SOLUTION: find the exact value of the trigonometric expression given that tanA=3/4 and cosB=-4/5, A is in Quadrant 1 and pi < B < 3pi/2
The problem is sin(A+B)
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Question 1117332: find the exact value of the trigonometric expression given that tanA=3/4 and cosB=-4/5, A is in Quadrant 1 and pi < B < 3pi/2
The problem is sin(A+B)
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find the exact value of the trigonometric expression given that tanA=3/4 and cosB=-4/5, A is in Quadrant 1 and pi The problem is sin(A+B)
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Sin(A+B) = sin(A)cos(B)+sin(B)cos(A)
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Since tan(A)=(y/x) = 3/4), sin(A) = (y/r) = 3/sqrt(3^2+4^2) = 3/5
and cos(A) =(x/4) = 4/5
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Since cos(B)= (x/r) = -4/5, sin(B)=(y/r) = sqrt(5^2-4^2)/5 = 3/5
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Ans: sin(A+B) = (3/5)(-4/5) + (3/5)(4/5) = 0
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Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ =\ \frac{3}{4})
}{\cos(A)}\ =\ \frac{3}{4})
\ =\ 3\cos(A))
\ =\ 9\cos^2(A))
\ =\ 9\(1\ -\ \sin^2(A)\))
\ =\ 9)
\ =\ \pm\sqrt{\frac{9}{25}})
\ >\ 0\ \Right\ \sin(A)\ =\ \frac{3}{5})
Use a similar argument to demonstrate:
\ =\ \frac{4}{5})
And once more:
\ =\ -\frac{4}{5})
\ =\ \frac{16}{25})
\ =\ 1\ -\ \frac{16}{25}\ =\ \frac{9}{25})
\ =\ \pm\sqrt{\frac{9}{25}})
\ <\ 0\ \Right\ \sin(B)\ =\ -\frac{3}{5})
======================
\ =\ \sin(A)\cos(B)\ +\ \sin(B)\cos(A))
Plug in the numbers and do the arithmetic. Mind your signs
John
My calculator said it, I believe it, that settles it
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