Question

(cosx-cos3x)/(sinx+sin3x)=tanx. How do I prove this?

Answer 1

Answer:

Apply the trigonometric identities:

$cos\alpha-cos\beta=-2sin\frac{\alpha+\beta}{2}*sin\frac{\alpha-\beta}{2}$

$sin\alpha+sin\beta=2sin\frac{\alpha+\beta}{2}*cos\frac{\alpha-\beta}{2}$

$sin(-\alpha)=-sin(\alpha)\\cos(-\alpha)=cos(\alpha)$

$\frac{sin\alpha}{cos\alpha}=tan\alpha$

Step-by-step explanation:

By definition you have:

$cos\alpha-cos\beta=-2sin\frac{\alpha+\beta}{2}*sin\frac{\alpha-\beta}{2}$

$sin\alpha+sin\beta=2sin\frac{\alpha+\beta}{2}*cos\frac{\alpha-\beta}{2}$

$sin(-\alpha)=-sin(\alpha)\\cos(-\alpha)=cos(\alpha)$

$\frac{sin\alpha}{cos\alpha}=tan\alpha$

 

 Keeping the above on mind, you can rewrite the expression as following:

$=\frac{(-2sin\frac{x+3x}{2}*sin\frac{x-3x}{2})}{(2sin\frac{x+3x}{2}*cos\frac{x-3x}{2})}$

Simplify:

 $=\frac{(-2sin\frac{4x}{2}*sin\frac{-2x}{2})}{(2sin\frac{4x}{2}*cos\frac{-2x}{2})}$

(You can cancel out the like terms)

$=\frac{-(-sin\frac{2x}{2})}{cos\frac{-2x}{2}}\\=\frac{sinx}{cosx}\\\\=tanx$