Proving Trigonometric Identities: A Comprehensive Guide

In this article, we will explore a fascinating problem involving trigonometric identities. We aim to prove that if tanA sinA m and tanA - sinA n, then m^2 - n^2^2 16mn. This will be achieved using basic algebraic manipulations and trigonometric formulas, making it a valuable learning experience for students and advanced learners alike.

Introduction to the Problem

Given the equations m tan A sin A and n tan A - sin A, we will derive the relationship m^2 - n^2^2 16mn. This involves expressing m and n in terms of sine and cosine, then performing algebraic operations to derive the desired identity.

Step-by-Step Derivation

Step 1: Express tanA and sinA using trigonometric identities.

Recall that tan A sin A / cos A. We start by expressing m and n in terms of sine and cosine:

Expression for m:

Using the identity for tan A:

m tan A sin A sin A frac{sin A}{cos A} sin A left(frac{1}{cos A}right) sin A left(frac{1 cos A - cos A}{cos A}right) sin A left(frac{1 cos A - cos A}{cos A}right) sin A left(frac{1 cos A}{cos A}right)

Expression for n:

Similarly, for n:

n tan A - sin A frac{sin A}{cos A} - sin A sin A left(frac{1}{cos A} - 1right) sin A left(frac{1 - cos A}{cos A}right)

Derivation of m^2 and n^2

Next, we express m^2 and n^2 in terms of sine and cosine:

Expression for m^2:

m^2 left(sin A frac{1 cos A - cos A}{cos A}right)^2 sin^2 A frac{(1 cos A)(1 - cos A)}{cos^2 A} sin^2 A frac{1 - cos^2 A}{cos^2 A} sin^2 A frac{sin^2 A}{cos^2 A}

Expression for n^2:

n^2 left(sin A frac{1 - cos A}{cos A}right)^2 sin^2 A frac{(1 - cos A)^2}{cos^2 A} sin^2 A frac{1 - 2cos A cos^2 A}{cos^2 A}

Subtracting and Simplifying m^2 - n^2

Now we calculate m^2 - n^2:

m^2 - n^2 sin^2 A frac{1 - cos^2 A}{cos^2 A} - sin^2 A frac{(1 - cos A)^2}{cos^2 A} sin^2 A frac{1 - cos^2 A - (1 - 2cos A cos^2 A)}{cos^2 A} sin^2 A frac{1 - cos^2 A - 1 2cos A - cos^2 A}{cos^2 A} sin^2 A frac{-2cos^2 A 2cos A}{cos^2 A} 2sin^2 A frac{cos A(1 - cos A)}{cos^2 A} 2sin A left(frac{1 - cos A}{cos A}right) 2tan A sin A

Expressing (m^2 - n^2)^2 and 16mn

Next, we find (m^2 - n^2)^2:

(m^2 - n^2)^2 (2tan A sin A)^2 4tan^2 A sin^2 A 16tan^4 A sin^2 A

Now we calculate mn:

mn left(sin A frac{1 cos A - cos A}{cos A}right) left(sin A frac{1 - cos A}{cos A}right) sin^2 A frac{(1 cos A)(1 - cos A)}{cos^2 A} sin^2 A frac{1 - cos^2 A}{cos^2 A} sin^2 A frac{sin^2 A}{cos^2 A} tan^2 A sin^4 A

Verification of the Identity

To verify the identity, we need to show that 16tan^4 A sin^2 A 16tan^2 A sin^4 A:

16tan^4 A sin^2 A 16tan^2 A sin^4 A 16mn

Thus, we have demonstrated that 16tan^4 A sin^2 A 16mn. This completes the proof.

Alternative Calculation Method

Another method involves the following steps:

Step 1: Calculate mn:

mn tanA sinA (tanA - sinA) tan^2 A - sin^2 A sin^2 A left(frac{1}{cos^2 A} - 1right) sin^2 A frac{1 - cos^2 A}{cos^2 A} sin^2 A tan^2 A (tan A sin A) (tan A - sin A)

Step 2: Calculate the square root of mn:

sqrt{mn} sqrt{tanA sinA (tanA - sinA)} sinA tanA sqrt{tan^2A - sin^2A}

Step 3: Calculate m^2 - n^2:

m^2 - n^2 4(2 tan atimes 2 sin a) 4(2 tan a sin a) 4 tan a sin a 4√tan a sin a (tan a - sin a) 4√tan^2 a - sin^2 a 4 sin a √1 - cos^2 a / cos a 4 tan a sin a

Conclusion

In conclusion, we have demonstrated the identity m^2 - n^2^2 16mn by expressing m and n in terms of sine and cosine, and through algebraic manipulations. This problem not only tests the understanding of trigonometric identities but also enhances problem-solving skills in mathematics.