Question

Sendo a função f:R-- R definida por f(x)= 2sen x + sen 2x + cos 3x, calcule:
A) f ($\frac{ \pi }{2}$)
B) f ($\pi$)
C) $\frac{f(0) + f(2x)}{f( \frac{3 \pi }{2}) }$

Answer 1

f(x)= 2sen x + sen 2x + cos 3x

a)
$f(\frac{\pi}{2})= 2sen (\frac{\pi}{2})} + sen 2.(\frac{\pi}{2}) + cos 3.(\frac{\pi}{2})= \\ \\ f(\frac{\pi}{2})= 2sen (\frac{\pi}{2})} + sen \ \pi + cos (\frac{3\pi}{2})= \\ \\ f(\frac{\pi}{2})=2.1+0+0= \\ \boxed{f(\frac{\pi}{2})=2}$

b)
$f(\pi)= 2sen \pi + sen 2 \pi + cos 3 \pi = \\ \\ f(\pi)=2.0+0+(-1)= \\ \\ \boxed{f(\pi)=-1}$

c)

$f(0)= 2sen 0 + sen 2.0 + cos 3.0 \\ \\ f(0)=2.0+0+1 f(0)=1 \\ \\ f(2\pi)= 2sen 2\pi + sen 2.2\pi + cos 3.2\pi \\ \\ f(2\pi)=2.0+0+1 \\ \\ f(2\pi)=1 \\ \\ f(\frac{3\pi}{2})= 2sen \frac{3\pi}{2} + sen 2.\frac{3\pi}{2} + cos 3.\frac{3\pi}{2}= \\ \\ f(\frac{3\pi}{2})= 2sen \frac{3\pi}{2} + sen 3\pi + cos \frac{9\pi}{2}= \\ \\ f(\frac{3\pi}{2})=2.(-1)+0+0=-2$

Logo

$\boxed{\frac{f(0)+f(2\pi)}{f(\frac{3\pi}{2})}=\frac{1+1}{(-2)}=-1}$