How do you solve #sin (x + (π/4)) + sin (x - (π/4)) = 1#?

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Answer 1 · 2015-07-22T12:11:02

#x=(-1)^n(pi/4)+npi" ", n in ZZ#

We use the identity(otherwise called the Factor Formula) :

#sinA +sinB=2sin((A+B)/2)cos((A-B)/2)#

Like this :

#sin (x + (pi/4)) + sin (x - (pi/4)) = 2sin[((x+pi/4)+(x-pi/4))/2]cos[(x+pi/4-+(x-pi/4))/2]= 1#

#=>2sin((2x)/2)cos((2*(pi/4))/2)=1#

#=>2sin(x)cos(pi/4)=1#

#=>2*sin(x)*sqrt(2)/2=1#

#=>sin(x)=1/sqrt(2)=sqrt(2)/2#

#=>color(blue)(x=pi/4)#

The General Solution is : #x= pi/4 + 2pik# and #x=pi-pi/4 +2pik=pi/4 + (2k+1)pi" ",k in ZZ#

You can combine the two sets of solution into one as follows :

#color(blue)(x=(-1)^n(pi/4)+npi)" ",n in ZZ#