Question

trong tam giác ABC ta có

sin2A + sin2B + sin2C = 4.sinA.sinB.sinC

Answer 1

\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)

\(=2sinC.cos\left(A-B\right)+2sinC.cosC=2sinC\left[cos\left(A-B\right)+cosC\right]\)

\(=4sinC.cos\left(\frac{A+C-B}{2}\right).cos\left(\frac{A-B-C}{2}\right)\)

\(=4sinC.cos\left(\frac{\pi-2B}{2}\right).cos\left(\frac{2A-\pi}{2}\right)=4sinC.cos\left(\frac{\pi-2B}{2}\right).cos\left(\frac{\pi-2A}{2}\right)\)

\(=4sinC.cos\left(\frac{\pi}{2}-B\right).cos\left(\frac{\pi}{2}-A\right)\)

\(=4sinA.sinB.sinC\)

Answer 2

https://i.imgur.com/Iy6thtc.png