How do you solve #Cos(theta) - sin(theta) = sqrt 2#?

1 Answer
Bdub
Apr 18, 2016

#S={-45^@+360^@n} #

Explanation:

Use the following formulas to transform the equation then solve
#Acosx+Bsinx=C cos(x-D)#
#C=sqrt(x^2+y^2), cosD=A/C, sinD=B/C#

#A=1, B=-1, C=sqrt2#
#cosD=1/sqrt2->D=cos^-1(1/sqrt2)=-45^@#

Note that cosine is positive but sine is negative therefore the triangle is in quadrant four so our angle D will either be #-45^@ or 315^@#. I pick the #-45^@#

#sqrt2 cos(theta+45^@)=sqrt2#

#cos(theta+45^@)=1#

#theta+45^@+=cos^-1 1#

#theta+45^@=0^@+360^@n#

#theta=-45^@+360^@n#

#S={-45^@+360^@n} #

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