# Formula obtained by using Trignometric approximation for a triangle with a very small side

I am reading a paper on the force between hooft polyakov monopoles, but I am completely baffled by one of the 'elementary trignometric' equation they have got using an approximation. Consider a triangle say triangle ABC. The author says that when A is very close to B, i.e. when $\cos{C}\approx 1$, we get $\cos{C}-1=-\frac{1}{2a^2}c^{2}\sin^2{B}$. Please can anyone tell me, how has this been done. I am extremely sorry if this is a silly question.

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EDIT Hmm, the question changed while I was answering. See the 2. revision.

When $A$ is close to $B$, we get a small $\theta_2$. So a truncated Taylor series would look like $$\cos{\theta_2}-1=-\frac{1}{2}\theta_2^2\; .$$ Now $\theta_2\approx \tan \theta_2=\frac{r_1\cos \theta_1}{s}$, where $r_1\cos \theta_1$ is the projection of $r_1$ on the opposite side in a right triangle $AB'C$, with $B'$ being in $\overline{BC}$, such that $AB' \perp BC$.

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Hey, thanks. This solves my question. Actually, the project of $r_1$ on the right angled triangle formed by dropping the perpendicular is $r_1 \sin{\theta_1}$. SO I get the required answer. Didnt think of using taylor expansion. How could I oversee it? – user23238 Jul 17 '12 at 11:06
@ram: anytime somebody's talking about an "approximation for small (something)", it will likely involve the use of a Taylor/Maclaurin expansion. – J. M. Jul 17 '12 at 11:16

From the Law of Sines we get $c\sin B=b\sin C$, so the mystery conclusion can also be written $$\cos C -1 = -\frac12 \left(\frac ba\right)^2 \sin^2 C$$ Note it is true in general that $\cos \theta -1 \approx -\frac12 \sin^2\theta$ for small angles $\theta$ (this is a matter of comparing the Taylor approximations of the two sides).

So the important premise is not just that angle $C$ is small, but that points $A$ and $B$ are close to each other and therefore $\frac ba\approx 1$.

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