Three methods :

1) By using the general (recurrence) formula :

$$J_{\nu-1}(x)+J_{\nu+1}(x)=\frac{2 \nu}{x}J_{\nu}(x)$$

(formula 2.4 p. 13 of this excellent document)

Taking $\nu=\tfrac12$:

$$J_{3/2}(x)=\frac{1}{x}J_{1/2}(x)-J_{-1/2}(x)$$

Knowing that (formulas (2.16) of the document):

$$\begin{align}
J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sin{(x)}\\
J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}} \cos{(x)}
\end{align}$$

we get your result. I assume that $\phi$ is a typo for $\pi$...

2) By using Laplace Transform. Let us prove the equivalent formula :

$$\sqrt{\frac {\pi}{2}} x^{3/2} J_\frac 32 (x) = \sin x - x \cos x \tag{1}$$

As indicated for example in this table,

$$\mathfrak{L}{(x^{\nu}J_{\nu}(x))}=2^{\nu+1}\dfrac{1}{\sqrt{\pi}}\Gamma(\nu+\tfrac12)(s^2+1)^{-\nu-\tfrac12}$$

Knowing the Laplace Transforms:

$$\mathfrak{L}(\sin x)=\frac{1}{s^2+1} \ \ \text{and} \ \ \mathfrak{L}(x \cos x)=\frac{s^2-1}{(s^2+1)^2},$$

it is easy to conclude to the exactness of (1).

3) by taking $n=1$ in the (rather classical) formula :

$J_{p+{\frac{1}{2}}}(x)=\sqrt{\dfrac{2}{\pi}}
(-1)^p x^{p+{\frac{1}{2}}}
\left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^p\left(\frac{\sin x}{x}\right)\tag{2}$

(see this answer, knowing that the **spherical Bessel** function of order $n$ is defined by :

$$j_n(x):=\sqrt{\dfrac{\pi}{x}}J_{n+\tfrac12}(x)$$

(please note the lowercase $j$).

*Fig. 1 : Blue curve : $y=$$\text{sinc}$$(x)$, red curve : $y=\cos(x)$, magenta curve : $y=\sqrt{\frac12 \pi x} J_\frac 32 (x)$*
**Remarks :**

1) The roots $r_k$ of Bessel functions in general are important. Here, the roots of $J_{3/2}(x)=0$, are (except $0$) the same as the roots of:

$$\tan x = x \tag{3}$$

These roots have a nice property (please note that we deal only with positive roots):

$$\sum_{k=1}^{\infty} \dfrac{1}{r_k^2}=\dfrac{1}{10}.\tag{4}$$

See proofs of (4) here or here.

2) Connected : the second and third integral of this question.