Problem (1) For what value of #x#, #sinx+cosx=2# Problem (2) For what value of #x#, #sinxcosx=1#?

1 Answer
Mar 14, 2017

#(sinx+cosx)# can never be #2#

and #sinxcosx# can never be #1#.

Explanation:

Problem (1)

#sinx+cosx# can never be #2#, the maximum value of #sinx+cosx# can only be #sqrt2# as

#sinx+cosx=sqrt2(sinx xx 1/sqrt2+cosx xx 1/sqrt2)#

= #sqrt2(sinxcos(pi/4)+cosxsin(pi/4))#

= #sqrt2sin(x+pi/4)#

as maximum value of any sine ratio can only be #1#,

maximum value of #sqrt2sin(x+pi/4)# or #sinx+cosx# can only be #sqrt2#.

Hence #sinx+cosx# can never be #2#.

Problem (2)

As #sinxcosx=1/2(2sinxcosx)=1/2xxsin2x#

and again as maximum value of any sine ratio can only be #1#,

maximum value of #sinxcosx# or #1/2sin2x# can only be #1/2#

and it can never be more than #1/2# and hence cannot be #1#.