if cos(A-B) + cos(B-C) + cos(C-A) = -3/2 Prove cos A + cos B + cos C = sin A + sin B + sin C = 0

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should convert the sum of two cosine functions into a product such that:

`cos(A-B) + cos(C-A) = 2cos (A-B+C-A)/2*cos (A-B-C+A)/2 `

`cos(A-B) + cos(C-A) = 2cos (B-C)/2*cos (2A-B-C)/2`

You may write `cos(B-C) = cos 2((B-C) /2)`  such that:

`cos 2((B-C) /2) = 2cos^2 ((B-C) /2) - 1`

Hence `cos(A-B) + cos(B-C) +cos(C-A) = 2cos (B-C)/2*cos (2A-B-C)/2 + 2cos^2 ((B-C) /2) - 1.`

But `cos(A-B) + cos(B-C) + cos(C-A) = -3/2` , hence `2cos (B-C)/2*cos (2A-B-C)/2 + 2cos^2 ((B-C) /2) - 1 = -3/2`

Factoring out `2cos (B-C)/2`  yields:

`2cos (B-C)/2*(cos (2A-B-C)/2 + cos ((B-C) /2)) = 1 - 3/2`

You may convert `cos (2A-B-C)/2 + cos (B-C) /2`  into a product such that:

`cos (2A-B-C)/2 + cos (B-C) /2 = 2cos (2A - B - C + B - C)/4*cos (2A - B - C - B + C)/4`

`cos (2A-B-C)/2 + cos (B-C) /2 = 2cos (2A-2C)/4*cos (2A-2B)/4`

`cos (2A-B-C)/2 + cos (B-C) /2 = 2cos (C-A)/2*cos (A-B)/2`

Hence, substituting `2cos (C-A)/2*cos (A-B)/2`  for `(cos (2A-B-C)/2 + cos ((B-C) /2))`   yields:

`2cos (B-C)/2*(2cos (C-A)/2*cos (A-B)/2) = -1/2 `

`4cos (A-B)/2*cos (B-C)/2*cos (C-A)/2 = -1/2 `

`cos (A-B)/2*cos (B-C)/2*cos (C-A)/2 = -1/8`

You need to prove that `cos A + cos B + cos C = 0 =gt cos A + cos C = -cos B`

Converting the sum into product yields:

`cos A + cos C = -cos B`

`2cos (A+C)/2*cos(A-C)/2 = -cos B`

Multiplying by `cos (A-B)/2*cos (B-C)/2`  both sides yields:

`-(2cos (A+C)/2)/8 = -cos B *cos (A-B)/2*cos (B-C)/2 `

`cos (A+C)/2 = 4cos B *cos (A-B)/2*cos (B-C)/2`

`cos^2(A+C)/2 = 16cos^2 B *cos^2 (A-B)/2*cos^2 (B-C)/2`

`sin A + sin C = -sinB`

`2sin(A+C)/2*cos(C-A)/2 = -sinB`

`sin(A+C)/2 = 4sin B *cos (A-B)/2*cos (B-C)/2`

Raising to square both sides yields:

`sin^2(A+C)/2 = 16sin^2 B *cos^2 (A-B)/2*cos^2 (B-C)/2`

`sin^2(A+C)/2 + cos^2(A+C)/2 = 16cos^2 (A-B)/2*cos^2 (B-C)/2*(sin^2 B + cos^2 B )`

`1 = 16cos^2 (A-B)/2*cos^2 (B-C)/2 =gt cos (A-B)/2*cos (B-C)/2 = -1/4`

Since, the problem provides the information that `cos (A-B)/2*cos (B-C)/2*cos (C-A)/2 = -1/8 =gt -1/4*cos (C-A)/2 = -1/8`

Hence, `cos (C-A)/2 = 1/2 =gt (C-A)/2 = pi/3 =gt C-A=2pi/3,`  that is correct since `cos(A-B) + cos(B-C) + cos(C-A) = -3/2 =gt cos A + cos B + cos C = sin A + sin B + sin C = 0.`

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

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cos(A-B) + cos(B-C) + cos(C-A) = -3/2

➯ cosAcosB+sinAsinB+cosBcosC+sinBsinC+cosCcosA

+sinCsinA=-3/2

➯ 2cosAcosB+2sinAsinB+2cosBcosC+2sinBsinC+2cosCcosA+

2sinCsinA+3=0

➯ 2cosAcosB+2sinAsinB+2cosBcosC+2sinBsinC+2cosCcosA+

2sinCsinA+sin²A+cos²A+sin²B+cos²B+sin²C+cos²C=0

➯ (sin²A+sin²B+sin²C+2sinAsinB+2sinBsinC+2sinCsinA)+(cos²A+cos²B+cos²C+2cosAcosB+2cosBcosC+2cosCcosA)=0

➯ (sinA+sinB+sinC)²+(cosA+cosB+cosC)²=0

∴ (sinA+sinB+sinC)² = 0 AND (cosA+cosB+cosC)²=0

(sinA+sinB+sinC) = 0 AND (cosA+cosB+cosC) = 0

∴  sinA+sinB+sinC = cosA+cosB+cosC = 0