Trig Homework Help??

Question

I've spent about an hour and a half on this problem and i can't figure it out! Please help!


show that:

1/2*b*c*sinA= (a^2sin(2B)+b^2sin(2A))/4

were are suppose to show that the area of a triangle which is the first part equals the second part!

Answer 1

I will preface this by saying it's been a while since I've done trig. That said, I think here it is. Notice, this is shorthand, so some steps might be missing. Not the major ones though.

Start by multiplying your original equation by 4, both sides:
2bc sinA = a2 sin2B + b2 sin2B
2bc sinA = 2a2 sinBcosB + 2b2 sinAcosA :converting double angle
bc sinA = a2 sinBcosB + b2 sinAcosA: divided by 2
bc sinA = a2 sinBsinA + b2 sinAsinB : cosB=sinA in right triangle
bc sinA = (a2 + b2) (sinBsinA)
bc sinA = c2 sinBsinA : pythagorean theorem
bc = c2 sinB : sinB = b/c in a right triangle
bc = c2 b/c
bc = bc

Major Assumption: we are dealing with right-angled triangle. Check the reference below, relationships in right triangles - if this looks familiar, we are on the right path.