Prove this identity trigonometry?
(cos A - cos B) / (sin A + sin B) = (sin B - sin A) / (cos A + cos B)
4 Answers
- TomVLv 72 years agoFavorite Answer
(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
- cidyahLv 72 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
- PrakashLv 52 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
- Wayne DeguManLv 72 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
:)>
The "multiply by sinA -sinB" step could more efficiently be accomplished by factoring (sin²B - sin²A) into (sinB + sinA)(sinB - sinA). 20-20 hindsight.