MHB How Do You Calculate the Sum of Cubed Tangents for Angles from 0 to 89 Degrees?

[anemone]

Here is this week's POTW:

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Evaluate $$\sum_{k=0}^{89} \frac{1}{1+\tan^{3} (k^{\circ})}$$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. castor28
2. kaliprasad

Solution from castor28:

Spoiler

Let us write $f(k) = \dfrac{1}{1+\tan^3(k\mbox{°})}$ and $S$ for the sum. We have:
$$
S = f(0) + \sum_{k=1}^{44} \left(f(k)+f(90-k)\right) + f(45)
$$
Now,
\begin{align*}
f(k) + f(90-k) &= \frac{1}{1+\tan^3(k\mbox{°})} + \frac{1}{1+\cot^3(k\mbox{°})}\\
&= 1
\end{align*}
where we use the fact that $\cot(x)=\dfrac{1}{\tan(x)}$ and the identity:
$$
\frac{1}{1+x} + \frac{1}{1+\dfrac{1}{x}}=1
$$
This gives:
\begin{align*}
S &= f(0) + \sum_{k=1}^{44}(1) + f(45)\\
&= 0 + 44 + \frac12\\
&= {\bf 45.5}
\end{align*}

Alternate solution from kaliprasad:

Spoiler

We have $$\sum^{89}_{n=0}\frac{1}{1+\tan^3(k^\circ)}$$
$$
= \sum^{89}_{n=0}\frac{\cos^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$$

$$=\frac{\cos^3(0^\circ)}{\cos^3(0^\circ)+\sin ^3(0^\circ)} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{\sin^3(45^\circ)}{\cos^3(45^\circ)+\sin ^3(45^\circ)}+ \sum^{89}_{n=46}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$$ splitting into 4 parts

$$=\frac{1}{1+0} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{\sin^3(45^\circ)}{\sin^3(45^\circ)+\sin ^3(45^\circ)}+ \sum^{89}_{n=46}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$$ as $\sin\,45^\circ = \cos \, 45^\circ$

$$=\frac{1}{1+0} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{1}{2}+ \sum^{89}_{n=46}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$$ putting values and using $\sin\,x^\circ = \cos \,(90^\circ-x)$

$$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \sum^{89}_{n=46}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$$

$$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \sum^{46}_{n=89}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$$ reversing order of sum of cos terms

$$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \sum^{44}_{n=1}\frac{\cos^3( k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$$

$$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)+\cos^3( k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$$

$$=\frac{3}{2} + \sum^{44}_{n=1} 1= 1.5 + 44 \\= 45.5$$