SOLUTION: if cosA=(-7/25) and sinB=(4/5), with A in QIII and B in QII, find the exact value of the following expresions please and thank you!! sin(B/2) sin(A+B) cos^2(B/2) tan(A-B)

Question 679209: if cosA=(-7/25) and sinB=(4/5), with A in QIII and B in QII, find the exact value of the following expresions please and thank you!!
sin(B/2)
sin(A+B)
cos^2(B/2)
tan(A-B)
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
if cosA=(-7/25) and sinB=(4/5), with A in QIII and B in QII, find the exact value of the following expressions please and thank you!!
sin(B/2)
sin(A+B)
cos^2(B/2)
tan(A-B)
**
In QIII, cos and sin<0, tan>0
cosA=-7/25=adj side/hypotenuse
opp side=-√(25^2-7^2)=-√(625-49)=-√576=-24
sinA=opp side/hypotenuse=-24/25
tanA=opp side/adj side=-24/-7=24/7
..
In QII, cos and tan<0, sin>0
sinB=4/5 (you are working with a 3-4-5 right triangle)
cosB=-3/5
tanB=-4/3
..
Identity: sin(x/2)=�√[(1-cosx)/2]
sin(B/2)=√[(1-cosB)/2]
=√[(1-(-3/5))/2]
=√[(1+3/5)/2]
=√[8/5)/2]
=√[8/10]
=√.8
..
Identity: sin(A+B)=sinA cosB+cosA sinB
=(-24/25)*(-3/5)+(-7/25)*(4/5)
=(72/125)-28/125
=44/125
..
B/2=�√[(1+cosB)/2]
cos^2(B/2)=(1+cosB)/2=(1+(-3/5))/2=(2/5)/2=2/10=0.2
..
tan(A-B)=(tanA-tanB)/(1+tanA tanB)
I will let you do this one
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