Proving the Trigonometric Identity: sinB - sinC / sinBsinC cot(BC / 2)tan(B - C / 2)

Trigonometry is a fundamental branch of mathematics with a plethora of identities that are often used in solving complex problems in various fields, including engineering, physics, and mathematics. This article focuses on one such identity, proving the following:

sinB - sinC / sinBsinC cot(BC / 2)tan(B - C / 2)

Introduction to Trigonometric Identities

Trigonometric identities are equations that relate different trigonometric functions and angles. They are useful in simplifying trigonometric expressions and solving equations involving trigonometric functions. One of the most common sets of identities involved are the sum-to-product formulas and the angle-sum identities, which we will use in this proof.

The Sum-to-Product Formulas

The sum-to-product formulas are a set of trigonometric identities that allow us to express the sum or difference of sines and cosines as products. These formulas are particularly useful in transforming expressions into more manageable forms. The formulas are stated as follows:

sinB - sinC 2cos(B C / 2)sin(B - C / 2)

sinBsinC 2sin(B C / 2)cos(B - C / 2)

The Proof Using Sum-to-Product Formulas

Let's start with the left-hand side (LHS) of the identity:

LHS: (frac{sin B - sin C}{sin B sin C})

Using the sum-to-product formulas, we can rewrite the numerator and the denominator:

Numerator: (sin B - sin C 2cosleft(frac{B C}{2}right)sinleft(frac{B - C}{2}right))

Denominator: (sin B sin C 2sinleft(frac{B C}{2}right)cosleft(frac{B - C}{2}right))

Now, substituting these into the LHS:

LHS: (frac{2cosleft(frac{B C}{2}right)sinleft(frac{B - C}{2}right)}{2sinleft(frac{B C}{2}right)cosleft(frac{B - C}{2}right)})

Notice that the factor of 2 cancels out, leaving us with:

LHS: (frac{cosleft(frac{B C}{2}right)sinleft(frac{B - C}{2}right)}{sinleft(frac{B C}{2}right)cosleft(frac{B - C}{2}right)})

Which can be further simplified to:

LHS: (frac{1}{tanleft(frac{B C}{2}right)}tanleft(frac{B - C}{2}right))

Recognizing that (frac{1}{tanleft(frac{B C}{2}right)} cotleft(frac{B C}{2}right)), we get:

LHS: (cotleft(frac{B C}{2}right)tanleft(frac{B - C}{2}right))

Thus, the left-hand side (LHS) is equal to the right-hand side (RHS) of the identity, proving the given trigonometric identity:

(frac{sin B - sin C}{sin B sin C} cotleft(frac{BC}{2}right)tanleft(frac{B - C}{2}right))

The Proof Using Angle-Sum Identities

Let's also prove the identity using the angle-sum identities. Let's denote:

B C / 2 P

B - C / 2 Q

Therefore:

B C 2P

B - C 2Q

Adding the two equations gives:

2B 2P 2Q

B P Q

Subtracting the two equations gives:

2C 2P - 2Q

C P - Q

Now, substituting these into the original identity:

(frac{sin P sin Q - sin P sin(-Q)}{sin P sin Q})

Using the sine rules:

(sin P sin Q 2 sinleft(frac{P Q}{2}right)cosleft(frac{P-Q}{2}right))

(sin P sin(-Q) 2 sinleft(frac{P Q}{2}right)cosleft(frac{P Q}{2}right))

The identity now simplifies to:

(frac{2 sinleft(frac{P-Q}{2}right) cosleft(frac{P Q}{2}right) - 2 sinleft(frac{P Q}{2}right) cosleft(frac{P-Q}{2}right)}{2sinleft(frac{P Q}{2}right)cosleft(frac{P-Q}{2}right)})

This simplifies to:

(frac{2 left(sinleft(frac{Q-P}{2}right) - sinleft(frac{P Q}{2}right)right) cosleft(frac{P Q}{2}right)}{2sinleft(frac{P Q}{2}right)cosleft(frac{P-Q}{2}right)})

And further simplifies to:

(frac{tanleft(frac{Q-P}{2}right) - tanleft(frac{P Q}{2}right)^{-1}}{1})

Which is:

(cotleft(frac{P Q}{2}right)tanleft(frac{P-Q}{2}right))

Conclusion

In conclusion, both methods have shown that the given trigonometric identity, sin B - sin C / sin B sin C cot(BC / 2)tan(B - C / 2), holds true. Understanding and applying these identities is crucial for problem-solving in various fields, including trigonometry, physics, and engineering. By mastering these fundamental identities, one can navigate complex problems with ease.

References:

Fraleigh, J.B. (2003), A First Course in Abstract Algebra, 7th Edition, Addison Wesley. Stewart, J. (2015), Calculus: Early Transcendentals, 8th Edition, Brooks/Cole. Thomas, G.B., Weir, M.D., Hass J. (2016), Thomas' Calculus, 14th Edition, Pearson.