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 and verify the identity `\(\sin 2t = 2\sin t\cos t\)` using differentiation and check if it holds for the identities `\(\cos 2t = \cos^2t - \sin^2t\)` and `\(\cos 2t = 2\cos^2t - 1\)`. Answer: By differentiating the given identities and following the necessary steps, we have proven that the identity `\(\sin 2t = 2\sin t\cos t\)` holds for both the cases.

Step-by-step solution

Differentiate both sides with respect to t

We need to differentiate both sides of the identity \(\cos 2t = \cos^2t - \sin^2t\) with respect to t.

Apply chain rule and differentiate both sides

We will apply the chain rule to differentiate both sides. Differentiate \(\cos 2t\) with respect to t: \[-\sin(2t) \cdot 2 = -2\sin(2t)\] Differentiate \(\cos^2 t - \sin^2 t\) with respect to t: \[-2\sin t\cos t - 2\sin t\cos t = -4\sin t\cos t\] Now we have, \[-2\sin(2t) = -4\sin t\cos t\]

Proving the identity

To prove the identity \(\sin 2t = 2\sin t\cos t\), we need to find a common factor and simplify the equation. Divide both sides of the equation by -2: \[\sin(2t) = 2\sin t\cos t\] The identity \(\sin 2t = 2\sin t\cos t\) is now proven. #b. Verify that you obtain the same identity for \(\sin 2t\) as in part (a) if you differentiate the identity \(\cos 2t=2\cos ^{2} t-1\)

Differentiate both sides with respect to t

We need to differentiate both sides of the identity \(\cos 2t = 2\cos^2t - 1\) with respect to t.

Apply chain rule and differentiate both sides

We will apply the chain rule again to differentiate both sides. We already know that the derivative of \(\cos 2t\) is \(-2\sin(2t)\). Differentiate \(2\cos^2 t - 1\) with respect to t: \[4\cos t(-\sin t) = -4\sin t\cos t\] Now we have, \[-2\sin(2t) = -4\sin t\cos t\]

Verifying the same identity for \(\sin 2t\)

To verify the same identity for \(\sin 2t\), we need to find a common factor and simplify the equation. Divide both sides of the equation by -2: \[\sin(2t) = 2\sin t\cos t\] The same identity for \(\sin 2t\) has been verified as in part (a). #c. Differentiate both sides of the identity \(\sin 2t = 2\sin t\cos t\) to prove that \(\cos 2t = \cos^2t - \sin^2t\)

Differentiate both sides with respect to t

We need to differentiate both sides of the identity \(\sin 2t = 2\sin t\cos t\) with respect to t.

Apply chain rule and differentiate both sides

We will apply the chain rule again to differentiate both sides. Differentiate \(\sin 2t\) with respect to t: \[\cos(2t) \cdot 2 = 2\cos(2t)\] Differentiate \(2\sin t\cos t\) with respect to t: \[\cos^2 t - \sin^2 t\] Now we have, \[2\cos(2t) = \cos^2 t - \sin^2 t\]

Proving the identity

To prove the identity \(\cos 2t = \cos^2t - \sin^2t\), we need to simplify the equation. Divide both sides of the equation by 2: \[\cos(2t) = \cos^2 t - \sin^2 t\] The identity \(\cos 2t = \cos^2t - \sin^2t\) is now proven.

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus, especially when dealing with composite functions. It allows you to differentiate a function that contains another function inside of it. For example, if you have a function of the form \(f(g(x))\), the chain rule helps you find the derivative of this composite function.

• Mathematically, the chain rule is stated as: if \( y = f(u) \) and \( u = g(x) \), then \( \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \).
• This means you differentiate the outer function (keeping the inner function unchanged) and then multiply it by the derivative of the inner function.

In the context of trigonometric identities, the chain rule is applied while differentiating terms like \(\cos(2t)\) and \(2\sin t\cos t\).

• For example, when differentiating \(\cos(2t)\), the outer function is \(\cos(u)\), where \(u = 2t\). The derivative is \(-\sin(2t)\cdot 2\).
• The easy memorization and application of the chain rule help you in smoothly solving complex differentiation problems involving nested functions like these.

Differentiation

Differentiation is a critical tool used in calculus to find the rate at which a function is changing at any point. In simple terms, it is finding the slope of the tangent line to the function at any given point.

• This slope is known as the derivative of the function, which provides information on how the function behaves at various points.
• For a function \(f(t)\), the derivative is denoted as \(f'(t)\) or \(\frac{df}{dt}\).

The process involves applying rules like the power rule, product rule, and importantly, for our example, the chain rule.
With trigonometric functions, such as \(\sin(t)\) and \(\cos(t)\), the differentiation becomes slightly more nuanced.

• The derivative of \(\sin(t)\) is \(\cos(t)\).
• The derivative of \(\cos(t)\) is \(-\sin(t)\).
• When dealing with composite trigonometric functions, both the basic differentiation rules and the chain rule are combined to find derivatives accurately.

Double Angle Formulas

Double angle formulas are essential identities in trigonometry that express trigonometric functions of double angles (like \(2t\)) in terms of single angle trigonometric functions (like \(t\)). These formulas simplify expressions and solve equations involving double angles.

• The double angle formula for sine is \(\sin 2t = 2 \sin t \cos t\).
• The double angle formula for cosine has two common variants: \(\cos 2t = \cos^2 t - \sin^2 t\) and \(\cos 2t = 2 \cos^2 t - 1\).

These formulas are derived from using the angle addition formulas for sine and cosine and are used to establish identities or simplify problems in trigonometry.
When proving identities, like in the provided exercise, the double angle formulas help translate between forms and verify equalities efficiently.

• For instance, verifying \(\sin 2t = 2 \sin t \cos t\) used both direct differentiation and the application of derived identities.
• The mastery of these formulas allows you to handle various trigonometric manipulations smoothly and confidently.