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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