Question #ed504

2 Answers
Nov 23, 2016

We start by squaring both sides to get rid of the square root.

#(costheta- sin theta)^2 = (sqrt(cos(pi/4 + theta))^2#

#cos^2theta - 2costhetasintheta + sin^2theta = cos(pi/4 + theta)#

We now use the identities #cos^2beta + sin^2beta = 1# and #cos(A + B) = cosAcosB - sinAsinB# to simplify.

#1 - 2costhetasintheta = cos(pi/4)costheta - sin(pi/4)sin theta#

#1 - 2costhetasintheta = costheta/sqrt(2) - sintheta/sqrt2#

As you can see, the two sides aren't equal, so the identity is false.

Hopefully this helps!

Nov 23, 2016

I think the question should be-
Prove that

#costheta-sintheta=sqrt2cos(pi/4+theta)#

LHS

#=costheta-sintheta#

#=sqrt2(1/sqrt2costheta-1/sqrt2sintheta)#

#=sqrt2(cos(pi/4)costheta-sin(pi/4)sintheta)#

#=sqrt2cos(pi/4+theta)=RHS#

Proved