Question #d76e9

2 Answers

Answer:

#0#

Explanation:

#sin2x=1#, #2x=pi/2# or #x=pi/4#

Hence

#u=cos(8*pi/4)+2cos(6*pi/4)+cos(4*pi/4)#

#=cos(2pi)+2cos((3pi)/2)+cos(pi)#

#=1+2*0-1#

#=0#

1) I found #x# for #sin2x=1# condition.

2) I substituted to #x=pi/4#.

Jul 24, 2017

Answer:

#0#.

Explanation:

#sin(2x)=1#.

In general, equations of the form #sin(theta)=K# and solved by #theta = arcsin(K)+2npi# and #pi-theta=arcsin(K)+2npi#, #n in ZZ#, because of the periodicity of the sine function#.

Note that #arcsin(1)=pi/2#. This means that if #sin(2x)=1#,

#2x=pi/2+2npi#,
#x=pi/4+npi#.

#pi-2x=pi/2+2npi#,
#2x=pi/2-2npi#,
#x=pi/4-npi#.

As #n in ZZ# these solution sets are identical.
So if #sin(2x)=1# then #x=pi/4+npi#, #n in ZZ#.

Then

#cos(8(pi/4+npi)+2cos(6(pi/4+npi))+cos(4(pi/4+npi))#,
#cos(2pi+4(2npi))+2cos((3pi)/2+3(2npi))+cos(pi+2(2npi))#.

Multiplies of the periodicity can be removed and the values substituted to give,

#1+2*0-1=0#.

In this case, it was safe to assume #sin(2x)=1 => x=pi/4#. This will not always be the case! So it is good to solve more generally.