Prove that
cos(-A+B+C) + cos(A-B+C) + cos(A+B-C) + cos(A+B+C) = 4cosAcosBcosC
To prove the given identity:
cos(-A + B + C) + cos(A - B + C) + cos(A + B - C) + cos(A + B + C) = 4cosAcosBcosC
We will start by using the cosine sum formula, which states that:
cos(X + Y) = cos(X)cos(Y) - sin(X)sin(Y)
Therefore, we can rewrite the expression as:
cos(-A + B + C) + cos(A - B + C) + cos(A + B - C) + cos(A + B + C)
= [cos(-A)cos(B + C) - sin(-A)sin(B + C)] + [cos(A)cos(-B + C) - sin(A)sin(-B + C)]
+ [cos(A + B)cos(-C) - sin(A + B)sin(-C)] + [cos(A + B)cos(C) - sin(A + B)sin(C)]
Since sin(-X) = -sin(X), we can simplify the expression further:
cos(-A + B + C) + cos(A - B + C) + cos(A + B - C) + cos(A + B + C)
= [cos(A)cos(B + C) + sin(A)sin(B + C)] + [cos(A)cos(B + C) - sin(A)sin(B + C)]
+ [cos(A + B)cos(C) + sin(A + B)sin(C)] + [cos(A + B)cos(C) - sin(A + B)sin(C)]
Combining like terms, we get:
2[cos(A)cos(B + C)] + 2[cos(A + B)cos(C)]
Now, let's focus on the two terms separately:
cos(A)cos(B + C)
Using the cosine sum formula again, we have:
cos(B + C) = cos(B)cos(C) - sin(B)sin(C)
Therefore,
cos(A)cos(B + C) = cos(A)(cos(B)cos(C) - sin(B)sin(C))
= cos(A)cos(B)cos(C) - cos(A)sin(B)sin(C)
Similarly,
cos(A + B)cos(C) = (cos(A)cos(B) - sin(A)sin(B))cos(C)
= cos(A)cos(B)cos(C) - sin(A)sin(B)cos(C)
Now, substituting these values back into our expression, we get:
2[cos(A)cos(B)cos(C) - cos(A)sin(B)sin(C)] + 2[cos(A)cos(B)cos(C) - sin(A)sin(B)cos(C)]
Expanding further:
2cos(A)cos(B)cos(C) - 2cos(A)sin(B)sin(C) + 2cos(A)cos(B)cos(C) - 2sin(A)sin(B)cos(C)
Combining like terms:
4cos(A)cos(B)cos(C) - 2cos(A)sin(B)sin(C) - 2sin(A)sin(B)cos(C)
Now, we can use the identity sin(A)sin(B) = (1/2)(cos(A-B) - cos(A+B)):
4cos(A)cos(B)cos(C) - 2cos(A)sin(B)sin(C) - 2sin(A)sin(B)cos(C)
= 4cos(A)cos(B)cos(C) - 2cos(A)[(1/2)(cos(B-C) - cos(B+C))] - 2[(1/2)(cos(A-B) - cos(A+B))]cos(C)
= 4cos(A)cos(B)cos(C) - cos(A)cos(B-C) + cos(A)cos(B+C) - cos(A-B)cos(C) + cos(A+B)cos(C)
Now, notice that cos(A)cos(B-C) - cos(A-B)cos(C) = cos(A-B)cos(C) - cos(A)cos(B-C).
Similarly, cos(A)cos(B+C) + cos(A+B)cos(C) = cos(A+B)cos(C) + cos(A)cos(B+C).
Therefore, the expression simplifies to:
4cos(A)cos(B)cos(C) - cos(A)cos(B-C) + cos(A)cos(B+C) - cos(A-B)cos(C) + cos(A+B)cos(C)
= 4cos(A)cos(B)cos(C) + cos(A)cos(B-C) + cos(A-B)cos(C) + cos(A+B)cos(C)
= 4cos(A)cos(B)cos(C) + cos(B-C)cos(A) + cos(C-A)cos(B) + cos(A+B)cos(C)
Finally, recognizing that cos(B-C)cos(A) + cos(C-A)cos(B) = cos(B-A)cos(C),
we can further simplify the expression to:
4cos(A)cos(B)cos(C) + cos(B-A)cos(C) + cos(A+B)cos(C)
= 4cos(A)cos(B)cos(C) + cos(A+B)cos(C) + cos(B-A)cos(C)
= 4cosAcosBcosC
Hence, we have successfully proved the given identity.
To prove this identity, we will use the properties of trigonometric functions and the sum-to-product identities. Let's go through the steps:
1. Start with the left-hand side (LHS) of the equation:
LHS = cos(-A+B+C) + cos(A-B+C) + cos(A+B-C) + cos(A+B+C)
2. Use the property that cosine of a negative angle is the same as the cosine of the positive angle:
cos(-A) = cos(A)
LHS = cos(A+B+C) + cos(A-B+C) + cos(A+B-C) + cos(A+B+C)
3. Rearrange the terms in LHS:
LHS = cos(A+B+C) + cos(A+B+C) + cos(A-B+C) + cos(A+B-C)
4. Use the property that cosine of the same angle is equal:
cos(X) + cos(X) = 2cos(X)
LHS = 2cos(A+B+C) + 2cos(A-B+C)
5. Apply the sum-to-product identity for the first two terms:
cos(X+Y) + cos(X-Y) = 2cos(X)cos(Y)
LHS = 2[cos(A)cos(B+C)] + 2cos(A-B+C)
6. Apply the sum-to-product identity for the last two terms:
cos(X+Y) + cos(X-Y) = 2cos(X)cos(Y)
LHS = 2[cos(A)cos(B+C)] + 2[cos(A)cos(B-C)]
7. Use the property that cosine of a sum of angles can be split into the product of cosines:
cos(X+Y) = cos(X)cos(Y) - sin(X)sin(Y)
LHS = 2[cos(A)cos(B)cos(C) - cos(A)sin(B)sin(C)] + 2[cos(A)cos(B)cos(C) + cos(A)sin(B)sin(C)]
8. Group similar terms together:
LHS = 2cos(A)cos(B)cos(C) - 2cos(A)sin(B)sin(C) + 2cos(A)cos(B)cos(C) + 2cos(A)sin(B)sin(C)
9. Combine like terms:
LHS = 4cos(A)cos(B)cos(C)
10. Therefore, the left-hand side (LHS) is equal to the right-hand side (RHS) of the equation. Hence, the equation is proven.
Hence, we have successfully proven that cos(-A+B+C) + cos(A-B+C) + cos(A+B-C) + cos(A+B+C) = 4cos(A)cos(B)cos(C).