# Prove this identity trigonometry?

(cos A - cos B) / (sin A + sin B) = (sin B - sin A) / (cos A + cos B)

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• TomV
Lv 7
2 years ago

(cos A - cos B) / (sin A + sin B) = (sin B - sin A) / (cos A + cos B)

LHS = (cosA - cosB)/(sinA + sinB)

{multiply by conjugate of numerator}

= (cos²A - cos²B)/[(sinA + sinB)(cosA + cosB)]

{apply Pythagorean Theorem to numerator and simplify}

= (sin²B - sin²A)/[(sinA + sinB)(cosA + cosB)]

{multiply by sinA - sinB}

= (sin²B - sin²A)(sinA - sinB)/[(sin²A - sin²B)(cosA + cosB)

= -(sin²A - sin²B)(sinA - sinB)/[(sin²A - sin²B)(cosA + cosB)

{cancel common term (sin²A - sin²B)}

= -(sinA - sinB)/(cosA + cosB)

{distribute negative}

= (sinB - sinA)/(cosA + cosB)

= RHS

QED

• TomV
Lv 7
2 years agoReport

The "multiply by sinA -sinB" step could more efficiently be accomplished by factoring (sin²B - sin²A) into (sinB + sinA)(sinB - sinA). 20-20 hindsight.

• cidyah
Lv 7
2 years ago

(cos A - cos B) / (sin A + sin B) = - 2 sin ((A+B)/2) sin ( (A-B)/2) / ( 2 sin ((A+B)/2) cos ((A-B)/2))

= - sin ((A-B)/2) / cos((A-B))/2)

= - tan ((A-B)/2)

(sin B - sin A) / (cos A + cos B) = 2 cos ((A+B)/2) sin ((B-A)/2) / (2 cos ((A+B)/2) cos ((A-B)/2)

= sin ((B-A)/2) / 3cos ((A-B)/2)

= - sin ((A-B)/2) / cos((A-B)/2)

= - tan((A-B)/2)

Left-hand side = Right-hand side

• 2 years ago

(cos A - cos B) / (sin A + sin B) = (sin B - sin A) / (cos A + cos B)

Cross-multiplying, we get

(cos A - cos B)*(cos A + cos B) = (sin B - sin A)*(sin A + sin B)

or (cos^2 A - cos^2 B) = (sin^2 B - sin^2 A)

or (cos^2 A + sin^2 A) = (sin^2 B + cos^2 B)

or 1 = 1

• 2 years ago

Multiplying to remove the denominators we have:

(cosA - cosB)(cosA + cosB) = (sinB - sinA)(sinA + sinB)

Expanding we get:

cos²A - cos²B = sin²B - sin²A

Now, sin²A => 1 - cos²A and sin²B => 1 - cos²B

i.e. cos²A - cos²B = 1 - cos²B - (1 - cos²A)

=> cos²A - cos²B = 1 - cos²B - 1 + cos²A

or, cos²A - cos²B = -cos²B + cos²A

so, LHS => RHS

:)>