Question

a. Differentiate both sides of the identity \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\) to prove that \(\sin 2 t=2 \sin t \cos t\). b. Verify that you obtain the same identity for sin \(2 t\) as in part (a) if you differentiate the identity \(\cos 2 t=2 \cos ^{2} t-1\). c. Differentiate both sides of the identity \(\sin 2 t=2 \sin t \cos t\) to prove that \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\).

Short answer

Question: Prove the following trigonometric identities using differentiation: a) \(\sin 2t = 2\sin t \cos t\) using the identity \(\cos 2t = \cos^2 t - \sin^2 t\) b) Verify that the identity for \(\sin 2t\) is the same as in part (a) using the identity \(\cos 2t = 2\cos^2 t -1\) c) Prove that \(\cos 2t=\cos^2 t - \sin^2 t\) using the identity \(\sin 2t=2\sin t\cos t\) Answer: a) Differentiate the given identity, equate the derivatives, and simplify to obtain \(\sin 2t = 2\sin t \cos t\). b) Differentiate the given identity, equate the derivatives, and simplify to confirm \(\sin 2t = 2\sin t \cos t\). c) Differentiate the given identity, equate the derivatives, and simplify to obtain \(\cos 2t = \cos^2 t - \sin^2 t\).

Step-by-step solution

Differentiate both sides of the given identity

To start, differentiate each side of the equation \(\cos 2t = \cos^2 t - \sin^2 t\) with respect to \(t\). On the left, we will apply the chain rule, and on the right, we'll use the sum/difference rule and chain rule. Left side differentiation: \[\frac{d}{dt}\cos 2t = -\sin 2t \cdot 2\] Right side differentiation: \[\frac{d}{dt}\left(\cos^2 t - \sin^2 t\right)= 2\cos t(-\sin t) - 2\sin t (\cos t) \]

Equate the derivatives

Now, set the derivatives from Step 1 equal to each other: \[-2\sin 2t = -(2\sin t \cos t + 2\sin t \cos t)\]

Simplify the equation and obtain the desired identity

Simplify the equation to get the desired identity: \[\sin 2t = 2\sin t \cos t\] b. Verify that the identity for \(\sin 2t\) is the same as in part (a) using the identity \(\cos 2t = 2\cos^2 t -1\)

Differentiate both sides of the given identity

Differentiate each side of the equation \(\cos 2t = 2\cos^2 t -1\) with respect to \(t\). Left side differentiation: \[\frac{d}{dt}\cos 2t = -\sin 2t \cdot 2\] Right side differentiation: \[\frac{d}{dt}\left(2\cos^2 t -1\right)= 4\cos t(-\sin t) \]

Equate the derivatives

Set the derivatives from Step 1 equal to each other: \[-2\sin 2t = -4\sin t \cos t\]

Simplify the equation and obtain the desired identity

Simplify the equation to get the desired identity: \[\sin 2t = 2\sin t \cos t\] c. Prove that \(\cos 2t=\cos^2 t - \sin^2 t\) using the identity \(\sin 2t=2\sin t\cos t\)

Differentiate both sides of the given identity

Differentiate each side of the equation \(\sin 2t = 2\sin t \cos t\) with respect to \(t\). Left side differentiation: \[\frac{d}{dt}\sin 2t = \cos 2t \cdot 2\] Right side differentiation: \[\frac{d}{dt}\left(2\sin t \cos t\right)= (2\sin t)(-\sin t) + (2\cos t)(\cos t) \]

Equate the derivatives

Set the derivatives from Step 1 equal to each other: \[2\cos 2t = -2\sin^2 t + 2\cos^2 t\]

Simplify the equation and obtain the desired identity

Simplify the equation to get the desired identity: \[\cos 2t = \cos^2 t - \sin^2 t\]

Key concepts

Chain Rule

Understanding the chain rule is essential when it comes to differentiating composite functions, which are functions made up of other functions. In calculus, the chain rule is a method for finding the derivative of a composite function. The formula for the chain rule states that to differentiate a composite function, you differentiate the outer function and multiply by the derivative of the inner function. For example, if you have a function like \( f(g(t)) \), and you want to find \( \frac{d}{dt}f(g(t)) \), you would compute it as \( f'(g(t)) \cdot g'(t) \).

In the exercise provided, the chain rule is applied to differentiate functions like \( \cos(2t) \) where \( 2t \) is the inner function. When you differentiate \( \cos(2t) \) with respect to \( t \), you first find the derivative of the outer function, which is \( \cos \) becoming \( -\sin \), then multiply it by the derivative of the inner function \( 2t \) which is \( 2 \). This gives you the derivative \( -2\sin(2t) \).

It's like unwrapping a present; you deal with the outer layer first (in this case, \( \cos \) ) and then handle what's inside (\( 2t \) ). This rule is incredibly powerful and shows up frequently in calculus, especially when dealing with trigonometric identities as in the exercise.

Sum/Difference Rule in Differentiation

When differentiating functions that are sums or differences of other functions, the sum/difference rule becomes a handy tool. This rule simply states that the derivative of a sum or difference of functions is the sum or difference of their individual derivatives. If you have functions \( u(t) \) and \( v(t) \) and you want to differentiate \( u(t) + v(t) \) or \( u(t) - v(t) \) with respect to \( t \) , you can differentiate \( u(t) \) and \( v(t) \) separately and add or subtract the results accordingly.

Applying this rule to the trigonometric identities \( \cos^2 t - \sin^2 t \) involves differentiating \( \cos^2 t \) and \( \sin^2 t \) separately and then subtracting the latter from the former. This approach simplifies the differentiation process of more complex expressions and is clearly demonstrated in the step-by-step solution for the exercise. It allows for a straightforward calculation without needing to manipulate the overall expression beforehand.

Double Angle Identities

Trigonometric identities can greatly simplify the process of differentiation. Double angle identities are specific trigonometric identities that express trigonometric functions of double angles, like \( 2t \) in terms of single angles, like \( t \). They are particularly useful in calculus for transforming complex expressions into simpler ones that are easier to differentiate.

There are several double angle identities for sine and cosine, such as \( \sin(2t) = 2\sin(t)\cos(t) \) and \( \cos(2t) = \cos^2(t) - \sin^2(t) \) or alternatively \( \cos(2t) = 2\cos^2(t) - 1 \) . In the exercise provided, these identities allow us to transform the double angle into terms that can be more easily differentiated using known derivative rules. After differentiation, invoking these identities again can help verify and prove other related trigonometric identities, reinforcing their interconnected nature and importance in calculus.

These identities are not just mathematical curiosities; they're essential tools for solving a wide range of problems in calculus, physics, and engineering. They show the remarkable interplay between different trigonometric functions and angles, providing an elegant way of simplifying complex expressions.