SOLUTION: Tan a=-4/3, condition pi/2<a<pi Cos b = 1/2, condition. 0<b<pi/2 Need exact value cos(a+b). Sin(a-b). Tan(a-b). I believe a would be in quadrant 2

Question 854271: Tan a=-4/3, condition pi/2 Cos b = 1/2, condition. 0 Need exact value cos(a+b).
Sin(a-b).
Tan(a-b).
I believe a would be in quadrant 2.
And b would bi in quadrant 1.
How do I get tan b. or do I need sin a and cos a. If so how do I find them.

Thanks
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Tan a=-4/3, condition pi/2 Cos b = 1/2, condition. 0 Need exact value cos(a+b).
Sin(a-b).
Tan(a-b).
I believe a would be in quadrant 2 where x is negative and y is positive.
And b would bi in quadrant 1 where x and y are both positive.
How do I get tan b. or do I need sin a and cos a. If so how do I find them.
----
tan(a) = -4/3, so y = 4 and x = -3
Then r = sqrt(16 + 9] = 5
------
cos(b) = 1/2, so x = 1 and r = 2
Then y = sqrt(2^2-1) = sqrt(3)
And tan(b) = y/x = sqrt(3)/1 = sqrt(3)
--------
Need exact value
cos(a+b) = cos(a)cos(b)-sin(a)sin(b) = (-3/5)(1/2)-(4/5)(sqrt(3)/2)
-----------------------------------------------------
Sin(a-b) = sin(a)cos(b)+cos(a)sin(b) = (4/5)(1/2)+(-3/5)(sqrt(3)/2)
===================================
Tan(a-b) = [tan(a)+tan(b)]/[1-tan(a)tan(b)]
= [(-4/3)+sqrt(3)]/[1-(-4/3)sqrt(3)]
==============================================
Cheers,
Stan H.
================
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