delta L$ to give $$ \Delta L = \frac{d_1 d_2}{L} (\cos\alpha_1-\cos\alpha_2) \quad (3). $$ These are my questions: Can the difference of cosines in (3) be rewritten using identities and/or small angle approximations to match (1)? Is there an algebra error? Or a brand new approach is needed?
We can use the following trigonometric identity of the sum of two angles. $$\cos\left(\alpha_2\right)=\cos\left(\alpha_1 + \Delta\alpha\right) = \cos\left(\alpha_1\right)\cos\left(\Delta\alpha\right)-\sin\left(\alpha_1\right)\sin\left(\Delta\alpha\right)$$
For $\Delta\alpha\rightarrow 0$, We know that $\space\cos\left(\Delta\alpha\right)\rightarrow 1\space$ and $\space\sin\left(\Delta\alpha\right)\rightarrow \Delta\alpha$.
Therefore, the above equation can be written as, $$\cos\left(\alpha_2\right)=cos\left(\alpha_1\right)-\sin\left(\alpha_1\right)\Delta\alpha.$$
Now, substitute this value of $\cos\left(\alpha_2\right)$ into your equation (3) to get,
$$\Delta L = \frac{d_1 d_2}{L} \Big(\cos\left(\alpha_1\right)-\cos\left(\alpha_2\right)\Big)=\frac{d_1 d_2}{L} \Big(\cos\alpha_1-cos\left(\alpha_1\right)+\sin\left(\alpha_1\right)\Delta\alpha\Big)$$ $$\qquad\quad\qquad=\frac{d_1 d_2}{L} \sin\left(\alpha_1\right)\Delta\alpha.$$