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Trig and Inverse Trig Problems


Date: 30 Jul 1995 02:15:56 -0400
From: Greg Sharpe
Subject: Inverse Trig, Trig Equations

G'day!

   I have three trig-related questions. Would it be possible to include
explanations for solutions and some general rules? I am using Eudora on a
Macintosh and can read PICT files sent as attachments.

1) state the domain and range of

    y = 2 sin^-1 (2x+1)     and sketch the graph (if you support PICT files)

2) if f(x) = cos^-1 (-x) + cos^-1 (x), show that f'(x) = 0
   Hence determine the value of f(x)

3) solve 3sin (2x) = 2tan x,
      a) for 0<=x<=360
      b) give the general solution

Thanks,

Greg.


Date: 30 Jul 1995 12:36:42 -0400
From: Dr. Ken
Subject: Re: Inverse Trig, Trig Equations

Hello there!

>1) state the domain and range of  y = 2 sin^-1 (2x+1)     

Probably the easiest way to do this problem is to piece it together from
what you already know about the Sin^-1 function.  The graph of the Sin^-1
function is only defined between -1 and 1, right?  That's because Sine can
only output values between -1 and 1.  The range of Sin^-1 is -Pi/2 to Pi/2.  

So let's just do this piece by piece.  Here are some steps that get you from
the basic Sin^-1 function to what you want:

 Sin^-1 (x)
 Sin^-1 (2x)
 Sin^-1 (2x + 1)
2Sin^-1 (2x + 1)


Think about what each step does to change the function.  For instance, to go
from the first function to the second, all values of x are multiplied by 2.  
That means that whereas the domain of the first function was -1 to 1, the
domain of the second function will be -1/2 to 1/2.  Essentially, the
function will be "smooshed."  What effect will adding that 1 in there have?
And then multiplying the whole thing by 2?  I'll leave it to you to actually
graph it, but if you're having trouble, please write us back and we'll give
another hand.

>2) if f(x) = cos^-1 (-x) + cos^-1 (x), show that f'(x) = 0
>   Hence determine the value of f(x)

Well, let's take the derivative of both sides.  Since the derivative of
Cos^-1 is -1/Sqrt{1-x^2}, we get --1/Sqrt{1-x^2} + -1/Sqrt{1-x^2}, which is
zero.  So yeah, the derivative of f is zero.  Neat.  That means f is
constant.  So to find the value of f, we can just plug in some number in its
domain.  Plugging in 0, we find that f(0) = Cos^-1 (0) + Cos^-1 (0) 
                                          = Pi/2 + Pi/2
                                          = Pi.  

So the f(x) = Pi everywhere.

>3) solve 3sin (2x) = 2tan x,
>      a) for 0<=x<=360
>      b) give the general solution

I'll set this one up, then see if you can finish it.  The first thing you
should usually do (not always though, some things require creativity) is
change everything into Sin(x) and Cos(x).  Here we'll use the double-angle
formula for Sine.  Since Sin(2x) = 2Sin(x)Cos(x), we have
   6Sin(x)Cos(x) = 2Sin(x)/Cos(x).

Now divide through by 2Sin(x), and see what happens.

-K
    
Associated Topics:
High School Trigonometry

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