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Trigonometry Determine the value of AC/BC.

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anemone

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Feb 14, 2012
3,753
In the triangle ABC, the following equality holds:

$\displaystyle \sin^{23}\left(\frac{A}{2}\right) \cos^{48} \left(\frac{B}{2}\right)=\sin^{23} \left(\frac {B}{2}\right) \cos^{48} \left(\frac {A}{2}\right)$

Determine the value of $\dfrac {AC}{BC}$.

Hi all, the huge values of the exponents make me cringe to try to solve the problem. I want to ask if this is obvious that one of the possibilities is that the given equality is true iff the measures of both the angles A and B are equal. Is this true? That is, we're dealing with an isosceles triangle where $\dfrac {AC}{BC}=1$?
 

Klaas van Aarsen

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Mar 5, 2012
8,856
In the triangle ABC, the following equality holds:

$\displaystyle \sin^{23}\left(\frac{A}{2}\right) \cos^{48} \left(\frac{B}{2}\right)=\sin^{23} \left(\frac {B}{2}\right) \cos^{48} \left(\frac {A}{2}\right)$

Determine the value of $\dfrac {AC}{BC}$.

Hi all, the huge values of the exponents make me cringe to try to solve the problem. I want to ask if this is obvious that one of the possibilities is that the given equality is true iff the measures of both the angles A and B are equal. Is this true? That is, we're dealing with an isosceles triangle where $\dfrac {AC}{BC}=1$?
Hey anemone! (Malthe)

Let's rewrite that equation a bit.
$$\frac{\cos^{48} \left(\frac {A}{2}\right)}{\sin^{23}\left(\frac{A}{2}\right)}=\frac{ \cos^{48} \left(\frac{B}{2}\right)}{\sin^{23} \left(\frac {B}{2}\right)}$$
(Note that the denominator will not be zero if the angles are between $0$ and $\pi$.)

The left hand side and right hand side both represent the same function.
If that function is invertible (strictly monotonous) on its domain, that means that A and B have to be the same.
When we analyze that function (or graph it with Wolfram) we see indeed that it is monotonous on $(0,\pi)$ which is the range of the angle in a triangle.

So yes, it is an isosceles triangle with AC=BC.
 
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anemone

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Feb 14, 2012
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Hey anemone! (Malthe)

Let's rewrite that equation a bit.
$$\frac{\cos^{48} \left(\frac {A}{2}\right)}{\sin^{23}\left(\frac{A}{2}\right)}=\frac{ \cos^{48} \left(\frac{B}{2}\right)}{\sin^{23} \left(\frac {B}{2}\right)}$$
(Note that the denominator will not be zero if the angles are between $0$ and $\pi$.)

The left hand side and right hand side both represent the same function.
If that function is invertible (strictly monotonous) on its domain, that means that A and B have to be the same.
When we analyze that function (or graph it with Wolfram) we see indeed that it is monotonous on $(0,\pi)$ which is the range of the angle in a triangle.

So yes, it is an isosceles triangle with AC=BC.
Thanks, I like Serena...I understand what you say and I appreciate your response to my post! :)