Let tan(x)=2/5 what is the value of tan(2-x)

Hello from MrBillDoesMath!

-2/5

Discussion:

Generally,

tan(2*PI - x ) = - tanx   =>          (*)

and in our case

tan(2*PI - x ) = - tanx   = - (2/5) = -2/5

Proof of (*)

tan (2*PI-x) = sin (2*Pi-x)/ cos(2*Pi-x)

Now

     sin(2*Pi) * cos(-x) - cos(2*Pi) sin(x) =>  ( sin(2*PI) = 0, cos(2*Pi) = 1)

                     = -sin(x)

     cos(2*Pi-x) =  cos(2*Pi) * cos(-x) + sin(2*Pi)* sin(-x) =>

                            1 * cos(-x)              +  0  =

                            1 * cos(x)  = cos(x)

so tan (2*PI-x) = (-sin(x)) / cos(x) = -tanx

Regards,  

MrB

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!