Trigonometric Identities for Sine and Cosine

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SUMMARY

The discussion centers on solving the equation \(4 \sin \theta = 3 \csc \theta\). The correct manipulation leads to \( \sin^2 \theta = \frac{3}{4} \), resulting in \( \sin \theta = \pm \frac{\sqrt{3}}{2} \). The angles corresponding to these sine values are 30 degrees (Q1), 150 degrees (Q2), 210 degrees (Q3), and 330 degrees (Q4). The participant initially confused sine and cosine values but clarified their understanding with the help of others in the forum.

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Homework Statement



[itex]4 \ sin \ \theta \ = \ 3 \ csc\ \theta[/itex]

The Attempt at a Solution



[itex]sin\ \theta \ = \ \frac {3}{4} \ csc \ \theta[/itex]

[itex]sin^2 \ \theta \ = \ \frac {3}{4}[/itex]

[itex]sin \ \theta \ = \ \pm \ \frac {\sqrt{3}}{2}[/itex]

[itex]30 \ \deg \ in \ QI, \ 150 \ \deg \ in \ QII, \ 210 \ \deg \ in \ QIII, \ 330 \ \deg \ in \ QIV[/itex]

Am I doing this correctly?
 
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Almost, but I think you have confused sine and cosine
sin(30 degrees)=1/2
sin(60 degrees)=sqrt(3)/2
cos(30 degrees)=sqrt(3)/2
cos(60 degrees)=1/2
 
lurflurf said:
Almost, but I think you have confused sine and cosine
sin(30 degrees)=1/2
sin(60 degrees)=sqrt(3)/2
cos(30 degrees)=sqrt(3)/2
cos(60 degrees)=1/2

I understand the difference between the two, I must have hit the wrong button in my calculator. Thanks brotha.
 

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