Trigonometry coursework question coursework - finding sin(v) if sin(2v) = 1/3 and 2v is on quadrant I

Question

The question is pretty much in the title, but it was:

Find the precise value of sin(v) if 2v is on quadrant I and sin(2v) = 1/3

I had a few similar questions preceding these, where you were given sin(v) and you had to find sin(2v), and for them I used precise rules that we had learned to figure out v, but I never had to use the quadrant information given since all it tells me is that sin(v) is positive, so I thought that it was pretty pointless information. But I cannot figure this question out for the life of me with the rules I have been taught - they are as follows:

sin²(v) + cos²(v) = 1

sin(v) = ± tan(v) / √(1+tan²(v)) - how do I find tan(v) in order to use this rule?

cos(v) = ± 1 / √(1+tan²(v)) - do I need to use this one?

cos(2v) = 1 - 2sin²(v)

sin(2v) = 2sin(v)cos(v)

tan(2v) = 2tan(v) / (1 - tan²(v))

These are all the rules that come to mind when it comes to solving this question - the others I have learned seem pretty irrelevant to the question.

Is there any way to solve the question from this information?

Answer 1

You can use cos(2v) = 1-2sin²(v).

Since we know the sin(2v) = 1/3, then, from this find cos(2v). Draw the triangle with the hypotenuse as 3 and the side opposite 2v is 1. This makes the adjacent side √(3²-1²).

cos(2v) = √(3²-1²) / 3 = √8/3 = 2√2 / 3

Now you can solve for sin(v) knowing that if 2v is in the first quadrant, so is v. So sin(v) will be positive and we can ignore -√().

sin(v) = √(1- (2√2)/3)

Answer 2

do a quick check on some purported answers

sin2v = 1/3

use a trig calculator

2v = about 19.47 degrees

v = about 19.47/2 = 9.74 degrees

sinv = about .17= an approximation to 2 decimal places

which shows v is in quadrant I, not II

one suggested answer:

sinv = sqr(1-2sqr2)/3)

=sqr(1-2.8/3)

=sqr(1-.93)

sqr.07

= about .27 which is too far from .17

another suggested answer

sinv=(1/3)sqr(3-6sqr3)

=(1/3)sqr(3-10.4)

=(1/3)sqr(-7.4) which is an imaginary number

another answer:

sinv=(1/6)sqr(18-12sqr2)

=about .17 which is the same as the answer using trig functions on a calculator to get an approximation to 2 digits

sinv = exactly [sqr(18-12sqr2)]/6

sin2v=1/3=opposite/hypotenuse

cos2v= adjacent/hypotenuse=(2sqr2)/3

sinv=sqr((1-cos2v)/2)

=sqr(1-2sqr2/3)/ t2)

=sqr(3-2sqr2)/3/2)

=sqr(3-2sqr2)/6)

=sqr6sqr(3-2sqr2)/6

=[sqr(18-12sqr2)]/6

=about .17

Answer 3

Hello Havar,

2 things:

1. among the formulas you learned, did you learn the half-angle formula sin (x/2) = ±√[(1-cos x)/2]? If so, that could be one way to do this
2. I believe Mr. Bradford is correct: I worked it out and because 2v is in Q1, so is v, therefore your answer should not be plus OR minus, but JUST plus. Also, my answer is √ [(1-2√2)/3] which you could re-write as √(1-2√2) / √3. Multiplying top and bottom by √3 would then get you √(3 - 6√2) / 3 [please note that the last 3 is NOT part of the square root]

So in my humble opinion: it is possible, but unlikely that both Mr. Bradford and I are incorrect at this point. Especially considering that the answer, again, cannot be plus OR minus, but just the "plus" because 2v is already in Q1 so v would also be there.