Answer
$\sin\angle D=\frac{\sqrt3}{3}$
$\cos\angle D = \frac{\sqrt6}{3}$
Work Step by Step
It's obvious that the triangle $\triangle FHC$ is the right triangle, where $\angle H$ is the right angle. We have to find sine and cosine of $\angle C$.
By the definition sine is equal to the ratio of the side opposite a given angle (in a right-angled triangle) to the hypotenuse.
$\sin\angle C=\dfrac{opposite}{hypotenuse}$.
In our case
$\sin\angle C=\dfrac{FH}{CF}$.
Similarly, cosine is equal to the ratio of the side adjacent to an acute angle (in a right-angled triangle) to the hypotenuse.
$\cos\angle C=\dfrac{adjacent}{hypotenuse}=\dfrac{CH}{CF}.$
We just have to find $CF$, $CH$ because we know that $FH=x $ inches. For that we will use Pythagorean theorem.
The triangle $\triangle CDH$ is the right triangle, where $\angle D$ is the right angle. So
$CH^2=CD^2+DH^2$.
For our cube $CD=CH=x$ inches. And CH is equal $CH=x\sqrt2$ inches.
The next right triangle is our well known $\triangle FCH$. Similarly,
$CF^2=CH^2+FH^2=(x\sqrt2)^2+x^2=3⋅x^2$.
So $CF=x\sqrt3$ inches.
Now we have everything to find sine and cosine.
$$\sin\angle D=\dfrac{FH}{CF}=\dfrac{x}{x\sqrt3}=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}$$
$$\cos\angle D=\dfrac{CH}{CF}=\dfrac{x\sqrt2}{x\sqrt3}=\dfrac{\sqrt2}{\sqrt3}=\dfrac{\sqrt6}{3}$$
