Trig should I start by squaring both side?

  • Thread starter jrjack
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  • #1
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Trig...should I start by squaring both side???

Homework Statement



Fing the solutions that are in the interval [itex][0,2\pi)[/itex]

[tex]\tan 4t-\tan 7t=1+\tan 7t\tan 4t[/tex]

Homework Equations



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

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.
 

Answers and Replies

  • #2
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You can use the addition/subtraction formula if you can make the left hand side 1 by the appropriate division.
 
  • #3
Integral
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Is tan odd or even?

What happens when you let b be negative?
 
  • #4
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I'm not sure I understand what to divide by to get 1 on the left.
 
  • #5
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What would you divide 5 by to get 1 as the result?
 
  • #6
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Oh... now I'm starting to get it, but would I be better served to make the right side 1?
By dividing by[tex]1+\tan 7t\tan 4t[/tex]

leaving me with the subtraction formula.
 
  • #7
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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
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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
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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
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Yes, I have [tex]\tan(4t-7t)=1[/tex][tex]\tan(-3t)=1[/tex]
For [tex]\tan(x)=1, x=\frac{\pi}{4}, \frac{5\pi}{4}[/tex]
then dividing by -3 ???
I get [tex]t=-\frac{\pi}{12}, and -\frac{5\pi}{12}[/tex]

something doesn't look right?
 
  • #11
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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
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Sorry the question was [0,pi), not 2pi.
and tan it is even. -tan(3t)
 
  • #13
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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
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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
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Sorry the question was [0,pi), not 2pi.
and tan it is even. -tan(3t)
No, the tangent function is odd, which makes tan(-3t) = -tan(3t), which is what you have.
 
  • #16
PeterO
Homework Helper
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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.

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
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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

Thanks.
 

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