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C3 trig addition formulae equation

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I am trying to solve for theta but have ended up with this. What am I supposed to do now?


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Original post by jessyjellytot14


I am trying to solve for theta but have ended up with this. What am I supposed to do now?


Posted from TSR Mobile


Divide both sides by cosine to get tanθ\tan \theta
Original post by jessyjellytot14
ImageUploadedByStudent Room1476640650.662647.jpg

I am trying to solve for theta but have ended up with this. What am I supposed to do now?


Posted from TSR Mobile

On line 2:
cosθcos60=sinθ(1sin60)\cos \theta \cos 60 = \sin \theta(1-\sin60)
cos601sin60=tanθ\frac{\cos 60}{1- \sin 60} = \tan \theta
I think you can take it from here
(edited 7 years ago)
Original post by RDKGames
Divide both sides by cosine to get tanθ\tan \theta

the way it was simplified at the end would give a 1cosθ \frac{1}{\cos \theta} as well
Original post by 123Master321
the way it was simplified at the end would give a 1cosθ \frac{1}{\cos \theta} as well


No it wouldn't. I assume OP notices the mistake of forgetting the brackets. Otherwise, that is something she would quickly realise.
Original post by RDKGames
No it wouldn't. I assume OP notices the mistake of forgetting the brackets. Otherwise, that is something she would quickly realise.


Oh yh my bad, I just assumed the working was right and it got simplified poorly
Original post by RDKGames
Divide both sides by cosine to get tanθ\tan \theta


Original post by 123Master321
On line 2:
cosθcos60=sinθ(1sin60)\cos \theta \cos 60 = \sin \theta(1-\sin60)
cos601sin60=tanθ\frac{\cos 60}{1- \sin 60} = \tan \theta
I think you can take it from here


I tried both of these methods and they both got the same answers
Original post by jessyjellytot14
I tried both of these methods and they both got the same answers


That's because these aren't different methods.
Original post by RDKGames
That's because these aren't different methods.


I know but the first method doesn't involve factorising so they look slightly different when written down.

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