trigonometria

Prove que:

Lembre-se que:

cos0º = 1
cos89º = sen1º
cos88º = sen2º
..................................
cos60º = sen30º = 1/2
...................................
cos46º = sen44º
cos45º = sen45º = √2/2

Olá Romanelo;

Suponha que seja verdade o proposto pelo enunciado:

[latex]\frac{1}{cos(0\degree) \cdot cos(1\degree)} + \frac{1}{cos(1\degree) \cdot cos(2\degree)} + ... + \frac{1}{cos(88\degree) \cdot cos(89\degree)} = \frac{cos1 \degree}{\text{[}sen(1\degree)\text{]}^2}[/latex]

[latex]sen(1\degree)\text{[}\frac{1}{cos(0\degree) \cdot cos(1\degree)} + \frac{1}{cos(1\degree) \cdot cos(2\degree)} + ... + \frac{1}{cos(88\degree) \cdot cos(89\degree)}) \text{]}= \frac{cos1 \degree}{sen(1\degree)}[/latex]

[latex]\frac{sen(1\degree)}{cos(0\degree) \cdot cos(1\degree)} + \frac{sen(1\degree)}{cos(1\degree) \cdot cos(2\degree)} + ... + \frac{sen(1\degree)}{cos(88\degree) \cdot cos(89\degree)} = \frac{cos1 \degree}{sen(1\degree)}[/latex]

[latex]\frac{sen(1\degree - 0\degree)}{cos(0\degree) \cdot cos(1\degree)} + \frac{sen(2 \degree -1\degree)}{cos(1\degree) \cdot cos(2\degree)} + ... + \frac{sen(89 \degree - 88\degree)}{cos(88\degree) \cdot cos(89\degree)} = \frac{cos1 \degree}{sen(1\degree)} [/latex]

[latex]\text{[}tg(1\degree) - tg(0\degree)\text{]} + \text{[}tg(2\degree) - tg(1\degree)\text{]} + ... + \text{[}tg(89\degree) - tg(88\degree)\text{]} = \frac{cos(1\degree)}{sen(1\degree)}

tg(89\degree) - tg(0\degree) = cotg(1\degree) = \frac{cos(1\degree)}{sen(1\degree)}[/latex]

Bons estudos