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Trigonometric Identities


Date: 11/13/97 at 00:13:07
From: Charlotte
Subject: Trigonometric identities

Dear Dr. Math,

I am currently taking Trig in my junior year and I am having trouble 
finding an easy way to understand identities for trigonometric 
functions. For example, when I am faced with a problem such as this 
one: tan1/2x+cot1/2x = 2cscx I usually know that I have to begin on 
the left side. Unfortunately I never know which formula is appropriate 
for solving these equations.. I myself would choose tan1/2x half 
argument for the first part, but after that I do not know where to go. 
Do you have any suggestions as to finding an easy solution to 
remember that I could apply to all identities? And possibly could you 
help me solve this problem?

Thank you.


Date: 11/13/97 at 10:18:17
From: Doctor Rob
Subject: Re: Trigonometric identities

Very good questions!

There is a trade-off you have to make between ease of remembering and
short and beautiful solutions. Yes, there is an easy-to-remember 
method of verifying any identity. See below for details. 

Unfortunately, it often produces a lengthy proof. The short, beautiful 
proofs are almost always found by a process of pattern-recognition and 
intuition that is hard to describe and make systematic. I will 
illustrate both methods with your problem.

The easy-to-remember method goes like this:

1. Rewrite all secants, cosecants, tangents, and cotangents in terms 
   of sines and cosines.

2. Take all the arguments of the sines and cosines and find the 
   greatest common divisor of them, say u.  Use multiple angle 
   identities to write everything in terms of sin(u) and cos(u).  
   When arguments appear whose greatest common divisor is 1, like 
   sin(x+y) and cos(y-2*x), use identities for the sine or cosine 
   of a sum or difference of angles to simplify them to functions of 
   sin(x), cos(x), sin(y), cos(y), and so on.

3. Put each side over a common denominator.

4. Whenever second or higher powers of cos(u) occur, use the identity
   cos^2(u) = 1 - sin^2(u) to convert them to powers of sin(u).

5. Simplify whenever you can by combining like terms and pulling out
   common factors.

6. When you have obtained an obvious identity, reverse all the steps 
   to start with the obvious identity and end with what you want to 
   show.  This will be the proof of your identity.

   Your identity:

   tan(x/2) + cot(x/2) = 2*csc(x),
   sin(x/2)/cos(x/2) + cos(x/2)/sin(x/2) = 2/sin(x),
   sin(u)/cos(u) + cos(u)/sin(u) = 2/sin(2*u)   [u = x/2],

   because both x and x/2 are multiples of x/2,

   [sin^2(u) + cos^2(u)]/[sin(u)*cos(u)] = 2/[2*sin(u)*cos(u)],
   1/[sin(u)*cos(u)] = 1/[sin(u)*cos(u)],
   1/[sin(x/2)*cos(x/2)] = 1/[sin(x/2)*cos(x/2)]   [u = x/2].

   Now reverse all the steps.

   The shorter, cleverer proof:

   tan(x/2) + cot(x/2) = 2*csc(x),
   sin(x)*[tan(x/2)+cot(x/2)] = 2,
   2*sin(x/2)*cos(x/2)*[tan(x/2)+cot(x/2)] = 2,
   sin(x/2)^2 + cos(x/2)^2 = 1,

   which you know.  Now reverse the steps.

How was the shorter proof obtained? I observed that the common
denominator I would have on the left would be sin(x/2)*cos(x/2), which
reminded me immediately of the sine double-angle formula sin(2*t) =
2*sin(t)*cos(t), with t = x/2. Then I saw that sin(2*t) = sin(x) was 
just the denominator on the right. It seemed a good idea to multiply 
through by the common denominator and use the double-angle formula on 
the left, which I did. After expanding and using both cos(t)*tan(t) = 
sin(t) and sin(t)*cot(t) = cos(t), the left side was simplified, too.  
After dividing out a 2 from both sides, I was left with a familiar 
identity.

The "reminded" part is a matter of pattern-matching with a list of
memorized identities. In this case, the pattern in the denominator of 
the left side when the two terms are added matched the double-angle 
sine formula I know. It is almost subconscious with me now, but you 
would ask yourself questions like, "Do I know any identities which 
look like sine of something times the cosine of the same thing?"

Sometimes there will be more than one subexpression in the equation 
that matches familiar identities, and then you have to explore, "What 
would happen if I used the such-and-such identity with this 
expression? How would it look after I applied it?" If the result is 
simpler than the start, it probably is a good idea to proceed. If the 
result is no simpler, or more complicated, it probably is not a good 
move!

I hope I have helped with this answer.  If not, write again.

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

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