Question

Let $f(x)={\cos }^{2}x+{\sin }^{2}x$. a. Use the Chain Rule to show that $f^{\prime }(x)=0$. b. Assume that if $f^{\prime }=0,$ then $f$ is a constant function. Calculate $f(0)$ and use it with part (a) to explain why ${\cos }^{2}x+{\sin }^{2}x=1$.

Short answer

Question: Prove that ${\cos }^{2}x+{\sin }^{2}x=1$ by showing that the derivative of $f(x)={\cos }^{2}x+{\sin }^{2}x$ is $0$, and then finding the value of $f(0)$. Answer: By applying the Chain Rule, we found that the derivative of $f(x)={\cos }^{2}x+{\sin }^{2}x$ is $f^{\prime }(x)=0$. We also computed that $f(0)=1$. Since $f^{\prime }(x)=0$ and $f(0)=1$, this implies that $f(x)$ is a constant function with a value of 1 for all $x$. Therefore, ${\cos }^{2}x+{\sin }^{2}x=1$ for all values of $x$.

Step-by-step solution

Find the derivative of ${\cos }^{2}x$

To find the derivative of ${\cos }^{2}x$, we must use the Chain Rule. The Chain Rule states that the derivative of a composite function is the product of the derivative of the outer function times the derivative of the inner function. The inner function is $u=\cos x$, and the outer function is $v(u)=u^2$. To find the derivative of the composite function $v(u(x))$, we will need to find the derivatives of $u(x)$ and $v(u)$ with respect to $x$, and then multiply them together: 1. Finding the derivative of $u(x)$: $\frac{du}{dx}=-\sin x$ 2. Finding the derivative of $v(u)$: $\frac{dv}{du}=2u$ Now, we apply the Chain Rule by plugging $u=\cos x$ into $\frac{dv}{du}$: $\frac{d({\cos }^{2}x)}{dx}=-2\cos x\sin x$

Find the derivative of ${\sin }^{2}x$

Similar to Step 1, to find the derivative of ${\sin }^{2}x$, we will again use the Chain Rule. The inner function is $u=\sin x$, and the outer function is $v(u)=u^2$. We will need to find the derivatives of $u(x)$ and $v(u)$ with respect to $x$, and then multiply them together: 1. Finding the derivative of $u(x)$: $\frac{du}{dx}=\cos x$ 2. Finding the derivative of $v(u)$: $\frac{dv}{du}=2u$ Now, we apply the Chain Rule by plugging $u=\sin x$ into $\frac{dv}{du}$: $\frac{d({\sin }^{2}x)}{dx}=2\sin x\cos x$

Compute the derivative of $f(x)$

Now, we can find the derivative of $f(x)$ by adding the derivatives of both terms: $f^{\prime }(x)=\frac{d({\cos }^{2}x)}{dx}+\frac{d({\sin }^{2}x)}{dx}=-2\cos x\sin x+2\sin x\cos x$ As we can see, the two terms cancel each other out: $f^{\prime }(x)=0$ This confirms that the derivative of $f(x)$ is 0.

Compute $f(0)$

We should now find the value of $f(0)$ by plugging in $x=0$: $f(0)={\cos }^2(0)+{\sin }^2(0)=1^2+0^2=1$

Confirm that $f(x)$ is a constant function

Since we have shown that $f^{\prime }(x)=0$ and $f(0)=1$, this means that $f(x)$ is a constant function with a value of 1 for every $x$. Therefore, we can conclude that ${\cos }^{2}x+{\sin }^{2}x=1$ for all values of $x$.

Key concepts

Pythagorean Identity

The Pythagorean Identity is a fundamental principle in trigonometry. It states that for any angle $x$, the sum of the squares of the sine and cosine of $x$ is always equal to one. Mathematically, it can be expressed as ${\cos }^{2}x+{\sin }^{2}x=1$. This identity stems from the Pythagorean theorem applied in a unit circle, where the hypotenuse of each right triangle formed is always 1. This makes understanding the relationship between sine, cosine, and the unit circle essential.The unit circle helps to visualize this identity. When any angle $x$ is placed on the unit circle, the $x$-coordinate of the point where the ray intersects the circle is $\cos x$ and the $y$-coordinate is $\sin x$. Squaring these coordinates and adding them together will always result in 1, hence demonstrating the Pythagorean Identity. This identity is not only vital in solving trigonometric equations but is frequently used in higher-level calculus and math problems.

Derivative Calculation

Calculating derivatives is a core part of calculus, and the Chain Rule is a powerful tool for doing so, particularly with composite functions. In the exercise, we used the Chain Rule to derive $f^{\prime }(x)$ for $f(x)={\cos }^{2}x+{\sin }^{2}x$. The Chain Rule states that if you have a function within a function, you differentiate the outer function and multiply it by the derivative of the inner function.For ${\cos }^{2}x$:

• The inner function is $u=\cos x$, so $\frac{du}{dx}=-\sin x$.
• The outer function is $v(u)=u^2$, leading to $\frac{dv}{du}=2u$.
• Thus, $\frac{d({\cos }^{2}x)}{dx}=-2\cos x\sin x$.

Similarly, for ${\sin }^{2}x$:

• The inner function is $u=\sin x$, so $\frac{du}{dx}=\cos x$.
• And for the outer function $v(u)=u^2$, $\frac{dv}{du}=2u$.
• This results in $\frac{d({\sin }^{2}x)}{dx}=2\sin x\cos x$.

Adding these derivatives, we find that the terms cancel each other out:\ f'(x) = -2 \cos x \sin x + 2 \sin x \cos x = 0.\ This confirms that the entire function's derivative is zero.

Constant Function

In calculus, a constant function is one whose output value does not change regardless of the input. That means its derivative is always zero, reflecting no change or slope.In the exercise, we established that $f^{\prime }(x)=0$, which suggests that $f(x)$ remains constant. This led us to understand $f(x)$ as a constant, specifically at the calculated value $f(0)=1$.For a function where its derivative is zero to be constant might seem straightforward, yet it highlights a critical characteristic of constant functions. Since the derivative represents the rate of change, a zero derivative across its domain implies no change, confirming constancy. That is why when $f^{\prime }(x)=0$ for the whole domain, the identity ${\cos }^{2}x+{\sin }^{2}x=1$ holds true. The value remains constant because of the Pythagorean Identity. This reinforces the nature of these trigonometric functions tied to the fundamental properties of the unit circle, eternally constant regardless of the angle $x$.'