3
$\begingroup$

Prove that: $$\sum^{45}_{k=1}\frac{1}{\cos1^\circ-\cos(87+4k)^\circ}=\frac{1}{2\sin 1^\circ}$$

Numerically, this is accurate comparing the lhs and rhs. Some ideas: We can transform the question into proving that:

$$\sum^{45}_{k=1}\frac{1}{\sin(43+2k)^\circ\sin(44+2k)^\circ}=\frac{1}{\sin 1^\circ}$$ Since clearly every term is positive, there is no hope in trying to cancel terms. I have also tried to transform this into a sum of inverses of roots of an equation so that I can use Vieta's formulae, but I have had no success.

$\endgroup$
  • 1
    $\begingroup$ Where did the "$n$" come from? $\endgroup$ – Edward Jiang Nov 21 '14 at 4:20
  • $\begingroup$ @EdwardJiang Sorry, it was a $1$.Edited! $\endgroup$ – chubakueno Nov 21 '14 at 4:23
2
$\begingroup$

HINT:

$$\sin1^\circ=\sin[(44+2k)^\circ-(43+2k)^\circ]$$ $$=\sin(44+2k)^\circ\cos(43+2k)^\circ-\cos(44+2k)^\circ\sin(43+2k)^\circ$$

$$\implies\frac{\sin1^\circ}{\sin(44+2k)^\circ\sin(43+2k)^\circ}=\cot(43+2k)^\circ-\cot(44+2k)^\circ$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.