An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum values of I

take the derivative with respect to t
[tex] - w \sin(wt) + \sqrt{8} w cos(wt)[/tex]
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
[tex]0 = -w \sin(wt) + \sqrt{8} w cos(wt)[/tex]
divide by w
[tex]0 =- \sin(wt) + \sqrt{8} cos(wt)[/tex]
we add sin(wt) to both sides

[tex]\sin(wt)= \sqrt{8} cos(wt)[/tex]
divide both sides by cos(wt)
[tex] \frac{sin(wt)}{cos(wt)}= \sqrt{8} \\ \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\ \\ wt=arctan(2 \sqrt{2)} [/tex] OR[tex] \\ [/tex] [tex]wt=arctan( { \frac{1}{ \sqrt{2} } ) [/tex]
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

[tex]I=cos(2n \pi -2arctan( \sqrt{2} ))[/tex]
since 2npi is just the period of cos
[tex]cos(2arctan( \sqrt{2} ))= \frac{-1}{3} [/tex]
substituting our second soultion we get
[tex]I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))[/tex]
since 2npi is the period
[tex]I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3} [/tex]
so the maximum value =[tex] \frac{1}{3} [/tex]
minimum value =[tex]- \frac{1}{3} [/tex]