Question

Let \(s(x)\) and \(c(x)\) be two functions satisfying \(s^{\prime}(x)=c(x)\) and \(c^{\prime}(x)=-s(x)\) for all \(x .\) If \(s(0)=0\) and \(c(0)=1,\) prove that \([s(x)]^{2}+[c(x)]^{2}=1\)

Short answer

Given that the derivative of the assumed identity \([s(x)]^{2}+[c(x)]^{2}=1\) is zero and this assumed identity holds at \(x=0\), it proves that this identity is true for all \(x\).

Step-by-step solution

Assume the Identity to Be True

Assume \([s(x)]^{2}+[c(x)]^{2}=1\) as the identity to be proven.

Differentiate the Assumed Identity

Differentiate the above expression with respect to \(x\) which gives \(2s(x)s'(x) + 2c(x)c'(x)=0\).

Substitute the Given Equations

Substitute given \(s^{\prime}(x)=c(x)\) and \(c^{\prime}(x)=-s(x)\) into the equation which results to \(2s(x)c(x) - 2c(x)s(x)=0\).

Simplify the Equation

Simplify the equation to \(0=0\), which indicates that the derivative of the assumed identity is equal to zero, a requirement for any constant.

Utilize Boundary Conditions

Use the boundary conditions \(s(0)=0\) and \(c(0)=1\) to find that our assumed \([s(x)]^{2}+[c(x)]^{2}=1\) identity indeed holds at \(x=0\). Hence, because its derivative is always zero and it holds at \(x=0\), it must hold for all x.

Key concepts

Derivatives of Trigonometric Functions

Understanding the derivatives of trigonometric functions is a pivotal aspect in calculus, particularly when dealing with problems involving rates of change and motion.
When we have a function like the sine function, denoted as \( s(x) \), its derivative with respect to \( x \) is represented as \( s'(x) \). Similarly, the cosine function, \( c(x) \), has a derivative denoted by \( c'(x) \).

For example, the standard derivatives of basic trigonometric functions are known: \( \frac{d}{dx}(\sin(x)) = \cos(x) \) and \( \frac{d}{dx}(\cos(x)) = -\sin(x) \). These relations express how the rate of change of the sine function equates to the value of the cosine function and, conversely, how the rate of change of the cosine function relates to the negative value of the sine function.

In the context of our exercise, think of \( s(x) \) and \( c(x) \) as analogous to the sine and cosine functions, respectively. This understanding allows us to approach the proof methodically, by verifying the properties of these functions through differentiation, as outlined in the exercise solution.

Proofs in Calculus

In the realm of calculus, a proof is a logical argument that demonstrates the truth of a mathematical statement. Proofs require a blend of creativity, intuition, and rigorous application of calculus techniques.
For instance, one common strategy is assuming the identity you wish to prove as in the exercise. This step, known as assuming the consequent, is not the proof itself but a step that leads to verifying the necessary conditions.

Differentiation in proofs

Following the assumption, differentiation is used to show that the identity's derivative results in a true statement - typically, zero equals zero. This is because constant values don’t change, hence their derivative is zero. If the derivative of an assumed identity always results in zero, it suggests that the original function could be a constant.

In applying derivatives to our trigonometric functions \( s(x) \) and \( c(x) \), we were able to simplify the derivative of the identity to zero. This method of proving by showing that the derivative of an assumed constant identity equates to zero is often used for identities, as shown in our solution.

Calculus Boundary Conditions

Boundary conditions refer to the values of a function at specific points, which are crucial in defining solutions to differential equations and in proof problems in calculus.
These conditions can be initial, describing the state of a function at the beginning of the interval of interest, or boundary, providing information at the extremes of the interval.

Function values and boundary conditions

With \( s(0) = 0 \) and \( c(0) = 1 \) at \( x = 0 \), these initial conditions allow us to confirm whether our assumed identity holds at this specific point. If the identity holds at \( x = 0 \), and we have previously shown that its derivative is zero for all \( x \), we can deduce that the assumed identity \( [s(x)]^2 + [c(x)]^2 = 1 \) is indeed true for all \( x \).

This method illustrates how boundary conditions serve as a critical checkpoint to verify the generality of an identity or solution. The exercise presented utilizes these conditions to finalize the proof, affirming that if an equation holds under its boundary conditions and has a constant derivative (in this case, zero), the equation must be valid everywhere, for all \( x \) in its domain.