Trigonometry question

  • #1
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

  • #2
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.....
 
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  • #3
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....
 
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  • #4
Thanks for taking the time to make my post readable chanvincent.
 
  • #5
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.)
 
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