Answer:
Apply the trigonometric identities:
[tex]cos\alpha-cos\beta=-2sin\frac{\alpha+\beta}{2}*sin\frac{\alpha-\beta}{2}[/tex]
[tex]sin\alpha+sin\beta=2sin\frac{\alpha+\beta}{2}*cos\frac{\alpha-\beta}{2}[/tex]
[tex]sin(-\alpha)=-sin(\alpha)\\cos(-\alpha)=cos(\alpha)[/tex]
[tex]\frac{sin\alpha}{cos\alpha}=tan\alpha[/tex]
Step-by-step explanation:
By definition you have:
[tex]cos\alpha-cos\beta=-2sin\frac{\alpha+\beta}{2}*sin\frac{\alpha-\beta}{2}[/tex]
[tex]sin\alpha+sin\beta=2sin\frac{\alpha+\beta}{2}*cos\frac{\alpha-\beta}{2}[/tex]
[tex]sin(-\alpha)=-sin(\alpha)\\cos(-\alpha)=cos(\alpha)[/tex]
[tex]\frac{sin\alpha}{cos\alpha}=tan\alpha[/tex]
Keeping the above on mind, you can rewrite the expression as following:
[tex]=\frac{(-2sin\frac{x+3x}{2}*sin\frac{x-3x}{2})}{(2sin\frac{x+3x}{2}*cos\frac{x-3x}{2})}[/tex]
Simplify:
[tex]=\frac{(-2sin\frac{4x}{2}*sin\frac{-2x}{2})}{(2sin\frac{4x}{2}*cos\frac{-2x}{2})}[/tex]
(You can cancel out the like terms)
[tex]=\frac{-(-sin\frac{2x}{2})}{cos\frac{-2x}{2}}\\=\frac{sinx}{cosx}\\\\=tanx[/tex]
