Why must angles be measured in radians when differentiating trigonometric functions from first principles?

The limit $\lim_{h \to 0} \frac{\sin h}{h} = 1$ is only valid in radians. In degrees, the limit would be $\frac{\pi}{180}$, which would change the resulting derivative formula.

What is the difference between the compound angle identities used for $\sin(x+h)$ and $\cos(x+h)$ in first principles?

The expansion for $\sin(x+h)$ uses a plus sign ($+$), leading to a positive $\cos(x)$ derivative. The expansion for $\cos(x+h)$ uses a minus sign ($-$), which results in the negative sign in $-\sin(x)$.

What common error occurs when students expand $\cos(x+h)$ during the derivation?

Students often incorrectly use a plus sign in the expansion. The identity is $\cos(A+B) = \cos A \cos B - \sin A \sin B$, and missing this minus sign leads to the incorrect derivative $\sin(x)$ instead of $-\sin(x)$.

State the two fundamental limit results required to differentiate $\sin(x)$ and $\cos(x)$ from first principles.

The two limits are $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$. These are derived from small angle approximations.

How does the small angle approximation for $\cos(h)$ explain why $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$?

Since $\cos(h) \approx 1 - \frac{1}{2}h^2$, the expression becomes $\frac{(1 - \frac{1}{2}h^2) - 1}{h} = -\frac{1}{2}h$. As $h \to 0$, this value clearly approaches zero.

When differentiating $\tan(x)$ from first principles, what is the specific small angle approximation used for the tangent function?

The approximation used is $\tan(h) \approx h$ as $h \to 0$. This allows the $\tan(h)$ term in the expanded tangent addition formula to be simplified during the limit process.

What is the result of $\lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$?

The result is $\cos(x)$. This is achieved by grouping $\sin(x)(\cos(h)-1)/h$, which goes to $0$, and $\cos(x)(\sin(h)/h)$, which goes to $\cos(x)(1)$.

What happens to the term $\sin(x) \cdot \frac{\cos(h)-1}{h}$ as $h$ approaches zero?

The term becomes zero. This is because $\sin(x)$ is independent of $h$, and the limit of the fraction $\frac{\cos(h)-1}{h}$ is zero.

Why is the 'limit' notation necessary throughout the derivation?

The notation indicates that we are looking at the behavior of the function as $h$ gets arbitrarily small, not evaluating it at $h=0$ (which would cause division by zero). It maintains mathematical rigor until the final substitution.

Compare the derivative of $\sin(kx)$ with the basic first principles result for $\sin(x)$.

While the first principles proof for $\sin(x)$ yields $\cos(x)$, the derivative of $\sin(kx)$ is $k\cos(kx)$. This involves an extra step of substituting $kx$ and $k(x+h)$ into the limit definition.

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Differentiation of Trigonometric Functions from First Principles

Summary

The differentiation of trigonometric functions from first principles uses the formal limit definition of a derivative combined with specific trigonometric identities and small-angle approximations. This process rigorously proves that the rate of change of $\sin(x)$ is $\cos(x)$ and the rate of change of $\cos(x)$ is $-\sin(x)$ , provided the angle is measured in radians.

1. Definition & Core Concepts

2. Small Angle Approximations

Geometric representation of small angle approximation showing that as the angle θ approaches zero, the vertical side (sin θ) and the arc length (θ) become nearly identical.

3. Methodology: Differentiating sin(x)

4. Methodology: Differentiating cos(x)

5. Key Distinctions

6. Exam Strategy & Common Pitfalls