SOLUTION: If tan α = − 4/3 and cot β = 15/8 for a second-quadrant angle α and a third-quadrant angle β, find the following. 1. sin (a+b) 2. cos (a+b) 3. tan (a+b) I got -5 for

I assume a is  and b is ...



The trig functions take ANGLES as inputs, you can not plug in the output of other trig functions as those are RATIOS.   As such, you need to find angles a and b first:



a =   (made use of "angle in Q2")

b =   (made use of "angle in Q3")



Now you can plug a and b into the equations:



1.  sin(a+b) = sin(126.870 + 208.072) = sin(334.942) = -0.424



etc.



 

We draw a picture of α in Quadrant II, and a picture of β in Quadrant III

     

From the ends of the terminal sides we draw vertical lines to the x-axis,
in green, forming right triangles:

   

We are given that tan α = − 4/3

Since we know that TANGENT = OPPOSITE/ADJACENT = y/x, we label the
opposite side (y) as the numerator of -4/3, which is -4, negative because
it goes left.  We label the adjacent (x) as the denominator of -4/3,
which is +3 because it goes upward from the x-axis. 

We are given that cot(β)= 15/8

Since we know that COTANGENT = ADJACENT/OPPOSITE = x/y, we label the
adjacent side (x) as the numerator of 15/8, thought of as (-15)/(-8) which is
-15, negative because
it goes downward.  We label the opposite (y) as the denominator of (-15)/(-8),
which is -8 because it goes left.


   

Next we find the missing sides, the terminal sides, using the Pythagorean theorem:

    

Note that we always take the terminal side r as positive.

   

Now we're finally ready to calculate

1. sin (a+b)
2. cos (a+b)
3. tan (a+b) 

where

 

and

 







Edwin

           

           As this is a 3-4-5 right-triangle, r = 5. This triangle exists in the 2nd quadrant 

Therefore,  



           

           As this is an 8-15-17 right-triangle, r = 17. This triangle exists in the 3rd quadrant, where cos and sin are < 0. 

Therefore, 

 

1.   

     

2.   

    

3.