product of cosine and sine functions into sum of sine functions transformation I'm required to transform this expression by using elementary trigonometric functions into sum of sine functions. Any help or hints would be welcome.

$$m_e = -2I_1I_2L_d\cos(\omega_1 t)\cdot\cos(\omega_2 t+\alpha)\cdot\sin(\omega_m t+\delta)$$

/// EDIT: RESULT 
 A: Note,
$$2\cos(\omega_1 t)\cos(\omega_2 t+\alpha)
=\cos[(\omega_1+\omega_2)t+\alpha]+\cos[(\omega_2-\omega_1)t+\alpha]$$
$$2\cos[(\omega_1+\omega_2)t+\alpha] \sin(\omega_m t+\delta)$$
$$=\sin[(\omega_1+\omega_2+\omega_m)t+\alpha+\delta]-\sin[(\omega_1+\omega_2-\omega_m)t+\alpha-\delta]$$
$$2\cos[(\omega_2-\omega_1)t+\alpha] \sin(\omega_m t+\delta)$$
$$=\sin[(\omega_2-\omega_1+\omega_m)t+\alpha+\delta]-\sin[(\omega_2-\omega_1-\omega_m)t+\alpha-\delta]$$
A: Hint:
Use the linearisation formulæ:
\begin{cases}\begin{aligned}
\cos a \cos b&=\tfrac12\bigl(\cos(a+b)+\cos(a-b)\bigr),\\
\sin a \cos b&=\tfrac12\bigl(\sin(a+b)+\sin(a-b)\bigr).
\end{aligned}\end{cases}
A: Let us write the expression to be transformed in the following way :
$$m_e = -2I_1I_2L_d\cos(\omega_1 t)\cdot\cos(\omega_2 t+\alpha)\cdot\cos(\omega_m t+\delta-\frac{\pi}{2})$$
Let us use the following formula :

$$\eqalign{&\cos x\cos y\cos z\cr &\qquad=\tfrac{1}{4}\bigl(\cos(x+y+z)+\cos(x+y-z)+\cos(y+z-x)+\cos(z+x-y)\bigr)\ .\cr}$$

(see for example (https://math.stackexchange.com/q/2640161)) :
$$m_e=-\tfrac{1}{2}I_1I_2L_d\bigl[\cos((\omega_1+\omega_2+\omega_m) t+\alpha+\delta-\tfrac{\pi}{2})+\underbrace{\cos((\omega_1+\omega_2-\omega_m)t+\alpha-\delta+\tfrac{\pi}{2})}_{=\cos(-(\omega_1+\omega_2-\omega_m)t-\alpha+\delta-\tfrac{\pi}{2}) \ \text{using parity}}+\cos((-\omega_1+\omega_2+\omega_m)t+\alpha+\delta-\tfrac{\pi}{2})+\cos((\omega_1-\omega_2+\omega_m)t-\alpha+\delta-\tfrac{\pi}{2})\bigr]$$
It suffices now to use the classical formula
$$\cos(A-\tfrac{\pi}{2})=\sin(A)$$
to obtain the result you gave.
