# TrigonometryReview my solution: Trigonometry proof

#### sweatingbear

##### Member
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

Staff member
I see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram: 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
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
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.