Answer
$$
f^{\prime \prime}(\theta)=\sin \theta+\cos \theta ,\quad f(0)=3 ,\quad f^{\prime}(0)=4
$$
The required function is
$$
f(\theta)=-\sin \theta-\cos \theta+5 \theta+4
$$
Work Step by Step
$$
f^{\prime \prime}(\theta)=\sin \theta+\cos \theta ,\quad f(0)=3 ,\quad f^{\prime}(0)=4
$$
The general anti-derivative of $
f^{\prime \prime}(\theta)=\sin \theta+\cos \theta
$ is
$$
f^{\prime}(\theta)=-\cos \theta+\sin \theta+C
$$
To determine C we use the fact that $f^{\prime}(0)=4$:
$$
f^{\prime}(0)=-\cos (0)+\sin (0)+C =4
$$
$ \Rightarrow $
$$
-1+C=4\quad \Rightarrow \quad C=5,
$$
so
$$
f^{\prime}(\theta)=-\cos \theta+\sin \theta+5
$$
Using the anti-differentiation rules once more, we find that:
$$
f(\theta)=-\sin \theta-\cos \theta+5 \theta+D
$$
To determine D we use the fact that $f(0)=3 $:
$$
f(0)=-\sin (0)-\cos (0)+5 (0)+D=3
$$
$ \Rightarrow $
$$
-1+D=3\quad \Rightarrow \quad D=4,
$$
so the required function is
$$
f(\theta)=-\sin \theta-\cos \theta+5 \theta+4
$$
