Answer:
cos(a + b) = 
Step-by-step explanation:
cos(a + b) = cos(a).cos(b) - sin(a).sin(b) [Identity]
cos(a) = 
cos(b) = 
Since, terminal side of angle 'a' lies in quadrant 3, sine of angle 'a' will be negative.
sin(a) =
[Since, sin(a) =
]
= 
= 
Similarly, terminal side of angle 'b' lies in quadrant 2, sine of angle 'b' will be negative.
sin(b) =
= 
= 
By substituting these values in the identity,
cos(a + b) = 
= 
= 
= 
Therefore, cos(a + b) = 
+ 19 on both sides, -19 becomes 0, 1 becomes 20 x=20
The graph should look the the one I attached to this answer
Answer:

So then the integral converges and the area below the curve and the x axis would be 5.
Step-by-step explanation:
In order to calculate the area between the function and the x axis we need to solve the following integral:

For this case we can use the following substitution
and we have 

And if we solve the integral we got:

And we can rewrite the expression again in terms of x and we got:

And we can solve this using the fundamental theorem of calculus like this:

So then the integral converges and the area below the curve and the x axis would be 5.