0

Question

An isosceles triangle has two sides of length $A$ that meet at a vertex angle of measure $2x$. Using two different methods for obtaining the area of the triangle, prove/demonstrate $\sin 2x=2\sin x\cos x$

What I have so far

I first get the area which is equal to the following:

$$\frac{1}{2}A^2\sin{2x}$$

Then I notice that $\sin{2x}$ is in the area of the triangle and that is similar to what we want to prove.

But I do not know how to go from there.

Mythomorphic
  • 6,008
John Rawls
  • 2,685

1 Answers1

1

Let the triangle be $PQR$ with angle $P=2x$ and $PQ=QR$. Now note that angle $Q=R=\frac{\pi}{2}-x$. Let PS be the perpendicular from P to QR, now as $PQ=A$, we have $PS=Asin(\frac{\pi}{2}-x)$ and $QS=SR=Acos(\frac{\pi}{2}-x)$, which gives us area as $\frac{1}{2}Asin(\frac{\pi}{2}-x).2Acos(\frac{\pi}{2}-x)=\frac{1}{2}A^2(2\sin{x}\cos{x})$, which is what we want.

user1131274
  • 1,678