Problem 16 - Calculus

Question

Show that \((\cos t+i \sin t)^{2}=\cos 2 t+i \sin 2 t,\) if \(i^{2}=-1\).

Step-by-Step Solution

Answer

The identity is proven using expansion and trigonometric identities.

Step 1: Expand the Left Side

Start by expanding \((\cos t + i \sin t)^2\). Use the identity \((a + b)^2 = a^2 + 2ab + b^2\). Here, \(a = \cos t\) and \(b = i \sin t\). This gives:\[(\cos t + i \sin t)^2 = \cos^2 t + 2(\cos t)(i \sin t) + (i \sin t)^2\]

Step 2: Simplify the Terms

Substitute \(i^2 = -1\) in \((i \sin t)^2\):\[(i \sin t)^2 = i^2 \sin^2 t = -\sin^2 t\]The expanded and simplified expression becomes:\[\cos^2 t - \sin^2 t + 2i \cos t \sin t\]

Step 3: Use Trigonometric Identities

Apply the double angle formulas for cosine and sine: \[\cos 2t = \cos^2 t - \sin^2 t\] \[\sin 2t = 2 \cos t \sin t\]This simplifies the expression to:\[\cos 2t + i \sin 2t\]

Step 4: Conclude the Equality

The expanded form \(\cos^2 t - \sin^2 t + 2i \cos t \sin t\) matches \(\cos 2t + i \sin 2t\), thus proving that \[(\cos t + i \sin t)^{2} = \cos 2t + i \sin 2t\]

Key Concepts

Euler's Formula

Euler's Formula is a fundamental bridge between the real exponential function and trigonometry. It's expressed as \(e^{ix} = \cos x + i\sin x\). This formula elegantly ties together exponential functions with sine and cosine in the complex plane.
This connection makes it essential for solving problems involving complex numbers and trigonometric identities.
Given any angle \(t\), by applying Euler's Formula, one can rewrite \(\cos t + i \sin t\) as the exponential \(e^{it}\).
Consequently, the exercise can alternatively express \((\cos t + i \sin t)^2\) as \((e^{it})^2\).
This leads straightforwardly to \(e^{i(2t)} = \cos 2t + i\sin 2t\).
It shows Euler's power in simplifying expressions in complex numbers.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variables.
In our context, two main identities are primarily used:

• The Pythagorean identity: \(\cos^2 t + \sin^2 t = 1\)
• The identity for imaginary square: \(i^2 = -1\)

The exercise rests on manipulating these identities to prove that combining cosine and sine in a complex form maintains the same equality as their double angle formula.
Trigonometric identities simplify the algebraic expression derived from the expansion.

Algebraic Expansion

Algebraic expansion involves multiplying out expressions, especially when dealing with binomials and polynomials.
When \((\cos t + i \sin t)^2\) is expanded, traditional binomial multiplication is employed to create
\(\cos^2 t + 2i \cos t \sin t - \sin^2 t\).
Here, recognizing \(i^2 = -1\) is crucial, as it converts the \((i \sin t)^2\) term to \(-\sin^2 t\).
This careful expansion and simplification are foundational to translate a mathematically rich expression into a more interpretable form.
The expansion paves the way to applying trigonometric identities effectively.

Double Angle Formulas

Double angle formulas are specialized trigonometric identities that express trigonometric functions of doubled angles \(2x\) in terms of \(x\).
They simplify the task of calculating trigonometric values for double angles using known angles.
In the exercise, the double angle formulas are:

• \(\cos 2t = \cos^2 t - \sin^2 t\)
• \(\sin 2t = 2 \cos t \sin t\)

These forms reveal that the complex expansion ultimately aligns with the well-known double angle identities.
This part of trigonometry is incredibly handy not just in simplifying problems, but also in converting trigonometric expressions into useful and solvable equations.
Thus, double angle formulas serve as both a tool and verification in confirming the correctness of the expression \((\cos t + i \sin t)^2 = \cos 2t + i \sin 2t\).