Trigonometric IdentitiesDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/