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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)

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

- Doctor Ian, The Math Forum

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, 


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

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 

   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 

   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

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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