Write #sin(x-(3pi)/4)# in terms of #sinx#?

Question

10573 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-11T14:56:58

#sin(x-(3pi)/4)=-1/sqrt2(sinx+sqrt(1-sin^2x))#

As #sin(A-B)=sinAcosB-cosAsinB#

Hence #sin(x-(3pi)/4)=sinxcos((3pi)/4)-cosxsin((3pi)/4)#

= #sinxcos(pi-pi/4)-cosxsin(pi-pi/4)#

= #-sinxcos(pi/4)-cosxsin(pi/4)#

= #-sinx xx1/sqrt2-cosx xx1/sqrt2#

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

= #-1/sqrt2(sinx+sqrt(1-sin^2x))#