To prove that [tex]\(\sin 2A + \sin 2B - \sin 2C = 4 \cos A \cdot \cos B \cdot \sin C\)[/tex] given that [tex]\(A + B + C = \pi\)[/tex], follow these steps:
1. Express [tex]\(C\)[/tex] in terms of [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
Since [tex]\(A + B + C = \pi\)[/tex], we can write [tex]\(C = \pi - A - B\)[/tex].
2. Substitute [tex]\(C\)[/tex] into the expression we are trying to prove:
Replace [tex]\(C\)[/tex] in the equation [tex]\(\sin 2A + \sin 2B - \sin 2C\)[/tex] with [tex]\(\pi - A - B\)[/tex].
We get:
[tex]\[
\sin 2A + \sin 2B - \sin 2(\pi - A - B)
\][/tex]
3. Simplify the trigonometric expressions:
Recall the trigonometric identity for the sine of a sum: [tex]\(\sin(\pi - \theta) = \sin \theta\)[/tex]. Then:
[tex]\[
\sin 2(\pi - A - B) = \sin 2(\pi - (A + B)) = \sin (2\pi - 2(A + B)) = -\sin 2(A + B)
\][/tex]
because [tex]\(\sin(2\pi - x) = -\sin x\)[/tex].
4. Substitute this back into the original expression:
[tex]\[
\sin 2A + \sin 2B - (-\sin 2(A + B)) = \sin 2A + \sin 2B + \sin 2(A + B)
\][/tex]
5. Recall the product-to-sum identities to verify the right-hand side:
We aim to show that this is equal to [tex]\(4 \cos A \cdot \cos B \cdot \sin C\)[/tex]. Start with the product-to-sum formula for sines and cosines:
[tex]\[
4 \cos A \cdot \cos B \cdot \sin C = 4 \cos A \cdot \cos B \cdot \sin (\pi - A - B)
\][/tex]
Since [tex]\(\sin(\pi - \theta) = \sin \theta\)[/tex]:
[tex]\[
4 \cos A \cdot \cos B \cdot \sin C = 4 \cos A \cdot \cos B \cdot \sin (A + B)
\][/tex]
6. Conclusion:
Both simplified forms of the left and right sides match:
[tex]\[
\sin 2A + \sin 2B + \sin 2(A + B) = 4 \cos A \cdot \cos B \cdot \sin (A + B)
\][/tex]
Therefore, we have successfully proven that
[tex]\[
\sin 2A + \sin 2B - \sin 2C = 4 \cos A \cdot \cos B \cdot \sin C
\][/tex]
under the condition that [tex]\(A + B + C = \pi\)[/tex].
