Trig should I start by squaring both side?

1. Jul 20, 2011

jrjack

Trig...should I start by squaring both side???

1. The problem statement, all variables and given/known data

Fing the solutions that are in the interval $[0,2\pi)$

$$\tan 4t-\tan 7t=1+\tan 7t\tan 4t$$

2. Relevant equations

Use an addition or subtraction formula.
$$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$

3. The attempt at a solution

Should I start by squaring both sides, or move everything to one side and set equal to 0?
I tried both ways and I'm stuck after that step either way.

2. Jul 20, 2011

Pi-Bond

Re: Trig...should I start by squaring both side???

You can use the addition/subtraction formula if you can make the left hand side 1 by the appropriate division.

3. Jul 20, 2011

Integral

Staff Emeritus
Re: Trig...should I start by squaring both side???

Is tan odd or even?

What happens when you let b be negative?

4. Jul 20, 2011

jrjack

Re: Trig...should I start by squaring both side???

I'm not sure I understand what to divide by to get 1 on the left.

5. Jul 20, 2011

Pi-Bond

Re: Trig...should I start by squaring both side???

What would you divide 5 by to get 1 as the result?

6. Jul 20, 2011

jrjack

Re: Trig...should I start by squaring both side???

Oh... now I'm starting to get it, but would I be better served to make the right side 1?
By dividing by$$1+\tan 7t\tan 4t$$

leaving me with the subtraction formula.

7. Jul 20, 2011

Pi-Bond

Re: Trig...should I start by squaring both side???

The left hand side you are left with will easily give a value if the arctan function is applied to it, which is ideal in this case since you can use the subtraction formula.

8. Jul 20, 2011

jrjack

Re: Trig...should I start by squaring both side???

We haven't made it to the arctan function yet. How does this flip my fraction on the left and what will this do to the right side?

9. Jul 20, 2011

Pi-Bond

Re: Trig...should I start by squaring both side???

So you made the right side 1? In doing so you must have made the left side into the form of the subtraction formula. In other words you can express that side in terms of just a single tan function. Did you get till here?

10. Jul 20, 2011

jrjack

Re: Trig...should I start by squaring both side???

Yes, I have $$\tan(4t-7t)=1$$$$\tan(-3t)=1$$
For $$\tan(x)=1, x=\frac{\pi}{4}, \frac{5\pi}{4}$$
then dividing by -3 ???
I get $$t=-\frac{\pi}{12}, and -\frac{5\pi}{12}$$

something doesn't look right?

11. Jul 20, 2011

Pi-Bond

Re: Trig...should I start by squaring both side???

The values you got are correct, but not in the interval the question asks for.

How else can tan(-3t) be written? (Is it an even or an odd function?)

12. Jul 20, 2011

jrjack

Re: Trig...should I start by squaring both side???

Sorry the question was [0,pi), not 2pi.
and tan it is even. -tan(3t)

13. Jul 20, 2011

Pi-Bond

Re: Trig...should I start by squaring both side???

So now you have to solve the equation tan(3t)=(-1). Do you know where tan has the value -1 on the given interval?

14. Jul 20, 2011

jrjack

Re: Trig...should I start by squaring both side???

Yes, I forgot to divide by -1 to move the neg to the right side and solve for 3pi/4 and 7pi/4. Got it now.
Thanks.

15. Jul 20, 2011

Staff: Mentor

Re: Trig...should I start by squaring both side???

No, the tangent function is odd, which makes tan(-3t) = -tan(3t), which is what you have.

16. Jul 29, 2011

PeterO

Re: Trig...should I start by squaring both side???

Don't forget that you are first calculating a value for 3t, which you will then divide by 3 to get your final answers for t, so some possible answers that at first may appear to be too big, will reduce into the required range once you have divided by 3.

Peter

17. Jul 29, 2011

jrjack

Re: Trig...should I start by squaring both side???

Thanks.