SOLUTION: Given that sin(A-B)/cos(A+B) = 1/3[tan(A-B)],
(i) show that the value of tanAtanB = -1/2
(ii) find the values of cosB if sinA = 1/2.
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-> SOLUTION: Given that sin(A-B)/cos(A+B) = 1/3[tan(A-B)],
(i) show that the value of tanAtanB = -1/2
(ii) find the values of cosB if sinA = 1/2.
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Question 1056313: Given that sin(A-B)/cos(A+B) = 1/3[tan(A-B)],
(i) show that the value of tanAtanB = -1/2
(ii) find the values of cosB if sinA = 1/2. Found 2 solutions by Edwin McCravy, htmentor:Answer by Edwin McCravy(20056) (Show Source):
There is something wrong with this problem.
Here's why I say this:
It is certainly true if A = B = 0
sin(0-0)/cos(0+0) = 1/3[tan(0-0)]
0/1 = 1/3[0]
0 = 0
Yet tanA*tanB = 0*0 = 0 , not -1/2.
Edwin
You can put this solution on YOUR website! (i) Using sum and difference identities, we can write sin(A-B)/cos(A+B) as:
(sinAcosB - cosAsinB)/(cosAcosB - sinAsinB)
If we divide top and bottom by cosAcosB we are left with:
(tanA - tanB)/(1 - tanAtanB)
And we are told this expression is equal to 1/3[tan(A-B)]
Using a difference identity for tan(A-B), and equating both sides we have:
(tanA - tanB)/(1 - tanAtanB) = (tanA - tanB)/(1-tanAtanB) = (1/3)[(tanA - tanB)/(1 + tanAtanB)]
Solve for tanAtanB:
3(1+tanAtanB) = 1 - tanAtanB
tanAtanB = -1/2
To solve (ii), we use the information already given
From (i) we know that tanAtanB = -1/2 -> tanB = -1/(2tanA)
since tanAtanB = sinAsinB/cosAcosB = -1/2, and we are given sinA = 1/2
[I'm stuck on this one at the moment...]
