Question

довести тождество sinα+sinβ+sin(α-β)=4 sin/2*cos/2*cos α-β/2

Answer 1

$sin alpha+sinbeta +sin(alpha-beta)=sinalpha+2sinfrac{beta +alpha-beta}{2}cosfrac{beta -alpha+beta}{2}=\ \ =2sinfrac{alpha}{2}cosfrac{alpha}{2}+2sinfrac{alpha}{2}cosfrac{2beta-alpha}{2}=2sinfrac{alpha}{2}(cosfrac{alpha}{2}+cos(beta -frac{alpha}{2}))=\ \ =2sinfrac{alpha}{2}cdot 2cosfrac{frac{alpha}{2}+beta-frac{alpha}{2}}{2} cosfrac{frac{alpha}{2}-beta+frac{alpha}{2}}{2}=4sinfrac{alpha}{2}cosfrac{beta}{2}cosfrac{alpha-beta}{2}$

Answer 2

sina+sinb+sin(a-b)=4*sina/2*cosb/2*cos(a-b)/2
sina+(sinb+sin(a-b))=2*sina/2*cosa/2+

2sin(b+a-b)/2*cos(b-a+b)/2=

2sina/2*(cosa/2+cos(b-a/2)=

2*sina/2*(2*cos(a/2+b-a/2)/2*cos(a/2-b+a/2)/2)
=4*sina/2*cosa/2*cos(a-b)/2