Question

20. Considere as funções reais de variável real
definidas
por
f(x)=sen 1+3 TT
g(x) = sen(1-
sen(1-2).
K = f(9)-g(9), então,
pode-se
corretamente que o valor de K é igual a
Se
a) 1.
b)-1.
c) 0.
d)-2.
e) 2.
e
afirmar

Answer 1

$\displaystyle \sf f(x) = sen\left(1+\frac{x}{2}\right)\pi \ \ \ ;\ \ \ g(x) =sen\left(1-\frac{x}{2}\right)\pi \\\\\\\ queremos : \\\\ K = f(9)\cdot g(9) \\\\ Da{\'i}}: \\\\ f(9) = sen\left(1+\frac{9}{2}\right)\pi \to f(9) = sen\left(\frac{11\pi }{2}\right) \\\\\\ g(9) = sen\left(1-\frac{9}{2}\right)\pi \to g(9)= sen\left(\frac{-7\pi }{2}\right) \\\\\\ Da{\'i}}: \\\\K = f(9)\cdot g(9) = sen\left(\frac{11\pi }{2}\right)\cdot sen\left(\frac{-7\pi }{2}\right)$

Façamos o seguinte :

$\displaystyle \sf sen\left(\frac{11\pi}{2}\right) = sen\left(\frac{12\pi}{2}-\frac{\pi}{2}\right) = sen\left(6\pi-\frac{\pi}{2}\right) = -sen\left(\frac{\pi}{2}\right) =-1 \\\\\\\\ -sen\left(\frac{7 \pi}{2}\right) = -sen\left(\frac{6\pi }{2}+\frac{\pi}{2}\right) \to -sen\left(3\pi+\frac{\pi}{2}\right) = -\left[-sen\left(\frac{\pi }{2}\right)\right] = 1$

Daí :

$\displaystyle \sf K = f(9)\cdot g(9) = -1\cdot 1 = -1 \\\\ \huge\boxed{\sf k = -1}\checkmark$

letra b

Outra maneira de resolver

$\displaystyle \sf \text{Usando prostaf{\'e}rese} : \\\\ cos(a)-cos(b) = -2\cdot sen\left( \frac{a+b}{2}\right)\cdot sen\left( \frac{a-b}{2}\right) \\\\\\ sen\left( \frac{a+b}{2}\right)\cdot sen\left( \frac{a-b}{2}\right) = \frac{-1}{2}\cdot \left[cos(a)-cos(b) \right] \\\\\\ Da{\'i}} : \\\\\ sen\left(\frac{11\pi}{2}\right)\cdot sen\left(\frac{-7\pi}{2}\right) = \frac{-1}{2}\cdot [cos(a)-cos(b)]$

$\displaystyle \sf \left \begin{array}{I}\displaystyle \sf \frac{a+b}{2} = \frac{11\pi }{2} \to a+b = 11\pi \\\\\\ \displaystyle \sf \frac{a-b}{2} = \frac{-7\pi }{2} \to a-b = -7\pi \end{array} \right\} 2a = 4\pi \to a = 2\pi \ ; \ b =9\pi$

$\displaystyle \sf sen\left(\frac{11\pi}{2}\right)\cdot sen\left(\frac{-7\pi}{2}\right) = \frac{-1}{2}\cdot [cos(2\pi)-cos(9\pi )] \\\\\\ sen\left(\frac{11\pi}{2}\right)\cdot sen\left(\frac{-7\pi}{2}\right) = \frac{-1}{2}\cdot [1-cos(8\pi + \pi )] \\\\\\ sen\left(\frac{11\pi}{2}\right)\cdot sen\left(\frac{-7\pi}{2}\right) = \frac{-1}{2}\cdot [1-cos(\pi )]$

$\displaystyle \sf sen\left(\frac{11\pi}{2}\right)\cdot sen\left(\frac{-7\pi}{2}\right) = \frac{-1}{2}\cdot [1-(-1) ] \\\\\\ \boxed{\sf sen\left(\frac{11\pi}{2}\right)\cdot sen\left(\frac{-7\pi}{2}\right) = -1 = K \ }\checkmark$