How to prove the value of cos a + sin a is greater than 1 ?

2 Answers
Dec 26, 2016

For #2kpi < a < pi/2+2kpi# for #k=0,1,2,cdots#

Explanation:

If #cos(a)+sin(a) > 1# then #(cos(a)+sin(a) )^2=1+2 cos(a)sin(a) > 1# or
#cos(a)sin(a) > 0#
Concluding it is sufficient #cos(a)sin(a) > 0# to verify
#cos(a)+sin(a) > 1#
or
#2kpi < a < pi/2+2kpi# for #k=0,1,2,cdots#

Attached a plot showing

#cos(a)+sin(a)# in blue
#max(cos(a)sin(a),0)# in red and
the constant #1# in green.

enter image source here

Dec 26, 2016

#sin a + cos a in [-sqrt 2, sqrt 2]#.

Explanation:

For # a in (-oo, oo)#,

#cos a + sin a=sqrt2 (1/sqrt2cos a+1/sqrt2 sin a)#

#=sqrt2(sin (pi/4)cos a+cos(pi/4)sin a#

#=sqrt2sin(a+pi/4) in [-sqrt2, sqrt2]#

As a is not stated to be in an interval, I can assign #a = -pi/4# to get

0 for sin a + cos a.