Question #05b88

2 Answers
Sunayan S
Feb 8, 2018

#(sin theta+cos theta)=1#
#=>(sin theta+cos theta)^2=1#
#=>sin^2 theta+cos^2theta+2sintheta costheta=1#
#=>1+2sintheta costheta=1#
#=>sin2theta=0#

General solution:-

#2theta=n pi#
#theta=(n pi)/2{n in I}#

Hope it helps....
Thank you...

P dilip_k
Feb 8, 2018

#sin theta+ cos theta =1#

#=>1/sqrt2sin theta+1/sqrt2 cos theta =1/sqrt2#

#=>sin(pi/4)sin theta+cos(pi/4) cos theta =1/sqrt2#

#=>sin(pi/4)sin theta+cos(pi/4) cos theta =1/sqrt2#

#=>cos(theta-pi/4)=cos(pi/4)#

#=>theta-pi/4=2npipmpi/4#

So #theta=2npi or 2npi+pi/2 "where "n in ZZ#

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