Question 923530: Find the exact value, given that cos A= 1/3, with A in quadrant I, and sin B= -1/2, with B in quadrant IV, and sin C= 1/4, with C in quadrant II.
sin(A-B)
I don't understand this problem. Why do they give sin C? Please help me.
Found 2 solutions by lwsshak3, jim_thompson5910:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the exact value, given that cos A= 1/3, with A in quadrant I, and sin B= -1/2, with B in quadrant IV, and sin C= 1/4, with C in quadrant II.
sin(A-B)
***
cosA=1/3
sinA=√(1-cos^2A)=√(1-1/9)=√(8/9)=√8/3
..
sinB=-1/2
cosB=√(1-sin^2B)=√(1-1/4)=√(3/4)=√3/2
..
Use sin addition formulas:
sin(A-B)=sinAcosB-cosAsinB=√8/3*√3/2-1/3*-1/2=√24/6+1/6=(√24+1)/6
..
Check:
cosA=1/3 (Q1)
A≈70.53˚
sinB=-1/2(Q4)
B=330˚
A-B=70.53-330=-259.47
sin(A-B)=sin(-259.47)≈0.9831(in quadrant II where sin>0)
Exact value as computed=(√24+1)/6≈0.9831
Note: sin C is not required for this problem.
Answer by jim_thompson5910(35256) (Show Source):
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