Question 77324
If you're trying to evaluate:

*[Tex \large f(\pi/2)] for the function

*[Tex \large f(x)=3sinx + 2cos2x ] then:

*[Tex \large f(\pi/2)=3sin(\pi/2) + 2cos(2(\pi/2)) ]

Since *[Tex \large 3sin(\pi/2)=3*1=3] we can reduce the first term

*[Tex \large f(\pi/2)=3 + 2cos(2(\pi/2)) ]

*[Tex \large f(\pi/2)=3 + 2cos(\not{2}(\pi/\not{2})) ] notice these terms cancel

*[Tex \large f(\pi/2)=3 + 2cos(\pi) ]

Since *[Tex \large 2cos(\pi)=2(-1)=-2]

Which reduces to


*[Tex \large f(\pi/2)=3 + (-2) ]

Which finally reduces to


*[Tex \large f(\pi/2)=3-2=1 ]

So *[Tex \large f(\pi/2)=1 ]