Question

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

Answer 1

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

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