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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.

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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