Simplifying Summation of Tan Functions

[dimensionless]

Find

$$\sum_{1}^{n} \tan(a f_{n} )$$

$$\cos x = 1 - {x^2 \over 2!} + {x^4 \over 4!} - \cdots$$
$$\sin\left( x \right) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
$$\tan(x) = \sin(x) / \cos(x)$$

There might be equations for the summation of a series of sine functions or an equation for the summation of a series of consine functions. I don't know what they are. I have no idea how to go about deriving this.

 

[arildno]

What are you assumed to find?

 

[HallsofIvy]

What is the f_n(x)? Everything depends on that doesn't it?

 

[dimensionless]

Sorry. That wasn't very clear.

Find t
$$B = \sum_{1}^{n} \tan( f_{n} t )$$


Right now I'm just trying to get rid of the tan function. Getting rid of the summation sign might help.

I wrote down $$f_{n}$$ incorrectly.
$$f_{n} = a n^{2}+c b_{n}^{2}$$

where $$b_{n}$$ is an arbitrary constant