5
$\begingroup$

$$\sum_{r=1}^{89} \frac{1}{1+\tan^3 r }$$ where $r$ is in degrees

I tried this a lot using the $a^3+b^3$ identity but I don't seem to be getting anything fruitful :(

Can someone please give me a hint? I don't think the $V_n$ method is working here. ($T_r+T_{r-1}$)

Is it related to the $\tan(A+B)$ identity? However I'm not getting anything through that too.

I would be grateful if someone helped. Thanks!

$\endgroup$
3
  • $\begingroup$ You could try to work out something with $$1+\tan^3r=\frac{\cos^3r+\sin^3r}{\cos^3r}=\frac{\cos^2r-\cos r\sin r+\sin^2r}{\cos^3r}=\frac{1-\cos r\sin r}{\cos^3r}$$ $\endgroup$
    – DonAntonio
    Commented Aug 7, 2021 at 8:33
  • $\begingroup$ @DonAntonio The second equality is false. $\endgroup$
    – jjagmath
    Commented Aug 7, 2021 at 9:00
  • $\begingroup$ @jjagmath Yep: $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ and the factor $\;(\cos r+\sin r)\;$ is missing. May the OP fulfill this. Thanks. $\endgroup$
    – DonAntonio
    Commented Aug 7, 2021 at 9:42

1 Answer 1

10
$\begingroup$

Let $S=\displaystyle\sum_{r=1}^{89} \cfrac{1}{1+\tan^3 r }$

$S=\displaystyle\sum_{r=1}^{89} \cfrac{\cos^3r}{\cos^3r+\sin^3 r }$

Now Since $\displaystyle\sum_{r=a}^{b}f(r)=\displaystyle\sum_{r=a}^{b}f(a+b-r)$

Using which we get $S=\displaystyle\sum_{r=1}^{89} \cfrac{\cos^3(90-r)}{\cos^3(90-r)+\sin^3(90-r) }=\displaystyle\sum_{r=1}^{89} \cfrac{\sin^3r}{\cos^3r+\sin^3 r }$

Therefore $S=\displaystyle\sum_{r=1}^{89} \frac{\cos^3r}{\cos^3r+\sin^3 r} =\displaystyle\sum_{r=1}^{89} \frac{\sin^3r}{\cos^3r+\sin^3 r }$

Therefore $2S=\displaystyle\sum_{r=1}^{89} \cfrac{\sin^3r+\cos^3r}{\cos^3r+\sin^3 r }=\displaystyle\sum_{r=1}^{89} 1=89$

Therefore $S=\displaystyle \boxed{{\frac{89}{2}}}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .