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Math Help - Trig Help.

  1. #1
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    Trig Help.

    Hi,

    I have a couple of questions. My prof is having us do a chart. It's of sin, cos, tan, cot, etc functions. You're given one and have to get another.

    One of the ones I am stuck on is as follows.

    You're given Sin theta and need to get Tan theta. I think I need to use a Pythagorean Identity to prove it. Here's what I did...

    Sin theta to Tan theta
    Using the ident. Cot squared theta + 1 = Csc squared theta
    Then I used the Reciprocal Idents. to get 1 / Tan^2 theta + 1 = 1 / Sin^2 theta.
    Then 1 / Tan^2 theta = 1 / Sin^2 theta - 1

    That's where I'm stuck... What do I do next?

    Thanks,


    -Tom
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  2. #2
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    Quote Originally Posted by TomCat View Post
    Hi,

    I have a couple of questions. My prof is having us do a chart. It's of sin, cos, tan, cot, etc functions. You're given one and have to get another.

    One of the ones I am stuck on is as follows.

    You're given Sin theta and need to get Tan theta. I think I need to use a Pythagorean Identity to prove it. Here's what I did...

    Sin theta to Tan theta
    Using the ident. Cot squared theta + 1 = Csc squared theta
    Then I used the Reciprocal Idents. to get 1 / Tan^2 theta + 1 = 1 / Sin^2 theta.
    Then 1 / Tan^2 theta = 1 / Sin^2 theta - 1

    That's where I'm stuck... What do I do next?

    Thanks,


    -Tom
    There's a much simpler way:
    You've got sin(theta). cos(theta) = (+/-)sqrt(1 - [sin(theta)]^2) then tan(theta) = sin(theta) / cos(theta)

    NOTE: You need to make sure cos(theta) has the appropriate sign! You can tell that by noting which quadrant theta is in. (ie. cos(theta) is positive in I and IV, cos(theta) is negative in II, III.)

    -Dan
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  3. #3
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    That's not the answer is it? That is not like what my Prof. was doing in class earlier today. He said that we should not involve a third function.

    Here's an example that I can do:

    Sin (theta) to Cos (theta).
    I use the Pathag. Ident of Sin^2 (theta) + Cos^2 (theta) = 1.
    I subtract the Sin^2 (theta) from both sides to get...
    cos^2 (theda) = 1 - Sin^2 (theta).
    Then I take the sqrt and get...
    Cos (theta) = (+/-) the sqrt of 1 - Sin^2 (theta).

    That's the answer he is looking for... Does this make sense? There are 36 different things I need to find. 12 of them are extremely easy because the answers are sin (theta) or 1 / sin (theta) or Cos (theta) or 1 / Cos (theta... Etc...
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TomCat View Post
    That's not the answer is it? That is not like what my Prof. was doing in class earlier today. He said that we should not involve a third function.

    Here's an example that I can do:

    Sin (theta) to Cos (theta).
    I use the Pathag. Ident of Sin^2 (theta) + Cos^2 (theta) = 1.
    I subtract the Sin^2 (theta) from both sides to get...
    cos^2 (theda) = 1 - Sin^2 (theta).
    Then I take the sqrt and get...
    Cos (theta) = (+/-) the sqrt of 1 - Sin^2 (theta).

    That's the answer he is looking for... Does this make sense? There are 36 different things I need to find. 12 of them are extremely easy because the answers are sin (theta) or 1 / sin (theta) or Cos (theta) or 1 / Cos (theta... Etc...
    No matter what you do you are going to have to have either a cos(theta) or a sec(theta) in there to get tan(theta). So don't worry about introducing the third trig. function, its unavoidable.

    If I'm wrong about this, I'd love to see your professor's solution.

    -Dan
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  5. #5
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    Okay. Thanks for the help. I sent him an e-mail a little bit ago asking for some help to see what he says. Maybe when he was talking about not introducing another function he was being specific to a certain problem. I'm not sure since he never specified. Hopefully he replies to my e-mail... He seems to be the type that doesn't use computers and such, but he did provide an e-mail address on the syllabus so we'll see I guess.

    Thanks,


    -Tom
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  6. #6
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    Hello, TomCat!

    You're given sin θ and need to get tan θ.

    I assume this means: express tan θ in terms of sin θ (only).

    . . . . . . . sin θ . . . . . . sin θ
    tan θ .= .------- . = .-------------
    . . . . . . . cos θ] . . . √1 - sin²θ

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  7. #7
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    Quote Originally Posted by TomCat View Post
    Hi,

    I have a couple of questions. My prof is having us do a chart. It's of sin, cos, tan, cot, etc functions. You're given one and have to get another.

    One of the ones I am stuck on is as follows.

    You're given Sin theta and need to get Tan theta. I think I need to use a Pythagorean Identity to prove it. Here's what I did...

    Sin theta to Tan theta
    Using the ident. Cot squared theta + 1 = Csc squared theta
    Then I used the Reciprocal Idents. to get 1 / Tan^2 theta + 1 = 1 / Sin^2 theta.
    Then 1 / Tan^2 theta = 1 / Sin^2 theta - 1

    That's where I'm stuck... What do I do next?

    Thanks,


    -Tom
    cot^2(A) +1 = csc^2(A)
    If you get and equate their reciprocals,
    1/[cot^2(A) +1] = 1/[sin^2(A)]
    The lefthand side is not 1/[tan^2(A)] +1, so you cannot transpose the "1" to the righthand side.

    sinA in terms of tanA:
    tanA = sinA/cosA ----identity
    Multiply both sides by cosA,
    cosA*tanA = sinA
    Express the cosA in terms of sinA,
    [sqrt(1 -sin^2(A)]tanA = sinA ------------(i)
    There, only sinA and tanA are involved.
    Solve for sinA in terms of tanA and you're done.

    Here is one way of doing that,
    Square both sides of (i),
    (1 -sin^2(A)[tan^2(A)] = sin^2(A)
    Expand,
    tan^2(A) -[sin^2(A)][tan^2(A)] = sin^2(A)
    Collect the sinA terms,
    tan^2(A) = sin^2(A) +[sin^2(A)][tan^2(A)]
    Factor out the sin^2(A),
    tan^2(A) = [sin^2(A)][1 +tan^2(A)]
    Isolate the sin^2(A),
    [tan^2(A)] /[1 +tan^2(A)] = sin^2(A)
    So, take the square roots of both sides,
    sinA = (tanA) /[sqrt(1 +tan^2(A))] -------------answer.
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  8. #8
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    Quote Originally Posted by ticbol View Post
    :
    :
    sinA in terms of tanA:
    tanA = sinA/cosA ----identity
    Multiply both sides by cosA,
    cosA*tanA = sinA
    Express the cosA in terms of sinA,
    [sqrt(1 -sin^2(A)]tanA = sinA ------------(i)
    There, only sinA and tanA are involved.
    Solve for sinA in terms of tanA and you're done.

    Here is one way of doing that,
    Square both sides of (i),
    (1 -sin^2(A)[tan^2(A)] = sin^2(A)
    Expand,
    tan^2(A) -[sin^2(A)][tan^2(A)] = sin^2(A)
    Collect the sinA terms,
    tan^2(A) = sin^2(A) +[sin^2(A)][tan^2(A)]
    Factor out the sin^2(A),
    tan^2(A) = [sin^2(A)][1 +tan^2(A)]
    Isolate the sin^2(A),
    [tan^2(A)] /[1 +tan^2(A)] = sin^2(A)
    So, take the square roots of both sides,
    sinA = (tanA) /[sqrt(1 +tan^2(A))] -------------answer.
    This is OK in the first and third quadrants, but in the second and third
    sin and tan have different signs; so in full this should be:

    Code:
    sinA=(tanA) /[sqrt(1 +tan^2(A))]  for 0<=A<90 or 270<pi<=360
        =-(tanA) /[sqrt(1 +tan^2(A))] for 90<A<270
    also as tan is undefined at 90, 270 there is strictly no expression for sin in
    terms of tan at these points.

    RonL

    (Note: It's a bit late to comment on it as Jameson has locked this thread and
    I will not unlock something another has locked, but Soroban's post also suffers
    from this problem)
    Last edited by CaptainBlack; September 24th 2006 at 04:09 AM. Reason: add comment about Soroban's post
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    This is OK in the first and third quadrants, but in the second and third
    sin and tan have different signs; so in full this should be:

    Code:
    sinA=(tanA) /[sqrt(1 +tan^2(A))]  for 0<=A<90 or 270<pi<=360
        =-(tanA) /[sqrt(1 +tan^2(A))] for 90<A<270
    also as tan is undefined at 90, 270 there is strictly no expression for sin in
    terms of tan at these points.

    RonL
    I have deleted the discussion between ticbol and myself about this,
    basically he does not agree that the final identity in his work on this
    is deficient in quadrants 2 and 3.

    RonL
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