The website Physicsforums.com contains a question entitled "Verifying a Hard Trigonometric Identity". The problem asked in this forum is answered here, as well as a discussion of why those who replied to the question failed to answer it. The article "Trigonometric Identities Lesson and Manipulating Trig Functions" describes the basic trigonometric identities, and lists some tips to help solve them. Thjs article shows how to use the tips in that article to solve the problem at Physicsforums.com
Tips To Solve and Simplify Trigonometric Expressions
This list of tips can be found at "Trigonometric Identities Lesson and Manipulating Trig Functions" (link above). The number of each tip will be shown with the proof.
- Questions that require a trigonometric identity to be proven are usually of the form Left Hand Side = Right Hand Side. The objective is not to find a value of θ for which it is true. The objective is to prove that the equation is true for ALL values of θ.
- Leave one side of the statement alone, and manipulate the other side to prove that it is equivalent. Very often, it is easier to do the proof by leaving the most simple expression alone, and simplify the other.
- It is useful to remember the formula for the difference of two squares: (a² - b²) = (a + b)(a - b). This is used in many trigonometric identity proofs.
- It is often the tendency of students to simplify expressions. Sometimes the opposite needs to be done e.g. Convert "1" into sin²x + cos²x
- Recognize key "pairs" like 1 + sin2x. Expanding both of these into (sin²x + cos²x + 2sinxcosx) allows the term to become (sinx + cosx)².
- Try to eliminate denominators, where possible. For example, the term 1 / (1 + sinx) may be simplified by multiplying numerator and denominator by 1 - sinθ. The denominator then becomes 1 - sin²x, or cos²x, which may then be divided through the numerator.
- This tip is by far the most important: Practice, practice, and practice. It is only with copious amounts of practice that trigonometric proofs and algebraic manipulation become fluent and easy.
Solution To Prove That cos²5x - cos²x = - sin4x . sin6x
cos²5x - cos²x = -sin4x . sin6x
Tip #1, LHS = RHS
Noting that RHS =
Tip #4, De-simplify
- sin4x.sin6x = - (sin5x - x).(sin(5x + x)
= - [(sin5x.cosx - cos5x.sinx).(sin5x.cosx - cos5x.sinx)]
Tip #3, Difference of two squares
= - (sin²5x.cos²x - cos²5x.sin²x)
= - [(1 - cos²5x).cos²x - cos²5x.(1 - cos²x)]
= - [cos²x - cos²5x.cos²x - cos²5x + cos²5x.cos²x]
= - (cos²x - cos²5x)
= cos²5x - cos²x
= LHS
Q.E.D.
Summary To Proving That cos²5x - cos²x = - sin4x . sin6x
This is a fairly straightforward proof, once the key step of de-simplifying sin6x and sin4x into sin(5x + x) and sin(5x - x) respectively is done. The attempts to prove this identity were unsuccessful purely because this key step was not taken. It is essential with proving trigonometric identities that the student keeps the objective in mind i.e. To make functions of 5x and x comparable to functions of 6x and 4x. This "high level" strategy is all that was missed by the students who attempted this proof.
References For Trigonometric Proof Strategy
The example used in this article was taken from the forum "Verifying a Hard Trigonometric Identity", and the tips and tricks described here come from the article "Trigonometric Identities Lesson and Manipulating Trig Functions".
Martin Bell - Martin holds a B.Sc. degree in chemical engineering, and an M.Sc. degree in electronics and computing. He has spent more than 25 years ...
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