Question 1046265
.
what is the minimum & maximum value of sin(x)+cos(x)
~~~~~~~~~~~~~~~~~~~~~~~~~~~


The maximum is {{{sqrt(2)}}}.


The minimum is {{{-sqrt(2)}}}.


There are many different ways to prove it.


Below is one of the ways.


Let y = sin(x) + cos(x).


Then 


y^2 = (sin(x) + cos(x))^2 = sin^2(x) + 2sin(x)cos(x) + cos^2(x) = 1 + 2sin(x)cos(x) = 1 + sin(2x).       (1)


So,  y,  and hence,  sin(x) + cos(x)  is maximal  when sin(2x) = 1.
It happens when  2x = {{{pi/2}}},  or  x = {{{pi/4}}}.


At this value of  x,  y^2 = 2  and hence  y = {{{sqrt(2)}}}  is the maximum of  sin(x) + cos(x).


From (1),  &nbsp;{{{y^2}}} <= 2.  &nbsp;Hence, &nbsp;|y| <= {{{sqrt(2)}}},  or  {{{-sqrt(2)}}} <= y <= {{{sqrt(2)}}}.


Finally,  &nbsp;sin(x) + cos(x) = {{{-sqrt(2)}}} &nbsp;at &nbsp;x = {{{5pi/4}}}.


It proves that the maximum is &nbsp;{{{sqrt(2)}}} &nbsp;and the minimum is &nbsp;{{{-sqrt(2)}}}.

<TABLE> 
  <TR>
  <TD> 

{{{graph( 330, 330, -6.5, 6.5, -2.5, 2.5,
          sin(x) + cos(x)
)}}}


Plot y = sin(x) + cos(x)

  </TD>
  </TR>
</TABLE>