Mastering "The 't' Method": Solving Trig Equations w/ Pavadrin

In summary, "The "t" Method" is a method used to determine the value of x for trigonometric equations. It involves using the key components of tan A, sin A, and cos A, which can also be expressed in terms of t. This method is also known as the Weierstrass substitution. When using this method, it is important to be careful with signs and to check for all possible solutions.
  • #1
pavadrin
156
0
“The “t” Method”

Hey
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,
Pavadrin
 
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  • #2
pavadrin said:
Hey
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]
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.
 
  • #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 ?
 
  • #4
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.
 
  • #5
oh okay, ill post my wokring for the otherss when i have more time.
 
  • #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, I am not sure if what i am do is right so i would appreciate it iof somebody helped me out. if anybody knows of alternative ways of solving these equations, please tell me how

many thanks
Pavadrin.
 
  • #7
This is also known as the Weierstrass substitution.
 
  • #8
pavadrin said:
Hey
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,
Pavadrin
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
 
  • #9
okay,many thanks once again for the replies
 

1. What is "Mastering The 't' Method"?

"Mastering The 't' Method" is a method for solving trigonometric equations using the substitution t = tan(x/2). This method was developed by mathematician Pavadrin and is commonly used in trigonometry courses.

2. How does "Mastering The 't' Method" work?

The first step in using the t method is to rewrite the trigonometric equation in terms of t. Then, use algebraic techniques to solve for t. Once t is found, the original variable can be solved for using the inverse tangent function.

3. What are the advantages of using "Mastering The 't' Method"?

"Mastering The 't' Method" is a helpful tool for solving complicated trigonometric equations, as it can simplify the equations and make them easier to solve. It also allows for a systematic approach to solving trigonometric equations.

4. Are there any limitations to "Mastering The 't' Method"?

This method may not work for all types of trigonometric equations, particularly those involving multiple trigonometric functions. It is important to understand the underlying concepts of trigonometry in order to use this method effectively.

5. How can I practice and improve my skills in "Mastering The 't' Method"?

The best way to improve in "Mastering The 't' Method" is through practice. There are many online resources and textbooks that provide practice problems and solutions using this method. It is also helpful to review and understand the concept of t = tan(x/2) and its applications in solving trigonometric equations.

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