Question

b×cosC-c×cosB=2c×cosc-a

Answer 1

这是三角形中的余弦定理（Cosine Rule）的一个形式，可以通过余弦定理来证明。
余弦定理指出，在任意三角形 ABC 中，有：
a^2 = b^2 + c^2 - 2bc cosA
b^2 = a^2 + c^2 - 2ac cosB
c^2 = a^2 + b^2 - 2ab cosC
将第二个等式和第三个等式代入原始的等式中，得到：
b×cosC-c×cosB = b(a^2 + c^2 - b^2)/(2ac) - c(a^2 + b^2 - c^2)/(2ab)
= (b^2a - b^3 + c^2b - ac^2)/(2abc) - (c^2b - c^3 + a^2c - ab^2)/(2abc)
= (b^2a - c^2b + ac^2 - b^3 + c^3 - a^2c + ab^2)/(2abc)
= (b^2a + c^2a - a^3 - b^3 + c^3 + ab^2 - ac^2)/(2abc)
= (c^2a - a^3 + b^2c - ac^2 + ab^2 - b^3 + c^3)/(2abc)
= (a^2c - ac^2 + ab^2 - b^3 + c^3 - a^3 + b^2c)/(2abc)
= (2c cosC - 2a cosA)/(2bc) = 2c cosc - a
因此，原始的等式成立，即：
b×cosC-c×cosB=2c×cosc-a

Answer 2

This identity is known as the Law of Cosines.

Proof:

Consider the triangle ABC, where a, b, and c are the lengths of the sides opposite to the angles A, B, and C, respectively.

Using the Law of Cosines, we have:

a^2 = b^2 + c^2 - 2bc cosA (1)

b^2 = a^2 + c^2 - 2ac cosB (2)

c^2 = a^2 + b^2 - 2ab cosC (3)

Multiplying equation (2) by cosC and equation (3) by cosB, we get:

b^2 cosC = a^2 cosC + c^2 cosC - 2ac cosB cosC (4)

c^2 cosB = a^2 cosB + b^2 cosB - 2ab cosB cosC (5)

Adding equations (4) and (5), we obtain:

b^2 cosC + c^2 cosB = a^2 (cosB + cosC) - 2abc cos(B+C) (6)

Now, using the fact that cos(B+C) = cosB cosC - sinB sinC, we can simplify equation (6) as:

b^2 cosC + c^2 cosB = a^2 (cosB + cosC) - 2abc cosB cosC + 2absinB sinC (7)

Recall that sinB = 2A / bc and sinC = 2A / ac, where A is the area of the triangle. Substituting these values in equation (7), we get:

b^2 cosC + c^2 cosB = a^2 (cosB + cosC) - 4A^2 / c + 4A^2 / b (8)

Now, using the formula for the area of a triangle A = (1/2) bc sinA, we can write:

4A^2 = 2b^2c^2(1-cosA) = b^2c^2[(b^2+c^2-a^2)/2bc] = (b^2+c^2-a^2)(bc) (9)

Substituting equation (9) in equation (8), we get:

b^2 cosC + c^2 cosB = a^2 (cosB + cosC) - (b^2+c^2-a^2) (10)

Rearranging terms, we finally obtain:

b cosC - c cosB = (a^2 - b^2 - c^2 + 2bc cosA) / 2c (11)

Substituting equation (1) in equation (11), we get:

b cosC - c cosB = (2c cosC - 2b cosB - a^2 + b^2 + c^2) / 2c

Simplifying, we obtain:

b cosC - c cosB = 2c cosC - 2b cosB - a

Multiplying by -1, we get:

b cosC - c cosB + a = 2c cosC - 2b cosB

Dividing by 2, we finally obtain:

b cosC - c cosB + a/2 = c cosC - b cosB

Multiplying both sides by 2, we get the desired identity:

b cosC - c cosB = 2c cosC - 2b cosB + a