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Date: 08/01/2003 at 09:37:49
From: Umair
Subject: Trigonometry

I couldn't solve this question and hoped that you could provide me 
with some help. Thanks.

   2tan x + sin^2 x * sec x = 1 + sec x


Date: 08/01/2003 at 17:17:49
From: Doctor Ian
Subject: Re: Trigonometry

Hi Umair,

Usually the first thing to do is to replace all other functions with
sines and cosines. For example, tan(x) is the same as sin(x)/cos(x),
so I would rewrite the equation as

   2 sin(x)
  ---------- + sin^2(x) sec(x) = 1 + sec(x)
     cos(x)

Try that with sec(x), and let me know if you're still stuck. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 08/01/2003 at 18:27:47
From: Umair
Subject: Trigonometry

Thanks Dr. Ian. I applied your advice to the equation:

   2tan(x) + sin(2x)sec(x) = 1 + sec(x)

I ended up with: 

   2sin(x) + sin(2x) = cos(x) + 1

Could you please provide me with some more advice?


Date: 08/01/2003 at 21:50:32
From: Doctor Ian
Subject: Re: Trigonometry

Hi Umair,

The next step is usually to see if you can use identities like the
ones in our Trigonometry Formulas FAQ, 

   http://mathforum.org/dr.math/faq/formulas/faq.trig.html 

In particular, you usually want to reduce everything to sines and
cosines of a single variable. In this case, there is an identity that
you can use to change the argument from (2x) to (x).  It is

  sin(2x) = 2 sin(x)cos(x),

Substituting this gives

  2sin(x) + 2sin(x)cos(x) = 1 + cos(x)

Can you take it from here? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 08/01/2003 at 22:02:52
From: Umair
Subject: Trigonometry

I solved it and got sin(x) = 1/2, so therefore the solutions are pi/3 
and 5*pi/6 (30 and 150 degrees). Which according to the question book 
are right, but there is one more answer according to the book that 
I can't get, which is also correct but is not a value of x in 
sin(x) = 1/2. It is pi (180 degrees), so logically the only way I can 
get this if cos(x) = -1.

Once again, I need your help!


Date: 08/01/2003 at 23:28:59
From: Doctor Ian
Subject: Re: Trigonometry

Your solution is nicely done. But let's look at the original equation 
again:

   2tan(x) + sin(2x) sec(x) = 1 + sec(x)

We converted to sines and cosines to get

   2sin(x)   sin(2x)   cos(x) + 1
   ------- + ------- = ----------
    cos(x)   cos(x)      cos(x)

and expanded sin(2x) to 2sin(x)cos(x) to get

   2sin(x)   2sin(x)cos(x)   cos(x) + 1
   ------- + ------------- = ----------
    cos(x)   cos(x)          cos(x)

and multiplied both side of the equation by cos(x) to get

   2sin(x) + 2sin(x)cos(x) = cos(x) + 1

which we factored to get

   2sin(x)(1 + cos(x)) = (1 + cos(x))

Now, we _can_ divide both sides by (1 + cos(x)), which gives us a
simpler equation. But it also eliminates one of the possible 
solutions, which occurs when (1 + cos(x)) is equal to zero... which
happens when x is equal to pi. 

Look at an equation like

  x^2 - 3x = 0

One way to 'solve' this is to divide both sides by x, giving you

  x - 3 = 0

      x = 3

This is sort of like dividing

   2sin(x)(1 + cos(x)) = (1 + cos(x))

on both sides by (1 + cos(x)) to get

   2sin(x) = 1

    sin(x) = 1/2

But in each case, you've eliminated a solution by dividing instead of
factoring. Let's look at the first one. If instead of dividing, we
do this:

  x^2 - 3x = 0

  x(x - 3) = 0

we can see that there are _two_ solutions: x=3, and x=0. We 'lost' the 
second solution when we divided both sides by x.  

In the second case, if we hold off dividing immediately, we can look 
at 

   2sin(x)(1 + cos(x)) = (1 + cos(x))

and see that this fits the pattern

    xy = y

which has _two_ solutions: x = y, and y = 0.  Dividing by y 'loses'
this second solution. 

Does that make sense? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 08/02/2003 at 10:00:26
From: Umair
Subject: Thank you (Trigonometry)

That makes perfect sense. I do appreciate all this help you have given 
me. Thanks, doc.
Associated Topics:
High School Trigonometry

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