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Math Help - Expressing x in terms of k where x radians is an acute angle. tan and pi involved.

  1. #1
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    Expressing x in terms of k where x radians is an acute angle. tan and pi involved.

    Greetings to all.

    I having difficulty solving this straight-forward question. Basically, I just don't know what to do next. Here it is:

    Q. The acute angle x radians is such that tan x = k, where k is a positive constant. Express, in terms of k,

    Now, this comes in 3 parts which are:

    (i) tan(pi − x),
    (ii) tan[(1/2)pi − x],
    (iii) sin x.

    I'm stuck because I don't know how to proceed. Like in part (i), I write:

    tan x=k,
    x=tan^-1 (k),
    tan (pi-tan^-1(k) ),
    tan.pi - k.

    I just end up with the question, the only difference being that instead of x, there's k. How should I satisfy the question's demand?
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by SolCon View Post
    Greetings to all.

    I having difficulty solving this straight-forward question. Basically, I just don't know what to do next. Here it is:

    Q. The acute angle x radians is such that tan x = k, where k is a positive constant. Express, in terms of k,

    Now, this comes in 3 parts which are:

    (i) tan(pi − x),
    (ii) tan[(1/2)pi − x],
    (iii) sin x.

    I'm stuck because I don't know how to proceed. Like in part (i), I write:

    tan x=k,
    x=tan^-1 (k),
    tan (pi-tan^-1(k) ),
    tan.pi - k.

    I just end up with the question, the only difference being that instead of x, there's k. How should I satisfy the question's demand?
    For i) and ii), You can use:
    Tan(a-b)=\dfrac{Tan(a) - Tan(b)}{1 + (Tan(a)Tan(b))}

    For (iii) You'll have to use your identities.

    Tan(x)=\frac{Sin(x)}{Cos(x)}

    =\dfrac{Sin(x)}{\sqrt{1-Sin^2(x)}}

    Perhaps you could rearrange this. Or you could use other identities to help, such as:
    Tan^2(x)+1=Sec^2(x)
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  3. #3
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    (i) The function tg is periodical, which is its period?
    (ii) Draw a right-angled triangle which have a x angle and then see who is \frac{\pi}{2}-x. (Also use - tg x =\frac{opposite}{adjacent}, ctg x =\frac{adjacent}{opposite})
    (iii) How can be tg x write using sin x and cos x? (Also use that sin^2x+cos^2x=1)
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SolCon View Post
    Greetings to all.

    I having difficulty solving this straight-forward question. Basically, I just don't know what to do next. Here it is:

    Q. The acute angle x radians is such that tan x = k, where k is a positive constant. Express, in terms of k,

    Now, this comes in 3 parts which are:

    (i) tan(pi − x),
    (ii) tan[(1/2)pi − x],
    (iii) sin x.

    I'm stuck because I don't know how to proceed. Like in part (i), I write:

    tan x=k,
    x=tan^-1 (k),
    tan (pi-tan^-1(k) ),
    tan.pi - k.

    I just end up with the question, the only difference being that instead of x, there's k. How should I satisfy the question's demand?
    See here, for example:
    \displaystyle tan(a + b) = \frac{tan(a) + tan(b)}{1 - tan(a)~tan(b)}

    For the second one, since tan( \pi / 2) does not exist you are going to have to get a little more complicated. Use
    \displaystyle tan(a + b) = \frac{sin(a + b)}{cos(a + b)}

    The last one is a little trickier. You know that tan(x) = k is positive and you want sin(x). The tangent function is positive in QI and QIII, and the sine function is positive in QI and QII. Thus we are dealing with a QI angle and sin(x) will be positive.

    Now
    \displaystyle tan(x) = \frac{sin(x)}{cos(x)}

    and
    sin^2(x) + cos^2(x) = 1

    Can you finish from here?

    -Dan
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  5. #5
    Super Member Quacky's Avatar
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    Quote Originally Posted by topsquark View Post
    See here, for example:
    \displaystyle tan(a + b) = \frac{tan(a) + tan(b)}{1 - tan(a)~tan(b)}

    For the second one, since tan( \pi / 2) does not exist you are going to have to get a little more complicated. Use
    \displaystyle tan(a + b) = \frac{sin(a + b)}{cos(a + b)}
    Whoops, missed that entirely.
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  6. #6
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    For (i) it's easier to use that tg is a periodical and odd function.
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  7. #7
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    Thanks for the replies people.

    Unfortunately, I cannot examine the solutions at the moment and will not be able to post any possible questions/confirmations of understanding. I'll reply in a while with an EDIT. Hope that won't count as bumping.

    EDIT:

    Okay, I've looked at the solutions. I tried solving but got stuck in all 3 parts. This is what I've

    For (i):

    tan(pi-x) = tan(pi) - tan(x) / 1-tan(pi) tan(x)
    tan(pi) - tan(x) = tan(pi) -tan(x) / 1-tan(pi)tan(x)
    tan(pi) - k = tan(pi) - k / 1-tan(pi)(k)

    SWITCHING RHS DENOMINATOR and LHS NUMERATOR

    1-tan(pi)(k) = 1

    What to do next?

    For (ii):

    tan(0.5pi - x) = sin(0.5pi - x) / cos(0.5pi - x)

    CONFUSION: If tan(0.5pi) does not exist, what comes on the LHS? For now, I'll just use tan(0.5pi) as is:

    tan(0.5pi) - tan(x) = sin(0.5pi) - sin(x) / cos(0.5pi) - cos(x)
    tan(0.5pi) - k = 1 - sin(x) / 0 - cos(x)

    Stuck here.

    For (iii):

    Well, I really didn't understand this one. I understand the above formulas given by topsquark, but don't know how to apply them. With Quacky's solution, are we going to use sin(x) as k?

    Thanks for the help.
    Last edited by SolCon; April 12th 2011 at 07:40 AM.
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  8. #8
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    If you don't know the formulas learn them or don't use them.

    (i)

    tan is a periodic function with period pi. So, tan ( \pi-x) = tan (-x).
    tan is an odd function. So, tan (-x) = -tan x.

    (ii)







    (iii) sin x



    "The last one is a little trickier. You know that tan(x) = k is positive and you want sin(x). The tangent function is positive in QI and QIII, and the sine function is positive in QI and QII. Thus we are dealing with a QI angle and sin(x) will be positive." (topsquark)

    Last edited by veileen; April 18th 2011 at 12:35 AM.
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  9. #9
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    Appreciate the quick response.

    Okay, gonna go reverse from (iii) to (i), since I've just gotten (iii).

    So I looked at your solution and solved it for myself, step-by-step, for better understanding, using your solution as reference. Here's what I've done now:

    sinx:

    > sinx = cosxtanx [tanx = sinx/cosx (cross-multiply)]
    > sinx/tanx = cosx
    > sinx/k = cos x [tanx = k]
    > (sinx/k)^2 = (cosx)^2
    > sin^2x/k^2 = cos^2x
    > sin^2x = k^2 (1-sin^2x) [cos^2x+sin^2x=1]
    > sin^2x = k^2 - sin^2xk^2
    > k^2 = sin^2x+sin^2xk^2
    > k^2 = sin^2x (1+k^2)
    > K^2/(1+k^2) = sin^2x
    > [k^2/(1+k^2)] sqrt = (sin^2x) sqrt
    > k/(1+k^2) sqrt = sinx

    So, this one's down.
    Just one thing though. Was it absolutely necessary to have only sin and k. Why couldn't we just have sinx= cosx (k)?

    Now for (ii):

    I understand everything except the part where you flip cosx and sinx. Do you first cancel out the (pi/2-x) for sinx and cosx, then flip them? And if you do then how, seeing as sinx/cosx is tanx. How did you get 1/tanx instead of tanx?

    Lastly, (i):

    Doing this one step-by-step:

    Using formula: tan(A +/- B) = (tanA +/- tan B)/(1 -/+ tanAtanB)

    So here's what I did:

    >tan(pi-x) >>> tan(pi) - tan(x) / 1 - tan(pi)tan(x)
    > tan(pi)-k / 1- tan(pi-k)

    I get stuck here. How do we use the formula to solve this and get -k?

    Thanks for the help.
    >
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  10. #10
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    (iii) "Just one thing though. Was it absolutely necessary to have only sin and k." Yes, it was, your exercise says: "Express, in terms of k,".
    (ii) Uhm, I don't understand what you don't understand ^^'
    tan x is the ratio of sin x to cos x, so tan (pi/2-x) is the ratio of sin (pi/2-x) to cos (pi/2-x). I proved that sin (pi/2-x) is cos x and cos (pi/2-x) is sin x, so tan (pi/2-x) is the ratio of cos x and sin x.
    (i) "How do we use the formula to solve this and get -k?" Uhm, you don't use the correct formula, even if you wrote it well.

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  11. #11
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    Okay, I've finally got it now after reviewing the formulas and the solutions.

    For (i), it would have been solved much earlier if someone would have said that tan pi was allowed to be used as tan 180 which would give us 0 (I'm no good at maths and things like these tend to slip away from time to time from my mind ).

    For (ii), lol, I thought those sinx and cosx lines were examples, not part of the solutions. Looking at them again and seeing the list of formulas with me I realized that you were referring to the sin(A+/-B) = sinAcosB +/- cosAsinB and the similar cos formula. Got the cosx/sinx = 1/tanx bit also.

    Looks like this question is finally done.

    Thank you all.
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