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If tan theta = 2/3 then sin theta= ?

A) +- 0.55

C) 0.55

Why is the answer A, I am sure the solution is simple, and can you please not use CAST as an explanation.

Thank you.
Original post by fnatic NateDestiel
A) +- 0.55

C) 0.55

Why is the answer A, I am sure the solution is simple, and can you please not use CAST as an explanation.

Thank you.


How am I to explain it if I do not use the range in which tan is positive?

So - tan is positive between 0 and 90 - in this range sin is also positive

Also - tan is positive between 180 and 270 - in this range sin is negative


Given this - sin could be either positive or negative
Posted from TSR Mobile

Use the identity cot^2 + 1 = cosec^2

to find sin you will up having to take a square root that leads to +-
Reply 3
Original post by fnatic NateDestiel
A) +- 0.55

C) 0.55

Why is the answer A, I am sure the solution is simple, and can you please not use CAST as an explanation.

Thank you.


Draw a right angled triangle, note that tan theta= opposite/adjacent. Use pythag to find the hypotenuse
Original post by TenOfThem
How am I to explain it if I do not use the range in which tan is positive?

So - tan is positive between 0 and 90 - in this range sin is also positive

Also - tan is positive between 180 and 270 - in this range sin is negative


Given this - sin could be either positive or negative


I understand your first point and it is easily solvable on the calculator.

And I understand your second point, but I don't get how it is related to the question.

Because I only get -0.55 when tan theta is - 2/3.
Reply 5
That'll be useful in later life...
Original post by BankOfPigs
Posted from TSR Mobile

Use the identity cot^2 + 1 = cosec^2

to find sin you will up having to take a square root that leads to +-


I have tried this method, but only get the identities proving other identities or cancelling eachother out :frown:

Original post by Gome44
Draw a right angled triangle, note that tan theta= opposite/adjacent. Use pythag to find the hypotenuse


I did this initially, but I only get the positive answer..
Original post by fnatic NateDestiel
I have tried this method, but only get the identities proving other identities or cancelling eachother out :frown:





I'm not really too sure what you mean.

Cot^2 x + 1 = cosec^2 x

since tan = 2/3 cot = 3/2

so (3/2)^2+1 = 1/sin^2
13/4 = 1/ sin^2
4/13 = sin^2 x

Now this leads to 2 roots. If you aren't sure what I mean, do the substituion y = sin x

so 4/13 = y^2
This has 2 roots, one negative, one positive.
Original post by BankOfPigs
I'm not really too sure what you mean.

Cot^2 x + 1 = cosec^2 x

since tan = 2/3 cot = 3/2

so (3/2)^2+1 = 1/sin^2
13/4 = 1/ sin^2
4/13 = sin^2 x

Now this leads to 2 roots. If you aren't sure what I mean, do the substituion y = sin x

so 4/13 = y^2
This has 2 roots, one negative, one positive.


Oh I see thank you, this makes sense, one other guy on CC posted this; "You need to consider all the angles for which tan theta = 2/3. The first one is 33.69 degrees, but there are infinitely many, so find the next one and consider that." which also works.

Thanks though!

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