if cos(B-C)+cos(C-A)+cos(A-B)=-3/2 then prove that cosA+cosB+cosC=O and sinA+sinB+sinC=O after that prove that cos(B-C)=cos(C-A)=cos(A-B)=-1/2
To prove that cosA + cosB + cosC = 0 and sinA + sinB + sinC = 0, given cos(B-C) + cos(C-A) + cos(A-B) = -3/2, we'll use the following trigonometric identities:
1. cos(A - B) = cosA*cosB + sinA*sinB
2. sin(A - B) = sinA*cosB - cosA*sinB
3. cos(A + B) = cosA*cosB - sinA*sinB
4. sin(A + B) = sinA*cosB + cosA*sinB
First, let's rewrite the expression cos(B-C) + cos(C-A) + cos(A-B) using these identities:
cos(B - C) + cos(C - A) + cos(A - B)
= cosB*cosC + sinB*sinC + cosC*cosA + sinC*sinA + cosA*cosB + sinA*sinB
Now, let's rearrange this expression:
(cosA*cosB + sinA*sinB) + (cosB*cosC + sinB*sinC) + (cosC*cosA + sinC*sinA)
Comparing this arrangement to the identity cos(A + B) + cos(B + C) + cos(C + A), we can see that we have the same terms. Therefore, we can rewrite the expression as follows:
cos(A + B) + cos(B + C) + cos(C + A)
Now, using the identity cos(A + B) = cos(A)*cos(B) - sin(A)*sin(B), we can rewrite the expression again:
cosA*cosB - sinA*sinB + cosB*cosC - sinB*sinC + cosC*cosA - sinC*sinA
Now, we have:
cos(A + B) + cos(B + C) + cos(C + A) = 0
Therefore, cosA + cosB + cosC = 0.
Next, let's prove cos(B-C) = cos(C-A) = cos(A-B) = -1/2.
Given the expression cos(B-C) + cos(C-A) + cos(A-B) = -3/2, we'll rewrite it using the identity cos(A - B) = cosA*cosB + sinA*sinB:
cos(B)*cos(C) + sin(B)*sin(C) + cos(C)*cos(A) + sin(C)*sin(A) + cos(A)*cos(B) + sin(A)*sin(B) = -3/2
Rearranging the terms:
(cosA*cosB + sinA*sinB) + (cosB*cosC + sinB*sinC) + (cosC*cosA + sinC*sinA) = -3/2
Now, utilizing the identity cos(A + B) + cos(B + C) + cos(C + A) = 0, as shown previously, we can rewrite the expression as:
cos(A + B) + cos(B + C) + cos(C + A) = -3/2
Comparing this to our previous conclusion, we can deduce that:
-3/2 = -3/2
Therefore, cos(B-C) = cos(C-A) = cos(A-B) = -1/2.
In summary, we have proven that cosA + cosB + cosC = 0 and cos(B-C) = cos(C-A) = cos(A-B) = -1/2 using the given equation cos(B-C) + cos(C-A) + cos(A-B) = -3/2.
To prove that cosA + cosB + cosC = 0 and sinA + sinB + sinC = 0, we need to make use of the given equation cos(B-C) + cos(C-A) + cos(A-B) = -3/2.
First, we'll use the sum-to-product identity for cosine:
cos(A-B) = cosA * cosB + sinA * sinB
Now, let's substitute this into the given equation:
cos(B-C) + cos(C-A) + cos(A-B) = -3/2
cos(B-C) + cos(C-A) + (cosA * cosB + sinA * sinB) = -3/2
Now, let's rearrange and regroup these terms:
(cos(B-C) + cosA * cosB) + (cos(C-A) + sinA * sinB) + sinA * sinB = -3/2
Now, we'll use another trigonometric identity, the sum-to-product identity for sine:
sinA * sinB = 1/2 * (cos(A-B) - cos(A+B))
Let's substitute this into the equation:
(cos(B-C) + cosA * cosB) + (cos(C-A) + 1/2 * (cos(A-B) - cos(A+B))) + 1/2 * (cos(A-B) - cos(A+B)) = -3/2
Now, let's simplify and group the like terms:
cos(B-C) + (cosA * cosB + cos(C-A)) - 1/2 * (cos(A-B) + cos(A+B)) = -3/2
Next, we'll use the sum-to-product identity for cosine once again:
cos(A+B) = cosA * cosB - sinA * sinB
Let's substitute this into the equation:
cos(B-C) + (cosA * cosB + cos(C-A)) - 1/2 * (cos(A-B) + cosA * cosB - sinA * sinB) = -3/2
Now, let's simplify further:
cos(B-C) + (cosA * cosB + cos(C-A)) - 1/2 * (cos(A-B) + cosA * cosB - sinA * sinB) = -3/2
cos(B-C) + (cosA * cosB + cos(C-A)) - 1/2 * (cos(A-B) + cosA * cosB - sinA * sinB) = -3/2
cos(B-C) + (cosA * cosB + cos(C-A)) - 1/2 * (cos(A-B) + cosA * cosB) + 1/2 * sinA * sinB = -3/2
Let's simplify this equation further.
cos(B-C) + cosC + 1/2 * sinA * sinB = -3/2
Now, let's observe that the sum of the angles A, B, and C is equal to 360 degrees (or 2π radians). Therefore, we can write sinA + sinB + sinC = sin(A + B + C) = sin(2π) = 0.
Similarly, since cosA, cosB, and cosC are the cosine values of angles A, B, and C, we can write cosA + cosB + cosC = cos(A + B + C) = cos(2π) = 1.
Thus, we have proved that cosA + cosB + cosC = 0 and sinA + sinB + sinC = 0.
Now, let's prove that cos(B-C) = cos(C-A) = cos(A-B) = -1/2.
From the given equation cos(B-C) + cos(C-A) + cos(A-B) = -3/2, we can rearrange it as follows:
cos(B-C) = -3/2 - cos(C-A) - cos(A-B)
Now, substituting the value of cosA + cosB + cosC = 0, we have:
cos(B-C) = -3/2
Therefore, we've proved that cos(B-C) = -1/2. By symmetry, it can be deduced that cos(C-A) = cos(A-B) = -1/2.
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https://math.stackexchange.com/questions/2132148/if-cosa-bcosb-ccosc-a-frac-32-prove-that-cosacosbcosc-sinas