F.4 A.math 一題

sinA = 3/5

A is acute, then
cosA
= √[1 - sin²A]
= √[1 - (3/5)²]
= √16/25
= 4/5

cos(A + B) = -7/25
cosAcosB - sinAsinB = -7/25
(4/5)cosB - (3/5)sinB = -7/25
20cosB - 15sinB = -7
20√(1 - sin²B) = 15sinB - 7
[20√(1 - sin²B)]² = [15sinB - 7]²
400 - 400sin²B = 225sin²B - 210sinB + 49
625sin²B - 210sinB - 351 = 0
(125sinB - 117)(5sinB + 3) = 0
sinB = 117/125 or sinB = -3/5 (rejected)

Ans: sinB = 117/125

Refer to the question. Notice that both A and B are acute (< 90 degrees).

Therefore, all of sinA, cosA, sinB and cos B must be POSITIVE. All NEGATIVE sinB should be REJECTED.

sinA = 3/5 and cos(A+B)=－7/25
it take four case
(1)cosA=4/5,sin(A+B)=24/25
sinB=sin(A+B－A)=sin(A+B)cosA-sinAcos(A+B)
=(24/25)*(4/5)－(3/5)*(－7/25)=117/125

(2)cosA=4/5,sin(A+B)=－24/25
sinB=sin(A+B－A)=sin(A+B)cosA－sinAcos(A+B)
=(－24/25)*(4/5)－(3/5)*(－7/25)=(－96+21)/125=－3/4

(3)cosA=－4/5,sin(A+B)=24/25
sinB=Sin(A+B－A)=sin(A+B)cosA－sinAcos(A+B)
=(24/25)*(－4/5)－(3/5)*(－7/25)=(－96+21)/125=－3/4

(4)cosA=－4/5,sin(A+B)=－24/25
sinB=sin(A+B－A)=sin(A+B)cosA－sinAcos(A+B)
=(－24/25)*(－4/5)－3/5*(－7/25)=(96+21)/125=117/125



2009-03-05 21:28:55 補充：
SinA=3/5 ==>CosA=4/5
OR
SinA=3/5==>CosA=土4/5

Since sin A = 3/5, then cos A = 4/5.
cos ( A + B) = -7/25, then sin (A + B) = 24/25. That is
cos A cos B - sin A sin B = -7/25
4/5 cos B - 3/5 sin B = -7/25
4 cos B - 3 sin B = -7/5....................(1)
sin A cos B + cos A sin B = 24/25
3/5 cos B + 4/5 sin B = 24/25
3 cos B + 4 sin B = 24/5...................(2)
Solving the simultaneous equations (1) and (2) for sin B,
we get sin B = 117/125.