**Derivative of Sine Function**

The sine function, denoted as sin(x), is a periodic function that oscillates between -1 and 1, representing circular motion in mathematics and various physical phenomena in the real world. The derivative of the sine function is given by:

d/dx sin(x) = cos(x)

**Key Points**

- The derivative of sin(x) is cos(x), another fundamental trigonometric function. Understanding the basic differentiation rules is crucial for grasping these concepts.
- Both the sine function and its derivative are periodic, repeating their values every 2π radians.
- The maximum and minimum points of sin(x) occur where its derivative is zero, i.e., cos(x) = 0. This is an application of trigonometric identities in calculus.

**Example**

Consider the function f(x) = sin(x). Find the derivative and determine the slope of the tangent line at x = π/2.

**Solution**:

Using the derivative rule mentioned above, we have f'(x) = cos(x). At x = π/2, f'(π/2) = cos(π/2) = 0. Thus, the slope of the tangent line at x = π/2 is 0.

**Derivative of Cosine Function**

The cosine function, symbolised as cos(x), is crucial in mathematics, particularly in describing oscillatory and wave phenomena. Its derivative is:

d/dx cos(x) = -sin(x)

**Key Points**

- The derivative of cos(x) is -sin(x).
- The cosine function and its derivative are periodic, with a period of 2π radians.
- The points where cos(x) has local maxima and minima occur where its derivative is zero, i.e., -sin(x) = 0. Solving these kinds of problems efficiently often requires an understanding of solving trigonometric equations.

**Example**

Let's consider g(x) = cos(x). Find the derivative and evaluate the slope of the tangent line at x = 0.

**Solution**:

The derivative g'(x) = -sin(x). At x = 0, g'(0) = -sin(0) = 0. Therefore, the slope of the tangent line at x = 0 is 0.

**Derivative of Tangent Function**

The tangent function, expressed as tan(x), is essential in exploring the relationship between the sine and cosine functions. Its derivative is:

d/dx tan(x) = sec^{2}(x)

**Key Points**

- The derivative of tan(x) is sec
^{2}(x). - The tangent function and its derivative have a period of π radians.
- Critical points of tan(x) occur where its derivative is undefined, i.e., sec(x) = ±∞. For a deeper understanding of these concepts, exploring the integration of trigonometric functions can provide valuable insights.

**Example**

Given h(x) = tan(x), find the derivative and determine the slope of the tangent line at x = π/4.

**Solution**:

We have h'(x) = sec^{2}(x). At x = π/4, h'(π/4) = sec^{2}(π/4) = 2. Thus, the slope of the tangent line at x = π/4 is 2.

**Applications in Problems**

Differentiation of trigonometric functions finds extensive applications across various domains, including physics, engineering, and computer science. For instance, in wave mechanics, the velocity and acceleration of a particle in simple harmonic motion can be determined using derivatives of sine and cosine functions. Similarly, in electrical engineering, analysing alternating current circuits involves derivatives of sinusoidal functions. These applications often require combining trigonometric differentiation with differentiation of exponential and logarithmic functions for comprehensive analysis.

**Example**

A particle is moving along a path described by s(t) = 5sin(t) + 4cos(t), where s is in meters and t in seconds. Find the velocity of the particle at t = π/3 seconds.

**Solution**:

Velocity, being the rate of change of displacement with respect to time, is the derivative of s(t) with respect to t. Thus,

v(t) = d/dt s(t) = 5cos(t) - 4sin(t)

At t = π/3, v(π/3) = 5cos(π/3) - 4sin(π/3) = 5/2 - 2√3 m/s.

In these notes, we've explored the derivatives of sine, cosine, and tangent functions, providing a foundation for further studies in calculus and its applications in various fields. Through understanding and applying these basic differentiation rules, students can navigate through more complex problems and applications in their IB Mathematics journey. For further exploration of how trigonometric functions apply in different contexts, the study of applications of differentiation can expand your understanding of mathematical concepts in real-world scenarios.

## FAQ

Derivatives of trigonometric functions find extensive applications in various fields such as physics, engineering, and economics to analyse periodic phenomena. For example, in physics, the motion of pendulums and oscillating springs can be modelled using sine and cosine functions, and their derivatives are used to analyse velocities and accelerations. In electrical engineering, alternating current (AC) voltages can be described using trigonometric functions, and their derivatives help in understanding the rate of change of voltage and current. Moreover, in economics, trigonometric functions and their derivatives can be used to model and analyse cyclical behaviours in financial markets.

The points where the derivative of a trigonometric function is zero are significant because they indicate stationary points on the graph of the function. In the context of physical movement, such as a pendulum swinging, these stationary points represent moments where the object changes direction. Mathematically, for a function f(x), if f'(x) = 0, x is a critical point and may represent a local maximum, local minimum, or a point of inflection. In the realm of trigonometric functions like sin(x) and cos(x), these points often correspond to peaks and troughs in their wave-like graphs, providing vital information about their behaviour and properties.

Yes, the chain rule can be applied to the derivatives of trigonometric functions and is often used when the argument of the trigonometric function is not a simple x but a function of x. For instance, if we have a function f(x) = sin(g(x)), and we wish to find f'(x), we can apply the chain rule. The chain rule states that if you have a composite function, the derivative of the outer function should be evaluated at the inner function and multiplied by the derivative of the inner function. So, f'(x) = cos(g(x)) * g'(x). This application is crucial in solving problems where trigonometric functions are composed with other functions, providing a method to find rates of change and slopes of tangents in more complex scenarios.

The periodicity of trigonometric functions directly impacts their derivatives by also making them periodic. For instance, since sin(x) and cos(x) are periodic with period 2pi, their derivatives, cos(x) and -sin(x) respectively, are also periodic with the same period. This means that the rate of change of these functions repeats over intervals of 2pi. In practical applications, such as physics and engineering, this periodicity in the derivatives is crucial for modelling oscillatory and wave-like phenomena, ensuring that the behaviour of the system is consistent and repeats over regular intervals.

The derivative of sin(x) being equal to cos(x) is derived from the limit definition of the derivative. If we consider a unit circle, and take a point P that is an angle x radians from the positive x-axis, the x and y coordinates of P are cos(x) and sin(x) respectively. As we increase x by a small amount, say dx, the y-coordinate increases by a small amount dy. The ratio dy/dx as dx approaches 0 is the derivative of sin(x) with respect to x. Using trigonometric identities and the Pythagorean theorem, this ratio simplifies to cos(x). This relationship is fundamental in calculus and is used to determine rates of change and slopes of tangents for functions involving sin(x).

## Practice Questions

The derivative of the function y = 3sin(x) - 4cos(2x) with respect to x can be found by applying the basic trigonometric differentiation rules. The derivative of sin(x) is cos(x) and the derivative of cos(2x) is -2sin(2x). Applying these rules, we get dy/dx = 3cos(x) + 8sin(2x) To find the slope of the tangent to the curve at x = pi/3, we substitute this value into the derivative: dy/dx(pi/3) = 3cos(pi/3) + 8sin(2(pi/3)) = 3(1/2) + 8sin(2pi/3) = 3/2 + 8(sqrt(3)/2) = 3/2 + 4sqrt(3) Thus, the slope of the tangent to the curve at x = pi/3 is 3/2 + 4sqrt(3).

To find the velocity of the particle at a given time, we need to differentiate the displacement function s(t) = 2sin(t) + tcos(t) with respect to time t. Using the product rule for the second term (since it is a product of two functions, t and cos(t)) and basic trigonometric differentiation rules, we get v(t) = ds/dt = 2cos(t) + cos(t) - tsin(t) = 2cos(t) + cos(t) - tsin(t) = 3cos(t) - tsin(t) To find the velocity at t = pi/2, we substitute this value into the velocity function: v(pi/2) = 3cos(pi/2) - (pi/2)sin(pi/2) = 3(0) - (pi/2)(1) = -pi/2 Therefore, the velocity of the particle at t = pi/2 seconds is -pi/2 m/s.