Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

[tex]\[
\sin A + \sin B + \sin C = 4 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{B+C}{2}\right) \cdot \sin \left(\frac{C+A}{2}\right)
\][/tex]


Sagot :

Let's evaluate the given mathematical expression:

[tex]\[ \sin A + \sin B + \sin C = 4 \sin \frac{A+B}{2} \sin \frac{B+C}{2} \sin \frac{C A+C}{2} \][/tex]

We need to prove if the expression on the left-hand side equals the expression on the right-hand side.

1. Step 1: Understand the Given Expression

We are given three angles [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]. From the context, it's likely that these angles are the internal angles of a triangle. Hence, we know that:
[tex]\[ A + B + C = 180^\circ \quad \text{(or} \; \pi \; \text{radians)} \][/tex]

2. Step 2: Simplify the Angles

Let's assume an equilateral triangle for simplicity, where:
[tex]\[ A = B = C = 60^\circ \quad \text{(or} \; \frac{\pi}{3} \; \text{radians)} \][/tex]

3. Step 3: Calculate the Left Side

Now let's calculate the left side of the equation:
[tex]\[ \sin A + \sin B + \sin C \][/tex]
With [tex]\( A = B = C = 60^\circ \)[/tex], we get:
[tex]\[ \sin 60^\circ + \sin 60^\circ + \sin 60^\circ \][/tex]
[tex]\[ 3 \cdot \sin 60^\circ \][/tex]
And we know that:
[tex]\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \][/tex]
So the calculation becomes:
[tex]\[ 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \approx 2.598076211353316 \][/tex]

4. Step 4: Calculate the Right Side

Now we'll calculate the right-hand side of the equation:
[tex]\[ 4 \sin \frac{A+B}{2} \sin \frac{B+C}{2} \sin \frac{C A+C}{2} \][/tex]
For our equilateral triangle:
[tex]\[ \frac{A+B}{2} = \frac{60^\circ + 60^\circ}{2} = 60^\circ \][/tex]
and similarly:
[tex]\[ \frac{B+C}{2} = 60^\circ \][/tex]
and:
[tex]\[ \frac{C A+C}{2} = 60^\circ \][/tex]
Therefore:
[tex]\[ 4 \sin 60^\circ \sin 60^\circ \sin 60^\circ \][/tex]
Plugging in:
[tex]\[ 4 \cdot \left( \frac{\sqrt{3}}{2} \right) \cdot \left( \frac{\sqrt{3}}{2} \right) \cdot \left( \frac{\sqrt{3}}{2} \right) \][/tex]
[tex]\[ 4 \cdot \left( \frac{\sqrt{3}}{2} \right)^3 \][/tex]
[tex]\[ 4 \cdot \left( \frac{3\sqrt{3}}{8} \right) \][/tex]
[tex]\[ = \frac{3\sqrt{3}}{2} \approx 2.5980762113533156 \][/tex]

5. Conclusion

Both the left-hand side and right-hand side of the equation approximately equal [tex]\( \frac{3\sqrt{3}}{2} \approx 2.598076211353316 \)[/tex]. Thus, the expression holds true for the given angles of an equilateral triangle.

Therefore,
[tex]\[ \sin A + \sin B + \sin C = 4 \sin \frac{A+B}{2} \sin \frac{B+C}{2} \sin \frac{C A+C}{2} \][/tex]
is indeed verified.