SOLUTION: Hi, I need help with my homework. We are learning about the sum and difference of identities.
Here is the problem:
What is the value of sin(A-b) if tan A = 9/40 and A is in t
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Question 602906: Hi, I need help with my homework. We are learning about the sum and difference of identities.
Here is the problem:
What is the value of sin(A-b) if tan A = 9/40 and A is in the first quadrant, while cos B = -4/5 and B is in the third quadrant?
I really don't get this homework. What is up with A and B? Do we need to plot the graph as well? It's confusing me. I'm sorry if I'm asking stupid questions, I missed our class today, you see. I hope someone can explain this to me properly so I'll catch up with the lesson. Thank you!
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Here is the problem:
What is the value of sin(A-b) if tan A = 9/40 and A is in the first quadrant, while cos B = -4/5 and B is in the third quadrant?
I really don't get this homework. What is up with A and B? Do we need to plot the graph as well?
===========================
Here's the one I just did with the same angles.
------------
If tan A = 9/40, A is in the first quadrant, cos B = -4/5, B is in the third quadrant, find the value of cos(A+B)
--------------
Find the cos(A)
Tan = y/x
Plot the point (40,9)
--> hyp = sqrt(40^2 + 9^2) = 41
cos(A) = x/hyp = 9/41
-------------
cos(A + B) = cosA*cosB - sinA*sinB (from Wikipedia)
Find sin(A) and sin(B)
-------
sin(A) = y/hyp = 40/41
sin(B) = -3/5 in Q3 (it's the 3, 4, 5 triangle)
--------------
cos(A + B) = cosA*cosB - sinA*sinB
= (9/41)*(-4/5) - (40/41)*(-3/5)
= -36/205 + 120/205
= 84/205
=============================
Use some of the values found in the previous problem:
sin(A-b) = sin(A)cos(B) - cos(A)sin(B)
= (40/41)*(-4/5) - (9/41)*(-3/5)
= -200/205 + 27/205
= -173/205
------------
What is up with A and B?
Angles A & B are measured from the x-axis counter-clockwise. All angles on the x-y plane are measured that way.
------
Do we need to plot the graph as well? You don't have to if you don't want to.
Not much to graph, just a point for each angle.
For A draw a line from the Origin to (40,9)
For B it's to (-3,4)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
A is an angle and they tell you it is in the first quadrant and that the tangent is 
B is another angle. You are given that this angle is in the third quadrant and that the cosine is 
They want to subtract the measure of angle B from the measure of angle A and then take the sine of the difference.
For such an operation you need the difference formula for sine:
\ =\ \sin(A)\cos(A)\ -\ \cos(A)\sin(B))
Remember that tangent is sine over cosine, so:
}{\cos(A)}\ =\ \frac{9}{40})
So
\ =\ 9\cos(A))
Square both sides:
\ =\ 81\cos^2(A))
Use the Pythagorean Identity and substitute:
\ =\ 81\cos^2(A))
Collect like terms:
\ =\ \frac{1600}{1681})
Take the positive (remember, QI) square root:
\ =\ \frac{40}{41})
And
\ =\ \frac{81}{1681})
Again, the positive square root:
 =\ \frac{9}{41})
Similarly you can get from \ =\ -\frac{3}{5})
to \ =\ -\frac{4}{5})
. QIII, both sine and cosine are negative.
Now you have the four values you need to plug into:
You can do your own arithmetic.
Then, for futher study, here are the other three formulas you will need:
Sum and Difference Formulas
John
My calculator said it, I believe it, that settles it
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