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
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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:
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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
$$\sin\left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}$$
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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