Answer:
[tex]9\,\sin(60^\circ)[/tex], which is equal to [tex]\displaystyle \frac{9\sqrt{3}}{2}[/tex].
Step-by-step explanation:
An angle of [tex]45^\circ[/tex] corresponds to an isosceles right triangle: the length of the two legs (adjacent and opposite) would be equal. Accordingly:
[tex]\displaystyle \tan(45^\circ) = \frac{\text{Opposite Leg}}{\text{Adjacent Leg}} = 1[/tex].
Let [tex]A[/tex] denote the measure of an angle. Double-angle identity for sine:
[tex]2\, \sin(A) \cdot \cos(A) = \sin(2\, A)[/tex].
By this identity:
[tex]\begin{aligned}& (2\, \cos(30^\circ)) \cdot (3\, \sin(30^\circ)) \\ &= 3\, (2\, \cos(30^\circ) \cdot \sin(30^\circ)) \\ &= 3\, \sin(2 \times 30^\circ) \\ &= 3\, \sin(60^\circ)\end{aligned}[/tex].
([tex]A = 30^\circ[/tex] in this instance.)
Hence:
[tex]\begin{aligned}&(3\, \tan(45^\circ)) \cdot (4\, \sin(60^\circ)) - (2\, \cos(30^\circ)) \cdot (3\, \sin(30^\circ)) \\ &= 12\, \sin(60^\circ) - 3\, \sin(60^\circ) \\ &= 9\, \sin(60^\circ)\end{aligned}[/tex].
[tex]\displaystyle \sin(60^\circ) = \frac{\sqrt{3}}{2}[/tex]. Therefore, [tex]\displaystyle 9\, \sin(60^\circ) = \frac{9 \sqrt{3}}{2}[/tex].
