trigonometry===2

(a)
sinA + sinB = a
2sin[(A + B)/2]•cos[(A - B)/2] = a ... (1)

cosA - cosB = b
2cos[(A + B)/2]•cos[(A - B)/2] ... (2)

(1)/(2):
sin[(A + B)/2]/cos[(A + B)/2] = a/b
tan[(A + B)/2] = a/b


(b)
tan(A + B)
= 2tan[(A + B)/2]/(1 - tan²[(A + B)/2]
= 2(a/b)/[1 - (a/b)²]
= (2a/b)/[(b² - a²)/b²]
= (2a/b) x [b²/(b² - a²)]
= 2ab/(b² - a²)

tan²(A + B) = 4a²b²/(b² - a²)²
sin²(A+B)/cos²(A + B) = 4a²b²/[(a²)² - 2a²b² + (b²)²]
sin²(A+B)/[1 - sin²(A + B)] = 4a²b²/[(a²)² - 2a²b² + (b²)²]
sin²(A+B)[(a²)² - 2a²b² + (b²)²] = 4a²b²[1 - sin²(A + B)]
sin²(A+B)[(a²)² - 2a²b² + (b²)²] = 4a²b² - 4a²b²sin²(A + B)
sin²(A+B)[(a²)² - 2a²b² + (b²)²] + 4a²b²sin²(A + B) = 4a²b²
sin²(A+B){[(a²)² - 2a²b² + (b²)²] + 4a²b²} = 4a²b²
sin²(A+B)[(a²)² + 2a²b² + (b²)²] = 4a²b²
sin²(A+B)(a² + b²)² = 4a²b²
sin²(A+B) = 4a²b²/(a² + b²)²

Since both A and B are acute, then 0 <(A + B) < 180°
sin(A + B) = 2ab/(a² + b²) or sin(A + B) = -2ab/(a² + b²) (rejected)


(c)
cos(A+B)
= sin(A + B)/tan(A + B)
= [2ab/(a² + b²)]/[2ab/(b² - a²)]
= [2ab/(a² + b²)] x [(b² - a²)/2ab]
= (b² - a²)/(a² + b²)

a)sinA + sinB
= 2[sin (A+B)/2] cos (A-B)/2 = a......(1)
cosA + cosB
= 2[cos (A+B)/2] cos (A-B)/2 = b......(2)
(1) / (2) :
tan (A+B)/2 = a/b

b) By a) tan (A+B)/2 = a/b , so
sin(A+B) = (2 a/b) / (1 + (a/b)^2)
= (2a/b) / [(a^2 + b^2) / b^2]
= 2ab / (a^2 + b^2)

c) cos(A+B)
= [(1 - (a/b)^2] / (1 + (a/b)^2)
= [(1 - (a/b)^2] / [(a^2 + b^2) / b^2]
= (b^2 - a^2) / (a^2 + b^2)



2010-03-08 02:29:54 補充：
A and B are both acute angles means cos A > 0 , cos B >0 ,

so cosA + cosB = b is Not = 0 for tan (A+B)/2 = a/b