For the letters A, B and C, let's use the following triangle:
Using the Pythagoras theorem, let's find the value of y:
[tex]\begin{gathered} 4^2=x^2+y^2 \\ y^2=16-x^2 \\ y=\sqrt[]{16-x^2} \end{gathered}[/tex]
Now, let's find the equivalent for letter A:
[tex]\begin{gathered} \tan (\sin ^{-1}(\frac{x}{4})) \\ =\tan (k) \\ =\frac{x}{y} \\ =\frac{x}{\sqrt[]{16-x^2}} \end{gathered}[/tex]
So letter A is equivalent to number 1.
For letter B we have:
[tex]\begin{gathered} \cos (\sin ^{-1}(\frac{x}{4})) \\ =\cos (k) \\ =\frac{y}{4} \\ =\frac{\sqrt[]{16-x^2}}{4} \end{gathered}[/tex]
So letter B is equivalent to number 2.
For letter C we have:
[tex]\begin{gathered} \frac{1}{2}\sin (2\sin ^{-1}(\frac{x}{4})) \\ =\frac{1}{2}\sin (2k) \\ =\frac{1}{2}\cdot(2\sin (k)\cos (k) \\ =\sin (k)\cdot\cos (k)_{} \\ =\frac{x}{4}\cdot\frac{y}{4} \\ =\frac{x}{16}\cdot\sqrt[]{16-x^2} \end{gathered}[/tex]
So letter C is equivalent to number 5.
Now, for letters D and E, let's use this triangle:
Finding y, we have that:
[tex]\begin{gathered} y^2=x^2+4^2 \\ y=\sqrt[]{16+x^2^{}} \end{gathered}[/tex]
So for letter D we have:
[tex]\begin{gathered} \sin (\tan ^{-1}(\frac{x}{4})) \\ =\sin (k) \\ =\frac{x}{y} \\ =\frac{x}{\sqrt[]{16+x^2}} \end{gathered}[/tex]
So letter D is equivalent to number 3.
For letter E we have:
[tex]\begin{gathered} \cos (\tan ^{-1}(\frac{x}{4})) \\ =\cos (k) \\ =\frac{4}{y} \\ =\frac{4}{\sqrt[]{16+x^2}} \end{gathered}[/tex]
So letter E is equivalent to number 4.
