Proof of the trig identities for half-angles

AI Thread Summary
The discussion revolves around verifying the half-angle identity for sine using a right-angled triangle with specific side lengths. The initial calculation of sin(A/2) yielded an incorrect angle, prompting the user to suspect a mistake in applying the identity. They noted the importance of including the square root in the formula for sin(A/2) to ensure accurate results. The correct identity should be expressed as sin(A/2) = √((s-b)(s-c)/(bc)). The user plans to correct their LaTeX formatting to clarify the calculations.
chwala
Gold Member
Messages
2,827
Reaction score
415
Homework Statement
see attached
Relevant Equations
Trigonometry
1649053062522.png


I was just checking this out the sin##\frac {A}{2}## property, in doing so i picked a Right-Angled triangle, say ##ABC##, with ##AB=5cm##, ##BC=4cm## and ##CA= 3cm##. From this i have,
##s=6cm## now substituting this into the formula,
##sin\frac {A}{2}##= ##\frac {1×3}{5×3}##=##\frac {3}{15}##=##0.2##
giving us angle ##A=23.06^0## which does not look correct to me because,

angle ##A=53.13^0## ...using trigonometry directly...##sin{A}##= ##\frac {4}{5} =0.8##... i suspect a mix up in the indicated property...or i may have made a mistake.
i will amend my latex later using phone...aaaaaaaaargh these guys are missing the square root sign! I had to check that from google.
 
Last edited:
Physics news on Phys.org
The left hand sides should be squared or the right hand sides square-rooted
ie
$$\sin^2\left(\frac{A}{2}\right)=\frac{(s-b)(s-c)}{b c}$$
or
$$\sin\left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}$$
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top