Question

calcule a soma :
sen²1° + sen²2° + ... + sen²88° + sen²89

Answer 1

Bom, primeiramente lembremos da propriedade de arcos complementares:

$\mathrm{sen\,}\theta=\cos(90^{\circ}-\theta)\,,~~~~\text{para qualquer }\theta\in \mathbb{R}$


Logo,

$\mathrm{sen^2\,}\theta=\cos^2(90^{\circ}-\theta)$


De forma que

$\mathrm{sen^2\,}1^{\circ}=\cos^2 89^{\circ}\\\\ \mathrm{sen^2\,}2^{\circ}=\cos^2 88^{\circ}\\\\ \vdots\\\\ \mathrm{sen^2}43^{\circ}=\cos^2 47^{\circ}\\\\ \mathrm{sen^2}44^{\circ}=\cos^2 46^{\circ}$

Então, podemos rearrumar a nossa soma da forma a seguir, e usar a Relação Fundamental da Trigonometria:

$\mathrm{sen^2\,}1^{\circ}+\mathrm{sen^2\,}2^{\circ}+\ldots+\mathrm{sen^2\,}88^{\circ}+\mathrm{sen^2\,}89^{\circ}\\\\ =(\mathrm{sen^2\,}1^{\circ}+\mathrm{sen^2\,}89^{\circ})+(\mathrm{sen^2\,}2^{\circ}+\mathrm{sen^2\,}88^{\circ})+\ldots+(\mathrm{sen^2\,}44^{\circ}+\mathrm{sen^2\,}46^{\circ})+\mathrm{sen^2}45^{\circ}\\\\ =(\mathrm{sen^2\,}1^{\circ}+\cos^2 1^{\circ})+(\mathrm{sen^2\,}2^{\circ}+\cos^2 2^{\circ})+\ldots+(\mathrm{sen^2\,}44^{\circ}+\cos^2 44^{\circ})+\mathrm{sen^2}45^{\circ}\\\\ =\underbrace{1+1+\ldots+1}_{\text{44 parcelas}}+\mathrm{sen^2}45^{\circ}\\\\\\ =44+\mathrm{sen^2}45^{\circ}\\\\ =44+\Big(\dfrac{\sqrt{2}}{2} \Big)^{\!2}$

$=44+\dfrac{2}{4}\\\\\\ =44+\dfrac{1}{2}\\\\\\ =\dfrac{88}{2}+\dfrac{1}{2}\\\\\\ =\dfrac{88+1}{2}\\\\\\ =\dfrac{89}{2}\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathrm{sen^2\,}1^{\circ}+\mathrm{sen^2\,}2^{\circ}+\ldots+\mathrm{sen^2\,}88^{\circ}+\mathrm{sen^2\,}89^{\circ}=\dfrac{89}{2} \end{array}}$