How can I show that when r >> d, R1 - R2 is approximately equal to d cos theta? This was an intermediate step in my physics book when they derived some formula, and it just drives me crazy that I can't figure this out!

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- Mar 15th 2013, 02:10 PMgralla55Approximating difference between sides of triangle
How can I show that when r >> d, R1 - R2 is approximately equal to d cos theta? This was an intermediate step in my physics book when they derived some formula, and it just drives me crazy that I can't figure this out!

- Mar 15th 2013, 02:45 PMShakarriRe: Approximating difference between sides of triangle
Using the cosine rule for the triangle on the right hand side

$\displaystyle R^2_2=r^2+\frac{d^2}{4}-rd\cdot cos\theta $

And using the cosine rule on the triangle on the left side

$\displaystyle R^2_1=r^2+\frac{d^2}{4}-rd\cdot cos(180-\theta )$

$\displaystyle R^2_1=r^2+\frac{d^2}{4}-rd\cdot (cos(180)cos(\theta)+sin(180sin(\theta))$

cos180= -1, sin180=0

$\displaystyle R^2_1=r^2+\frac{d^2}{4}+rd\cdot cos\theta $

$\displaystyle R^2_2-R^2_1= r^2+\frac{d^2}{4}+rd\cdot cos\theta -(r^2+\frac{d^2}{4}-rd\cdot cos\theta)$

$\displaystyle R^2_2-R^2_1= 2rd\cdot cos\theta$

$\displaystyle (R_2-R_1)(R_2+R_1)= 2rd\cdot cos\theta$

As r becomes much larger than d R1 and R2 become nearly equal to r

$\displaystyle R_1+R_2 \approx 2r$

$\displaystyle (R_2-R_1)\frac{(R_2+R_1)}{2r}= d\cdot cos\theta$

$\displaystyle R_2-R_1= d\cdot cos\theta$