SOLUTION: 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 t

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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): You can put this solution on YOUR website!
You don't need to worry about C in this problem. My guess is that it applies to another problem.

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First we need to find sin(A) based off of the given cos(A) = 1/3

















A is in quadrant I, so sin(A) is positive








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Then we need to find cos(B) based on sin(B) = -1/2

















B is in quadrant IV, so cos(B) is positive







---------------------------------------------------------------

Expand out sin(A-B) using an identity and then do substitutions.





























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Let me know if that helps or not. Thanks.

If you need more help, feel free to email me at jim_thompson5910@hotmail.com

My Website: http://www.freewebs.com/jimthompson5910/home.html

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