How Does Trigonometry Relate Voltage Levels in Transmission Lines?

[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(x) + a)^2 + (e*sin(x) + b)^2, where a and b are constants. Expand the right-hand side of the equation and by expressing a*cos(x) + b*sin(x) in the form R*cos(x + alpha) show that the maximum and minimum values of R, as x varies, are e +/-sqr(a^2 + b^2)? On expanding I get the following:

E^2 = e^2*((cos(x))^2 + (sin(x))^2) + a^2 + b^2 + 2*e*(a*cos(x) + b*sin(x))
E^2 = e^2 + a^2 + b^2 + 2*e*sqr(a^2 + b^2)*(cos(alpha)*cos(x) + sin(alpha)*sin(x)); where,
cos(alpha) = a/sqr(a^2 + b^2), and sin(alpha) = b/sqr(a^2 + b^2), and tan(alpha) = b/a. Therefore
E^2 = e^2 + a^2 + b^2 +2*e*sqr(a^2 + b^2)(cos(x - alpha)), so I get R = 2*e*sqr(a^2 + b^2) not what it is claimed above. What is the next step? Many thanks.

 

[chanvincent]

Let me tidy up your equation.. otherwise no one is able to read that...

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=(ecos(x) + a)^2 + (esin(x) + b)^2, where a and b are constants. Expand the right-hand side of the equation and by expressing acos(x) + bsin(x) in the form Rcos(x + \alpha) show that the maximum and minimum values of R, as x varies, are e +/-\sqrt{a^2 + b^2}? On expanding I get the following:

E^2 = e^2(cos^2(x) + sin^2(x)) + a^2 + b^2 + 2e(acos(x) + bsin(x))
E^2 = e^2 + a^2 + b^2 + 2e\sqrt{a^2 + b^2}(cos(\alpha)cos(x) + sin(\alpha)sin(x)); where,
cos(\alpha) = a/\sqrt{a^2 + b^2}, and sin(\alpha) = b/\sqrt{a^2 + b^2} ,and tan(\alpha) = b/a. Therefore
E^2 = e^2 + a^2 + b^2 +2e\sqrt{a^2 + b^2}(cos(x - \alpha)), so I get R = 2e\sqrt{a^2 + b^2} not what it is claimed above. What is the next step? Many thanks.

Okay, I've finished my part... someone going to help him...

 

[chanvincent]

I believe you mis-understand what the question is asking...

The question is asking you to express acos(x) + bsin(x) in term of Rcos(x+\alpha), then show the maximum and minimum value of E is e+/-\sqrt{a^2+b^2}, That would be more make sense...

 

[John O' Meara]

Thanks for taking the time to make my post readable chanvincent.

 

[John O' Meara]

Expanding a cos(x) + b sin(x) [/]tex] I get :<br /> = \sqrt{a^2 + b^2}(a/\sqrt{a^2 + b^2}cos(x) + b/\sqrt{a^2 + b^2}sin(x))&lt;br /&gt; = \sqrt{a^2 + b^2}(sin(\alpha)cos(x) + cos(\alpha)sin(x)), where \(alpha)=tan(a/b), this expands out to: \sqrt{a^2 + b ^2}(1/2( sin(\(alpha) + x) + sin(\alpha - x)) + 1/2( sin(\alpha + x) - sin(\alpha -x))) <br /> , which reduces to \sqrt{a^2 + b^2}(sin(\alpha + x)).The question is how do you expand it to get R cos(x + \alpha)[\tex], and what do you do then.&lt;br /&gt; &lt;br /&gt; (Edited by HallsofIvy to fix tex. John, you put [ \tex ] when you should have [ /tex ] to end the tex.)