C3 Trig Watch

scientific222
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I came to get tan^2 0 = 49/576. So to get tan^0 as the question is asking, I square root it and end up with + or - 7/24. My question is how do I know which one is the right one? The answer in the textbook is tan 0 = - 7/24.Theta is meant to be reflex (180-360) but how do I apply this?
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TenOfThem
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(Original post by scientific222)
Name:  trig.PNG
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I came to get tan^2 0 = 49/576. So to get tan^0 as the question is asking, I square root it and end up with + or - 7/24. My question is how do I know which one is the right one? The answer in the textbook is tan 0 = - 7/24.Theta is meant to be reflex (180-360) but how do I apply this?
I am not sure why you have \tan ^2\theta but

You know that the angle is between 180 and 360

But you also know that \cos \theta >0 so this tells you if you are in 180-270 OR 270 -360
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scientific222
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(Original post by TenOfThem)
I am not sure why you have \tan ^2\theta but

You know that the angle is between 180 and 360

But you also know that \cos \theta >0 so this tells you if you are in 180-270 OR 270 -360
I got tan^2 0 by turning cos 0 into sec 0 then sec^2 0 which is identical to (1 + tan^0) then I eventually had tan 0 on its own. It gives the right answer; but looking at the answer this was all stupid because they simply drew up a right angled triangle (with the known 24 and 25 values from cos theta) and simply worked out the tan value from there. But I understand now, cos 0 > 0 means it would have to be in the 4th quadrant and the solution for tan 0 would be negative. Thank you very much for your help
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TenOfThem
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(Original post by scientific222)
I got tan^2 0 by turning cos 0 into sec 0 then sec^2 0 which is identical to (1 + tan^0) then I eventually had tan 0 on its own. It gives the right answer; but looking at the answer this was all stupid because they simply drew up a right angled triangle (with the known 24 and 25 values from cos theta) and simply worked out the tan value from there. But I understand now, cos 0 > 0 means it would have to be in the 4th quadrant and the solution for tan 0 would be negative. Thank you very much for your help
Ah I see, of course

I would have just drawn the triangle

But with trig there are often any approaches
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brianeverit
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(Original post by scientific222)
Name:  trig.PNG
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I came to get tan^2 0 = 49/576. So to get tan^0 as the question is asking, I square root it and end up with + or - 7/24. My question is how do I know which one is the right one? The answer in the textbook is tan 0 = - 7/24.Theta is meant to be reflex (180-360) but how do I apply this?
Surely the easiest way is to find \sin\theta first and noting that it must be negative if the angle is reflex.
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adelmoulton
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Would someone be able to help me with some C3 Trig?

I have 2 questions

a) √3 sin2(x) + 2sin^2(x) = 1

b) 4 tan2(x) tan(x) =1

Thank you.
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brianeverit
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(Original post by adelmoulton)
Would someone be able to help me with some C3 Trig?

I have 2 questions

a) √3 sin2(x) + 2sin^2(x) = 1

b) 4 tan2(x) tan(x) =1

Thank you.
(a) 2\sin^2 x=1-\cos2x does this help?
(b) Use double angle identity to get an equation in tan(x)
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