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Math Help - I need help with finding sin and cos please.

  1. #1
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    I need help with finding sin and cos please.

    The problem is tan= -\frac{2} {3}. I can't find sin or cos, How can I find them? I know Tangent is  \frac{x} {y} I know it can't be sin=-2 and cos=3, can it?


    Thanks

    Jason
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    First, you can't simply write \tan. It has to be tan of an angle, ex. \tan \theta, \tan x, etc. The trigonometric functions are there to determine the ratio between two sides based on an angle.

    Yes you are right that \sin \theta \neq -2 and that \cos \theta \neq 3. Both of them can only have values between -1 and 1.

    For your question, construct a right angle triangle and label an angle \theta. Since you know that \tan \theta = \frac{\text{opposite}}{\text{adjacent}}, label the opposite side -2 (extending below the x-axis) and the adjacent side +3. Now find the hypotenuse and by the Pythagorean theorem and you should be able to find \sin \theta and \cos \theta.
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    Quote Originally Posted by Darkhrse99 View Post
    The problem is tan= -\frac{2} {3}. I can't find sin or cos, How can I find them? I know Tangent is  \frac{x} {y} I know it can't be sin=-2 and cos=3, can it?


    Thanks

    Jason
    ok, depending on what quadrant we are in, either the sine or the cosine must be negative. so we are either in the second quadrant or the fourth quadrant. lets worry about that later.

    there are two main approaches to problems like this. there is the formula approach, where you manipulate trig identities to find the answers you want, and there is the more geometric approach where you use triangles. i will use the latter

    remember, tangent = opposite/adjacent

    we have tan(x) = -2/3

    forgetting the sign for a while, we think of the 2 as the opposite side and the 3 as the adjacent side. so if we draw a right triangle, and label an acute angle x. then opposite to that angle, the side length is 2 and adjacent to it, the side length is 3. we can use Pythagoras' theorem to find the hypotenuse. by Pythagoras, this would be \sqrt{13}

    now you can find sin(x) and cos(x)

    one of these must be negative while the other is positive. we must know what quadrant we are in to be sure.


    (remember, sine = opposite/hypotenuse and cosine = adjacent/hypotenuse)
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    My teacher hasn't tought us the latter way to solve the expressions. So the hyp= \sqrt{13}, which I see how it was done.Thanks . I'm working in quad. 2, I forgot to add that the 1st post.
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    So would sin (t) = \frac{2} {\sqrt13} and   cos (t) = \frac{3} {2} ?
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    Quote Originally Posted by Darkhrse99 View Post
    So would sin (t) = \frac{2} {\sqrt13} and   cos (t) = \frac{3} {2} ?
    first off, the cosine must be negative (we are in the second quadrant). secondly, the denominator of the ratios for sine and cosine is the hypotenuse, so you cannot have them being different. the length of the hypotenuse is \sqrt{13} as i said in my earlier post
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    So it is sin (t)= \frac{2} {\sqrt13} and cos (t)= \frac{-3}{\sqrt13}?
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    Quote Originally Posted by Darkhrse99 View Post
    So it is sin (t)= \frac{2} {\sqrt13} and cos (t)= \frac{-3}{\sqrt13}?
    Yes
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    Thanks a bunch. I plan to ask the teacher when we are going to learn about soa cah toa. It seems much easier to find the angles I'm looking for.
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    Quote Originally Posted by Darkhrse99 View Post
    Thanks a bunch. I plan to ask the teacher when we are going to learn about soa cah toa. It seems much easier to find the angles I'm looking for.
    actually, it's soh cah toa

    and we have used it already. that is where sine = opposite/hypotenuse, cosine = adjacent/hypotenuse and tangent = opposite/adjacent come from
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