# The t Method

1. Jun 23, 2006

“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:
$$t=tan \frac{A}{2}$$
$$tan A=\frac{1+t^2}{1-t^2}$$
$$sin A=\frac{2t}{1+t^2}$$
$$cos A=\frac{1-t^2}{1+t^2}$$
If this is famliar to a reader by another name, could you please post this methods other name.
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Problems which i need to solve:
1. $$2sin x + 3cos x = 5$$
2. $$3tan x + \sqrt{3}sec x=1$$
3. $$10cos (pi x) + 3sin (2pi x)=4$$
4. $$3sin 2x + 5cot 3x = 7$$
5. $$csc x + 2sec (pi x)$$
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My working for question 1. $$2sin x + 3cos x = 5$$
$$2\frac{2t}{1+t^2} + 3\frac{1-t^2}{1+t^2} = 5$$
$$\frac{4t + 3 - 3t^2}{1+t^2} = 5$$
$$4t + 3 -3t^2 = 5 +5t^2$$
$$4t + 2t^2 = 2$$
$$2t^2 + 4t - 2 = 0$$
$$t = 0.4142135624$$ or $$t = -2.414213562$$
$$t = \tan\frac{x}{2}$$
$$x = 2tan^-1 0.4142135624$$ or $$x = 2tan^-1 -2.414213562$$
$$x = 45$$ or $$x = -135.0005034$$
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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?

2. Jun 23, 2006

### HallsofIvy

You've lost track of a sign: adding 3t2 to both sides gives $8t^2- 4t+ 2= 0$ or $4t^2- 2t+ 1= 0$
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. Jun 24, 2006

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. Jun 24, 2006

### HallsofIvy

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. Jun 25, 2006

oh okay, ill post my wokring for the otherss when i have more time.

6. Jun 25, 2006

working for question number 2.

$$3tan x + \sqrt{3}sec x=1$$

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

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

7. Jun 25, 2006

### 0rthodontist

This is also known as the Weierstrass substitution.

8. Jun 25, 2006

### eglipo

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. Jun 26, 2006