If ABC is a triangle, then find \[{a^2}\sin 2C + {c^2}\sin 2A\].
A. \[\Delta \]
B. \[2\Delta \]
C. \[3\Delta \]
D. \[4\Delta \]

Hint: We will apply the double angle formula of sine to simplify the given expression. Then we will use the sine law, cosine law, and area of oblique triangle respectively to get the desired result.

Formula Used:
Double angle formula of sine:
\[\sin 2\theta = 2\sin \theta \cos \theta \]
Area of oblique triangle:
\[\Delta = \dfrac{1}{2}ab\sin C\]
\[\Delta = \dfrac{1}{2}bc\sin A\]
\[\Delta = \dfrac{1}{2}ac\sin B\]
Sine Law:
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]

Complete step by step solution:
Given expression is
\[{a^2}\sin 2C + {c^2}\sin 2A\]
Apply double angle formula \[\sin 2\theta = 2\sin \theta \cos \theta \]
\[ = {a^2}2\sin C\cos C + {c^2}2\sin A\cos A\]
\[ = 2{a^2}\sin C\cos C + 2{c^2}\sin A\cos A\] …(i)
The sine law is
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\left( {say} \right)\]
This implies \[\sin A = ak\], \[\sin B = bk\], and \[\sin C = ck\]
Substitute \[\sin A = ak\] and \[\sin C = ck\] in expression (i)
\[ = 2{a^2} \cdot ck\cos C + 2{c^2} \cdot ak\cos A\]
\[ = 2ack\left( {a\cos C + c\cos A} \right)\]
Apply cosine law:
\[ = 2ack\left( {a \cdot \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} + c \cdot \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)\]
\[ = 2ack\left( {\dfrac{{{a^2} + {b^2} - {c^2}}}{{2b}} + \dfrac{{{b^2} + {c^2} - {a^2}}}{{2b}}} \right)\]
\[ = 2ack\left( {\dfrac{{{a^2} + {b^2} - {c^2} + {b^2} + {c^2} - {a^2}}}{{2b}}} \right)\]
\[ = 2ack\left( {\dfrac{{2{b^2}}}{{2b}}} \right)\]
\[ = 2abck\]
Putting \[\dfrac{{\sin A}}{a} = k\]
\[ = 2abc \cdot \dfrac{{\sin A}}{a}\]
\[ = 2bc\sin A\]
\[ = 4 \cdot \dfrac{1}{2}bc\sin A\]
Apply the formula of area of oblique triangle \[\Delta = \dfrac{1}{2}bc\sin A\]
\[ = 4\Delta \]
Hence option D is the correct option.

Note: Students often make a mistake to apply the area of a triangle. They use a wrong formula that is \[\dfrac{1}{2}bc\]. But \[\dfrac{1}{2}bc\] is the formula of a right-angled triangle. The formula of the area of an oblique triangle is \[\dfrac{1}{2}bc\sin A\].