IDNLearn.com makes it easy to find accurate answers to your questions. Get accurate and comprehensive answers to your questions from our community of knowledgeable professionals.
Sagot :
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].
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].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.