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Trig Identities and Double Angles


Date: 01/04/98 at 19:30:45
From: Monique Sandiford
Subject: Trig Identities- Double Angles

Prove that 

                   cosA + sinA
  sec2A + tan2A =  ------------
                   cosA - sinA


Date: 01/05/98 at 11:41:52
From: Doctor Wilkinson
Subject: Re: Trig Identities- Double Angles

My first step in a problem like this would be to express everything in 
terms of sines and cosines, since you have sines and cosines on the 
right side. 

Also I know the identities for sines and cosines pretty well, but I 
tend to forget them for the other functions.

So

   sec2A + tan2A  =    1    +    sin2A   =   1 + sin2A
                     -----       -----       ---------
                     cos2A       cos2A          cos2A

Now apply the double angle formulas for the sin and cosine to get

            1 + 2sinA cosA
            -------------------
            cos^2(A) - sin^2(A)

Now it's time to get clever, because there are a number of directions 
we might consider going.  The denominator factors as a difference of 
squares:

            cos^2(A) - sin^2(A) = (cosA + sinA)(cosA - sinA)

This is kind of promising, because one of the factors is cosA - sinA, 
which is what we want to get in the denominator.  This tells us we 
probably don't want to substitute 

  cos^2(A) = 1 - sin^2(A)  or  sin^2(A) = 1 - cos^2(A) 

in the denominator. We already have the form we want in the 
denominator; let's look at what we can do with the numerator.  

At first glance things don't look so good. We could substitute for sin 
in terms of cosine or cosine in terms of sine, but that would bring 
square roots in, and would complicate things rather than simplify 
them.

A useful trick in trig identities is sometimes to make a substitution 
that looks crazy until you see how it works. Let's substitute

    sin^2(A) + cos^2(A) for 1

This turns the numerator into

    sin^2(A) + cos^2(A) + 2 sinAcosA 

which we can arrange to

    sin^2(A) + 2 sinAcosA + cos^2(A)

which factors into

    (sin(A) + cos(A))^2 = (cosA + sinA)^2

and the rest is easy.

-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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