Find theta angle in complex form

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Homework Help Overview

The discussion revolves around finding the angle theta in a complex equation involving trigonometric functions. Participants are exploring methods to solve for theta and comparing their results with a textbook answer.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss substituting trigonometric identities into the equation and simplifying to a quadratic form. There is also an exploration of plotting values to find roots and comparing results with expected answers.

Discussion Status

Some participants have provided alternative approaches and expressed differing results from the textbook answer. There is an ongoing examination of the validity of the textbook solution, with no clear consensus reached on the correct value of theta.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. There is mention of discrepancies between calculated values and those found in the textbook.

baby_1
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Homework Statement


Here is my equation that I want to find theta angle
gif.gif


Homework Equations

The Attempt at a Solution


I try to set different value of cos (theta) to find theta but it failed , I want to know the main solving strategy
Thanks
 
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What if you were to substitute cos(theta) = x and sin(theta) = y, and enforce the fact that y = sqrt(1-x^2)? Usually these eventually simplify to a quadratic equation to solve--after some work.
 
Thanks for your replay
I did with the same way:
%7B2%7D%5Ccos%28%5Ctheta%29%29%7D%7B%5Csin%28%5Ctheta%29%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D.gif

D%7B2%7D%5Ccos%28%5Ctheta%29%29%3D%5Csin%28%5Ctheta%29%3D%5Csqrt%7B1-%5Ccos%5E2%28%5Ctheta%29%7D.gif

gif.gif

gif.gif

as I plot these charts I have
file.jpg


It has two roots (+.63,-.63) but the both of them indicate that theta should be 53 degree that is opposite the my books answer that theta should be 12 degree

what is my fault?
 
No fault. I don't get 53 degrees but 50.9612 (cos = 0.629847 so like you found it).
Satisfies your equation.

Looks as if the book answer is wrong.
 
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I concur with BvU. 12 degrees does not satisfy the equation.
cos(pi/2*cos( 12 ))/sin( 12 ) = 0.1651,which does not equal sqrt(2)/2 = 0.7071.
 
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