Expressing sin2x in Terms of tanx: A Comprehensive Guide

In this article, we will explore the relationship between sin2x and tanx, using both algebraic manipulation and logarithmic techniques to derive the equation. Understanding the double-angle formula and trigonometric identities is crucial for solving such problems efficiently.

Introduction

Trigonometric identities form the backbone of trigonometry and are essential in various applications, from physics to engineering. One such identity is the expression of (sin{2x}) in terms of (tan{x}). This identity is particularly useful when working with angles and trigonometric functions in equations.

Algebraic Manipulation Method

Let's start by using the double-angle identity for sine and relating it to (tan{x}). The double-angle formula for sine is:

[sin{2x} 2 sin{x} cos{x}]

Next, we express (sin{x}) and (cos{x}) in terms of (tan{x}). Recall that:

[tan{x} frac{sin{x}}{cos{x}}]

From this, we can express (sin{x}) as:

[sin{x} tan{x} cos{x}]

Substituting this into the double-angle formula, we get:

[sin{2x} 2 tan{x} cos^2{x}]

Now, we need to express (cos^2{x}) in terms of (tan{x}). We use the identity:

[cos^2{x} frac{1}{1 tan^2{x}}]

Substituting this back into our equation for (sin{2x}), we get:

[sin{2x} 2 tan{x} frac{1}{1 tan^2{x}} frac{2 tan{x}}{1 tan^2{x}}]

Logarithmic Technique

An alternative approach involves using logarithmic functions to simplify the problem. Assume:

[x tan^{-1}{u} frac{1}{2i} ln{left(frac{1 eu}{1 - eu}right)}]

Then:

[tan{x} u]

Next, we want to express (sin{2x}) in terms of (u). Recall the double-angle formula for sine:

[sin{2x} frac{e^{2ix} - e^{-2ix}}{2i}]

Given that:

[x frac{1}{2i} ln{left(frac{1 eu}{1 - eu}right)}]

We have:

[2ix ln{left(frac{1 eu}{1 - eu}right)}]

Therefore:

[e^{2ix} frac{1 eu}{1 - eu}]

The multiplicative inverse:

[e^{-2ix} frac{1 - eu}{1 eu}]

Substituting these back into the formula for (sin{2x}), we get:

[sin{2x} frac{frac{1 eu}{1 - eu} - frac{1 - eu}{1 eu}}{2i}]

Combining the fractions:

[sin{2x} frac{frac{(1 eu)^2 - (1 - eu)^2}{(1 - eu)(1 eu)}}{2i}]

Expanding the numerator:

[(1 eu)^2 - (1 - eu)^2 (1 2eu e^{2}u^2) - (1 - 2eu e^{2}u^2) 4eu]

Thus:

[sin{2x} frac{frac{4eu}{(1 - eu)(1 eu)}}{2i} frac{2eu}{1 - e^{2}u^2}]

Using the fact that (tan{x} u), we have:

[sin{2x} frac{2 tan{x}}{1 - tan^2{x}}]

Conclusion

Using either the algebraic manipulation method or the logarithmic technique, we have shown that:

[sin{2x} frac{2 tan{x}}{1 tan^2{x}}]

This identity is particularly useful in various applications, especially when dealing with trigonometric functions in physics or engineering problems. Understanding and memorizing such identities can greatly simplify problem-solving processes in these fields.