Question

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

Answer 1

Apply the trigonometric identities:

`cos\alpha-cos\beta=-2sin(\alpha+\beta)/(2)*sin(\alpha-\beta)/(2)`

`sin\alpha+sin\beta=2sin(\alpha+\beta)/(2)*cos(\alpha-\beta)/(2)`

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

`(sin\alpha)/(cos\alpha)=tan\alpha`

Explanation:

By definition you have:

`cos\alpha-cos\beta=-2sin(\alpha+\beta)/(2)*sin(\alpha-\beta)/(2)`

`sin\alpha+sin\beta=2sin(\alpha+\beta)/(2)*cos(\alpha-\beta)/(2)`

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

`(sin\alpha)/(cos\alpha)=tan\alpha`

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

`=((-2sin(x+3x)/(2)*sin(x-3x)/(2)))/((2sin(x+3x)/(2)*cos(x-3x)/(2)))`

Simplify:

`=((-2sin(4x)/(2)*sin(-2x)/(2)))/((2sin(4x)/(2)*cos(-2x)/(2)))`

(You can cancel out the like terms)

`=(-(-sin(2x)/(2)))/(cos(-2x)/(2))\\=(sinx)/(cosx)\\\\=tanx`