How do you solve #Cos([θ/3]-[pi/4])=1/2#?

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Answer 1 · 2016-07-09T08:01:31

#theta=-45 ^o#

#"please remember that " cos (a-b)=cos a*cos b+sin a* sin b#

#cos (theta/3-pi/4)=cos theta*cos pi/4+sin theta*sin pi/4#

#cos pi/4=sin pi/4=sqrt2/2#

#cos((theta/3)-pi/4)=cos (theta/3)*sqrt2/2+sin( theta/3)*sqrt2/2#

#cos(theta/3)*sqrt2/2+sin (theta/3)*sqrt2/2=1/2#

#sqrt2/2(cos (theta/3) +sin (theta/3))=1/2#

#cos (theta/3)+sin (theta/3)=1/cancel(2)*cancel(2)/sqrt2#

#cos ( theta/3) +sin (theta/3)=1/sqrt2" ; "1/sqrt2=sqrt2/2#

#cos (theta/3)+sin (theta/3)=sqrt2/2#

#[cos (theta/3)+sin (theta/3)]^2=[sqrt2/2]^2#

#cos^2(theta/3)+2*sin(theta/3)*cos(theta/3)+cos^2(theta/3)=1/2#

#"so ;" cos^2 a+sin^2 a=1;#

#cos^2(theta/3)+sin^2(theta/3)=1#

#1+2*sin(theta/3)*cos(theta/3)=1/2#

#2*sin(theta/3)*cos(theta/3)=1/2-1#

#2*sin(theta/3)*cos(theta/3)=-1/2#

#"so ; "2*sin a*cos a=sin(2*a)#

#2*sin(theta/3)*cos(theta/3)=sin((2*theta)/3)#

#sin((2theta)/3)=-1/2#

#(2theta)/3=-30^o#

#2theta=-90#

#theta=-45 ^o#

Answer 2 · 2016-07-09T13:22:23

In #(0,2 pi), theta=(7pi)/4#.

The general value is #(6n+7/4)pi, n=0, +-1, +-2, +-3, ...#

The principal value of #(theta/3-pi/4)#

# =# the principal value of #cos^(-1)(1/2)=pi/3#.

So, this #theta# is given by#(theta/3-pi/4) =(pi/3#. And so, #theta=(7pi)/4#.

The general value is given by

#theta/3-pi/4=2npi+pi/3, n=0,+-1, +-2, =-3, ...#

S0, #theta = (6n+7/4)pi, n=0, +-1, +-2, +-3,...#