Trigonometric Identity Proof: v cosδ = V(1-cosβ) + u cos(α-β)

[SP90]

Homework Statement

Given the following two triangles:

Show that $v \cos{\delta} = V(1-\cos{\beta})+u\cos(\alpha - \beta)$

The Attempt at a Solution

Using the cosine law I've got:

$v^{2}=x^{2}+V^{2}-2xV\cos{(\theta + \beta)}$
and $u^{2}=x^{2}+V^{2}-2xV\cos{(\theta)}$

I figured maybe using the rule for $\cos{(A+B)}=\cos{(A)}\cos{(B})-\sin{(A)}\sin{(B)}$ would work, but that leads to introducing sines, which seems like it would get messy, especially since there are no sines in the solution.

I'm not sure how to proceed here. I'm confused where the 1 would come from unless going through $\cos^{2}{x}+\sin^{2}{x}=1$, but that makes no sense as the other terms aren't squared.

Any help or direction on this would be appreciated.

 

[SP90]

This is now solved. It can be solved by extending the left triangle to a right-angled triangle, with a base of V+k (hence where the single V comes from). Then $\cos{\delta}=V+k$

k is given by $x \cos{(\pi-(\theta + \beta))}=-x \cos({\theta + \beta})$.

Using the cosine sum rule I stated above, you can split this out. Then using the sine rule $u \sin{\alpha}=x \sin{\theta}$ you get $k=u\sin{\alpha}\sin{\beta}-x\cos{\theta}\cos{\beta}$

You then use $\sin{\alpha}{\beta}=cos({\alpha - \beta})-cos{\alpha} \cos{\beta}$

This leaves you with $v \cos{\delta} = V + u \cos({\alpha - \beta}) - \cos{\beta}(u\cos{\alpha}+x\cos{\theta})$

The right hand bracket defines V (draw a diagram) and this solves the problem.