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The t Method

  1. Jun 23, 2006 #1
    “The “t” Method”

    Recently I have been studying for an upcoming test where it requires me to use “The “t” Method”. In this method the value of x for trigonometric equations is determined through vair the key component of “The “t” Method” is:
    [tex]t=tan \frac{A}{2}[/tex]
    [tex]tan A=\frac{1+t^2}{1-t^2}[/tex]
    [tex]sin A=\frac{2t}{1+t^2}[/tex]
    [tex]cos A=\frac{1-t^2}{1+t^2}[/tex]
    If this is famliar to a reader by another name, could you please post this methods other name.
    Problems which i need to solve:
    1. [tex]2sin x + 3cos x = 5[/tex]
    2. [tex]3tan x + \sqrt{3}sec x=1[/tex]
    3. [tex]10cos (pi x) + 3sin (2pi x)=4[/tex]
    4. [tex]3sin 2x + 5cot 3x = 7[/tex]
    5. [tex]csc x + 2sec (pi x)[/tex]
    My working for question 1. [tex]2sin x + 3cos x = 5[/tex]
    [tex]2\frac{2t}{1+t^2} + 3\frac{1-t^2}{1+t^2} = 5[/tex]
    [tex]\frac{4t + 3 - 3t^2}{1+t^2} = 5[/tex]
    [tex]4t + 3 -3t^2 = 5 +5t^2[/tex]
    [tex]4t + 2t^2 = 2[/tex]
    [tex]2t^2 + 4t - 2 = 0[/tex]
    [tex]t = 0.4142135624[/tex] or [tex]t = -2.414213562[/tex]
    [tex]t = \tan\frac{x}{2}[/tex]
    [tex]x = 2tan^-1 0.4142135624[/tex] or [tex]x = 2tan^-1 -2.414213562[/tex]
    [tex]x = 45[/tex] or [tex]x = -135.0005034[/tex]
    I conclude my working here as I am not sure if it is correct. Is there an aleternaitve method I could use to check these final answers? And are there any more values of x for which the equation is true?
    Thanks well in advance,
  2. jcsd
  3. Jun 23, 2006 #2


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    You've lost track of a sign: adding 3t2 to both sides gives [itex]8t^2- 4t+ 2= 0[/itex] or [itex]4t^2- 2t+ 1= 0[/itex]
    That has only complex roots.

    In fact, since 2+ 3= 5 and sine and cosine are never larger than 1, the only way we could have 2sin x+ 3cos x= 5 is to have sin x= 1 and cos x=1 which is not true for any x.
  4. Jun 24, 2006 #3
    thanks HallofIvy.
    do they all have complex roots, or do i need to wokr them as i have done before to determine x when i have it in terms of 2tan^-1 t ?
  5. Jun 24, 2006 #4


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    Well, I don't know! I only looked at the first as that was the one for which you showed your work. I imagine that at least some of these have real solutions but you will have to do the algebra to see.
  6. Jun 25, 2006 #5
    oh okay, ill post my wokring for the otherss when i have more time.
  7. Jun 25, 2006 #6
    working for question number 2.

    [tex]3tan x + \sqrt{3}sec x=1[/tex]

    [tex]3\frac{2t}{1-t^2} + \sqrt{3}\frac{1+t^2}{1-t^2} = 1[/tex]
    [tex]\frac{6t + \sqrt{3} + t^2\sqrt{3}}{1 - t^2} = 1[/tex]
    [tex]6t + \sqrt{3} + t^2\sqrt{3} = 1 - t^2[/tex]
    [tex](1 + \sqrt{3})t^2 + 6t - (1 - \sqrt{3}) = 0[/tex]
    therefore: t = -0.1296640194 or t = -2.066488403
    for t = -0.1296640194
    [tex]t = tan^-1 \frac{x}{2}[/tex]
    [tex]x = 2tan^-1 -0.1296640194[/tex]
    [tex]x = -14.77596194[/tex]
    t = -2.066488403
    [tex]t = tan^-1 \frac{x}{2}[/tex]
    [tex]x = 2tan^-1 -2.066488403[/tex]
    [tex]x = -128.3541404[/tex]

    again, im not sure if what i am do is right so i would appriciate it iof somebody helped me out. if anybody knows of alternative ways of solving these equations, please tell me how

    many thanks
  8. Jun 25, 2006 #7


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    This is also known as the Weierstrass substitution.
  9. Jun 25, 2006 #8
    You’ve made a mistake between the third and forth lines. The moves to the left with a minus, so as a result you get , which has no real solutions use quadratic formula).

    Alternative approach to trig
  10. Jun 26, 2006 #9
    okay,many thanks once again for the replies
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