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I'm to visualize what 

f(t) = 2+cos(3t-6/pi) + 1/4cos(t/2+pi/3)+sin^2(t)

looks like, but I'm not sure how it's done by hand. I gathered that the function is periodic at 4pi and that the periods of the sin and cos -functions are 

T(cos(3t-6/pi)) = 2/3 * pi
T(1/4cos(t/2+pi/3)) = 4 * pi
T(sin(t)) = 2 * pi

Is the only way to do this by hand to draw each function here, evaluate how they act when they are added together and account for the +2 outside of the function?

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    I think the approach you have described is about the best you can do. Note that the period of $\sin^2 t$ is $\pi$, not $2 \pi$, but the overall period is still $4 \pi$ When the amplitudes are similar, as here, it is hard. Often the slowest wave has higher amplitude than the faster ones and you can just the fast ones as small perturbations. That makes it much easier. – 2017-01-31

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To have an idea of what the function looks like:

  • find the periodicity of the function (you seem to have already done that)

  • differentiate the function

  • find the sign of $f'$

  • you will then know the variations of the function.

It will look something like this:

enter image description here

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    What does differentiating here 'mean'? It seems to share the same period of 4pi with the original function, and the slopes are more steep – 2017-01-31