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Trigonometry question

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.
 

Answers and Replies

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

[tex] E^2 = e^2(cos^2(x) + sin^2(x)) + a^2 + b^2 + 2e(acos(x) + bsin(x))[/tex]
[tex] E^2 = e^2 + a^2 + b^2 + 2e\sqrt{a^2 + b^2}(cos(\alpha)cos(x) + sin(\alpha)sin(x))[/tex]; where,
[tex] cos(\alpha) = a/\sqrt{a^2 + b^2}[/tex], and [tex] sin(\alpha) = b/\sqrt{a^2 + b^2} [/tex] ,and [tex] tan(\alpha) = b/a[/tex]. Therefore
[tex] E^2 = e^2 + a^2 + b^2 +2e\sqrt{a^2 + b^2}(cos(x - \alpha))[/tex], so I get [tex]R = 2e\sqrt{a^2 + b^2}[/tex] not what it is claimed above. What is the next step? Many thanks.
Okay, I've finished my part... someone gonna help him.....
 
Last edited:
I believe you mis-understand what the question is asking......

The question is asking you to express [tex]acos(x) + bsin(x)[/tex] in term of [tex] Rcos(x+\alpha)[/tex], then show the maximum and minimum value of E is [tex] e+/-\sqrt{a^2+b^2}[/tex], That would be more make sense....
 
Last edited:
Thanks for taking the time to make my post readable chanvincent.
 
Expanding [tex] a cos(x) + b sin(x) [/]tex] I get :
[tex] = \sqrt{a^2 + b^2}(a/\sqrt{a^2 + b^2}cos(x) + b/\sqrt{a^2 + b^2}sin(x))
= \sqrt{a^2 + b^2}(sin(\alpha)cos(x) + cos(\alpha)sin(x))[/tex], where [tex] \(alpha)=tan(a/b)[/tex], this expands out to: [tex] \sqrt{a^2 + b ^2}(1/2( sin(\(alpha) + x) + sin(\alpha - x)) + 1/2( sin(\alpha + x) - sin(\alpha -x)))[/tex]
, which reduces to [tex] \sqrt{a^2 + b^2}(sin(\alpha + x))[/tex].The question is how do you expand it to get [tex] R cos(x + \alpha)[\tex], and what do you do then.

(Edited by HallsofIvy to fix tex. John, you put [ \tex ] when you should have [ /tex ] to end the tex.)
 
Last edited by a moderator:

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