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

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The evaluation of the sum f(91π/2002) + f(92π/2002) + ... + f(910π/2002) is simplified using the property that f(x) + f(π/2 - x) = 1. By applying this identity, pairs of terms can be grouped from x = 91π/2002 to x = 500π/2002, resulting in 410 pairs. Each pair sums to 1, leading to a total of 410 for the entire sum. Therefore, the final result of the evaluation is 410. This demonstrates the utility of symmetry in trigonometric functions for simplifying complex sums.
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If $$f(x)=\frac{1}{1+\tan^3 x}$$, evaluate $$f\left(\frac{91\pi}{2002} \right)+f\left(\frac{92\pi}{2002} \right)+\cdots+f\left(\frac{910\pi}{2002} \right)$$.
 
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Re: Evaluate f(91π/2002)+f(92π/2002)+...+(910π/2002)

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

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