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Trigonometry Review my solution: Trigonometry proof

sweatingbear

Member
May 3, 2013
91
For a triangle with sides \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\) and angle \(\displaystyle C\), where the angle \(\displaystyle C\) subtends the side \(\displaystyle c\), show that

\(\displaystyle c \geqslant (a+b) \sin \left( \frac C2 \right)\)
_______

Let us stipulate \(\displaystyle 0^\circ < C < 180^\circ\) and of course \(\displaystyle a,b,c > 0\). Consequently, \(\displaystyle 0^\circ < \frac C2 \leqslant 90^\circ \implies 0 < \sin^2 \left( \frac C2 \right) \leqslant 1\) (we include \(\displaystyle 90^\circ\) for the angle \(\displaystyle \frac C2\) in order to account for right triangles).

Law of cosines yield

\(\displaystyle c^2 = a^2 + b^2 - 2ab \cos ( C )\)

Using \(\displaystyle \cos (C) \equiv 1 - 2\sin^2 \left( \frac C2 \right)\), we can write

\(\displaystyle c^2 = a^2 + b^2 - 2ab + 4ab\sin^2 \left( \frac C2 \right)\)

Now since \(\displaystyle 0 < \sin^2 \left( \frac C2 \right) \leqslant 1\), \(\displaystyle a^2 + b^2 - 2ab + 4ab\sin^2 \left( \frac C2 \right)\) will either have to equal \(\displaystyle a^2 + b^2 - 2ab + 4ab\) or be greater than it. Thus

\(\displaystyle c^2 \geqslant a^2 + b^2 - 2ab + 4ab = a^2 + 2ab + b^2 = (a+b)^2 \)

A similar argument can be made for \(\displaystyle (a+b)^2\) versus \(\displaystyle (a+b)^2 \sin^2 \left( \frac C2 \right)\): \(\displaystyle (a+b)^2\) will either have to equal or be greater than the latter expression, due to the values \(\displaystyle \sin^2 \left( \frac C2 \right)\) can assume. Therefore

\(\displaystyle c^2 \geqslant (a+b)^2 \geqslant (a+b)^2 \sin^2 \left( \frac C2 \right) \)

We can finally conclude

\(\displaystyle c \geqslant (a+b) \sin \left( \frac C2 \right)\)

Thoughts?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram:

sweatingbear.jpg

I would postulate concerning the angle bisector $m$ of $\angle C$ and the altitude $h$, we must have:

(1) \(\displaystyle m\ge h\)

That is, the shortest distance between a point and a line is the perpendicular distance.

Now, the area $T$ of the triangle may be written in these ways:

\(\displaystyle T=\frac{1}{2}ch=\frac{1}{2}m(a+b)\sin\left(\frac{C}{2} \right)\)

Note: I have made use of the formulas (for a general triangle):

\(\displaystyle T=\frac{1}{2}bh\)

\(\displaystyle T=\frac{1}{2}ab\sin(C)\)

Thus, we have:

\(\displaystyle ch=m(a+b)\sin\left(\frac{C}{2} \right)\)

And from (1), we therefore conclude:

\(\displaystyle c\ge (a+b)\sin\left(\frac{C}{2} \right)\)
 

sweatingbear

Member
May 3, 2013
91
I see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram:

View attachment 1085

I would postulate concerning the angle bisector $m$ of $\angle C$ and the altitude $h$, we must have:

(1) \(\displaystyle m\ge h\)

That is, the shortest distance between a point and a line is the perpendicular distance.

Now, the area $T$ of the triangle may be written in these ways:

\(\displaystyle T=\frac{1}{2}ch=\frac{1}{2}m(a+b)\sin\left(\frac{C}{2} \right)\)

Note: I have made use of the formulas (for a general triangle):

\(\displaystyle T=\frac{1}{2}bh\)

\(\displaystyle T=\frac{1}{2}ab\sin(C)\)

Thus, we have:

\(\displaystyle ch=m(a+b)\sin\left(\frac{C}{2} \right)\)

And from (1), we therefore conclude:

\(\displaystyle c\ge (a+b)\sin\left(\frac{C}{2} \right)\)
Thank for the feedback and the excellent alternative solution!
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
A good rule of thumb in mathematics I live by is this:

If something is true, you should be able to provide two proofs, and one of them should have a picture. This is a perfect example of what I mean.