Question

Verify the following differential identity: $$ \left(\frac{d}{d \theta}+n \cot \theta\right) f(\theta)=\frac{1}{\sin ^{n} \theta} \frac{d}{d \theta}\left[\sin ^{n} \theta f(\theta)\right] \text { . } $$

Short answer

After simplifying the output by applying the product rule, we came across a term \(\sin^{n-1}\theta n \cot(\theta) f(\theta)\), which could be transformed into \(n \cos(\theta)\sin^{n-1}\theta f(\theta)\), and \(f'(\theta)\sin^{n-1}\theta\). These are the same terms as on the LHS, showing that the identity holds.

Step-by-step solution

Expanding the Right Hand Side

First, the Right Hand side (RHS) of the identity needs to be expanded. This is done by applying the product rule of differentiation, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. In this case, the two functions are \(\sin^{n}\theta\) and \(f(\theta)\). So, the RHS becomes \(\sin^{n-1}\theta n \cot(\theta) f(\theta) + \sin^{n-1}\theta f'(\theta)\).

Unfolding cotangent

We replace cotangent with the reciprocal of tangent which is cosine over sin. So, the first term becomes \(\sin^{n-1}\theta \frac{n \cos(\theta)}{\sin(\theta)} f(\theta)\) . This simplifies to \(n \cos(\theta)\sin^{n-1}\theta f(\theta)\).

Simplifying the RHS

The RHS now has the same terms: The derivative of \(f(\theta)\) and \(n \cos(\theta)\sin^{n-1}\theta f(\theta)\). This proves the identity as our Right Hand Side has become the same as our Left Hand Side.

Key concepts

Product Rule of Differentiation

The product rule of differentiation is an essential tool in calculus, used when finding the derivative of a product of two functions. According to the product rule, if you have two functions, say \( u(x) \) and \( v(x) \), their derivative is computed through the formula:

• \((uv)' = u'v + uv'\)

When expanding the right-hand side (RHS) of our given differential identity, we have two functions, \( \sin^{n}\theta \) and \( f(\theta) \). The product rule guide us to differentiate \( \sin^{n}\theta \) while holding \( f(\theta) \) constant, and vice versa.

• First compute the derivative of \( \sin^{n}\theta \) and multiply by \( f(\theta) \).
• Then compute the derivative of \( f(\theta) \) and multiply by \( \sin^{n}\theta \).

The final expression combines both these derivatives.

Trigonometric Identities

In mathematics, trigonometric identities are equalities involving trigonometric functions that are true for all values within their domains. One common trigonometric identity used here involves \( \cot \theta \), which equals \( \frac{\cos \theta}{\sin \theta} \). This is the cotangent function expressed in terms of cosine and sine, which is pivotal in simplifying expressions.

• By expressing \( \cot \theta \) this way, it helps integrate trigonometric identities in more complex equations.

In the exercise, \( n \cot \theta \) was incorporated into the derivative expression. By replacing \( \cot \theta \) with \( \frac{\cos \theta}{\sin \theta} \), the solution becomes much simpler and manageable.
Understanding and applying such identities are crucial steps for solving equations related to trigonometric functions.

Verification of Mathematical Proofs

Verification of mathematical proofs involves a systematic approach to confirming that a given equality or identity holds true. For the exercise at hand, both sides of the differential identity were simplified independently to check if they match.

• This process ensures that each transformation and deduction adheres to valid mathematical laws.
• It is important to keep the goal in sight: make both sides of the equation reflect the same expression.

In mathematical proofs, such verification not only provides the correctness but also encourages clarity in understanding the relationship between mathematical components.

Differential Equations

Differential equations involve equations containing derivatives, which represent rates of change. Solving differential equations often involves identifying and applying relevant mathematical techniques.

• This exercise highlights how differential operators such as \( \frac{d}{d\theta} \) work in orchestrating variable changes.
• The commingling of algebraic and trigonometric components within these equations requires careful analysis and understanding of calculus concepts.

Grasping differential equations helps model and solve many real-world phenomena, where change is constant. These equations are pivotal in physics, engineering, economics, and beyond, hence, mastering them is crucial for students tackling calculus and applied mathematics.