Evaluate f(91π/2002) + f(92π/2002) + .... + (910π/2002)

[anemone]

If $$f(x)=\frac{1}{1+\tan^3 x}$$, evaluate $$f\left(\frac{91\pi}{2002} \right)+f\left(\frac{92\pi}{2002} \right)+\cdots+f\left(\frac{910\pi}{2002} \right)$$.

 

[kaliprasad]

Re: Evaluate f(91π/2002)+f(92π/2002)+...+(910π/2002)

If $$f(x)=\frac{1}{1+\tan^3 x}$$, evaluate $$f\left(\frac{91\pi}{2002} \right)+f\left(\frac{92\pi}{2002} \right)+\cdots+f\left(\frac{910\pi}{2002} \right)$$.

Spoiler

f(x) = cos ^3 x/( sin ^3 x + cos^3 x)
f(π/2 -x) = sin ^3 x / ( sin ^3 x + cos^3 x)

So f(x) + f(π/2-x) = 1

So f(91π/2002) + f(910π/2002) = 1 as 91π/2002 + 910π/2002 = π/2
Similarly upto
f(500π/2002) + f(501π/2002) =1

we have 410 pairs and sum = 410

 

[juantheron]

Re: Evaluate f(91π/2002)+f(92π/2002)+...+(910π/2002)

Using the fact $\displaystyle f(x)+f\left(\frac{\pi}{2}-x\right) = \frac{1}{1+\tan^3 (x)}+\frac{1}{1+\cot^3(x)} = \frac{1+\tan^3 (x)}{1+\tan^3 (x)} = 1$

So we get $\displaystyle f(x)+f\left(\frac{\pi}{2}-x\right) = 1$

Now put $\displaystyle x = \frac{91\pi}{2002}$ to $\displaystyle x = \frac{500\pi}{2002}$

$\displaystyle \sum_{r=91}^{910} f\left(\frac{r\cdot\pi}{2002}\right) = 1+1+1+....+1)(410)$ -times

So $\displaystyle \sum_{r=91}^{910} f\left(\frac{r\cdot\pi}{2002}\right) = 410$