Simple minimum/maximum problem

[John O' Meara]

The voltage E applied to the sending end of a high-pressure transmitting line is connected to the voltage e at the receiving end by the equation E^2 = ( e cos \theta + a)^2 + ( e sin\theta +b)^2\\, where a and b are constants.
Expand the right-hand side of this equation and by expressing a cos\theta + b sin\theta \\ in the form R cos(\theta + \alpha) \\ show that the maximum and minimum values of R as \theta varies , are e +/- \sqrt{a^2 +b^2} \\. ( obtaining these results by differentation is a much more difficult method and we would remind our readers that the calculus is not always the best way of solving a problem.
Expanding this I get: E^2= e^2 cos^2\theta +a^2 +2ae cos\theta + e^2 sin^2\theta + 2eb sin\theta +b^2 \\, which gives E^2 = e^2 +2e(a cos\theta + b sin\theta ) +a^2 + b^2 \\, which can be expressed as
e^2 + 2e(a cos\theta + b cos(\frac{\pi}{2} - \theta) ) + a^2 + b^2. The thing is I do not know how to express R in terms of the coefficients a and b? Thanks for the help.

 

[D H]

You didn't do what you were asked to do.

You rewrote \sin\theta as \cos(\pi/2-\theta). That is not what the problem says to do. The problem specifically says to write a\cos\theta+b\sin\theta in the form R\cos(\theta+\alpha)[/tex]. Simply expand this latter form, equate to a\cos\theta+b\sin\theta, and solve for R and \alpha.

 

[Dick]

Check http://en.wikipedia.org/wiki/List_of_trigonometric_identities under the heading "Linear combinations".

 

[John O' Meara]

Expanding R cos(\theta + \alpha) \\ we getR( \cos\theta \sin\alpha - \sin\theta \sin\alpha) \\. Now R \cos\theta\cos\alpha - R \sin\theta\sin\alpha = a \cos\theta + b \sin\theta \\ When \theta = 0 R = \frac{a}{\cos\alpha} \\ When \theta = \frac{\pi}{2} R = \frac{-b}{\sin\alpha}\\ Therefore \tan\alpha = \frac{-b}{a}. I would appreciate help in getting R.

 

[Dick]

Ok, so a=R*cos(alpha) and b=-R*sin(alpha). Square and add both of those equations and use everybodies favorite trig identity.