can someone prove sin2x=2*sinx * cosx for me?Thank you

in the isosceles triangle ABC, segment AD is the height of triangle from point A,and segment BE is the height of the triangle from point Learn from point B.

because triangle ABC is a isosceles triangle ,so AB=AC ,and segment AD is the height of the triangle 

so angle ADB= Angle ADC=90 degrees

In the right triangle  ADB and right triangle ADC 

AB=AC,AD=AD, so right triangle ADB is congruent to right triangle ADC ( hypotenuse-leg)

so angle BAD= Angle CAD

If angle BAD=x degrees ,then angle BAC= 2x degrees

AD=AB*cosx  DB=AB*sinx   BE=AB*sin2x

the area of triangle = BD *AD*2/2=BD*AD=BE*AC/2=BE*AB/2

BD*AD=BE*AB/2

AB*cosx *AB*sinx=AB*sin2x*AB/2 (cancel AB*AB from both side and both side times 2)

2*cosx*sinx=sin2x

this is my method,Can someone prove it algebraicallly?

If you allow me to use the identity sin(x + y) = sin(x)cosy + cos(x)sin(y),

then sin(2x) = sin(x + x) = sin(x)cos(x) + cos(x)sin(x) = 2sin(x)cos(x).

However, this assumes that you have the original identity.

You can find the proof of the original identity at   faculty.atu.edu/mfinan/1203/Lecture19.pdf

Wow, quinn.....that's an impressive proof.....I don't think I've seen it before.....good job!!!