Question

Express \(6 \sin 2 t-3 \cos 2 t\) in the form \(A \cos (\omega t-\alpha), \alpha \geq 0\)

Short answer

Question: Express the trigonometric function \(6 \sin 2t - 3 \cos 2t\) in the form of \(A \cos(\omega t - \alpha)\), and find the values of A, ω, and α. Answer: The trigonometric function can be expressed as \(6 \sin 2t - 3 \cos 2t = 3\sqrt{5} \cos(2t - (\arctan(-2) + \pi))\). The values are \(A = 3\sqrt{5}\), \(\omega = 2\), and \(\alpha = \arctan(-2) + \pi\).

Step-by-step solution

Convert using sum-to-product formulas

To convert \(6 \sin 2t - 3 \cos 2t\) into the form \(A \cos(\omega t - \alpha)\), use the angle subtraction formula for cosine, which is: \(\cos (\omega t - \alpha) = \cos \omega t \cos \alpha + \sin \omega t \sin \alpha\) Let \(A \cos(\omega t - \alpha) = R \sin \omega t + S \cos \omega t\) Comparing with the given expression, we have \(R = 6\) and \(S = -3\)

Calculate A

The magnitude of A is given by the Pythagorean sum, i.e., \(A = \sqrt{R^2 + S^2}\) Substituting values, we have \(A = \sqrt{(6)^2 + (-3)^2} = \sqrt{45} = 3\sqrt{5}\)

Calculate α

To find \(\alpha\), we need to determine the tangent of the angle α, which can be done using the following ratio: \(\tan \alpha = \frac{R}{S}\) Substituting values, we have \(\tan \alpha = \frac{6}{-3} = -2\) Now we find \(\alpha\) using the inverse tangent function: \(\alpha = \arctan(-2)\) Since \(\alpha\) must be non-negative, we add \(\pi\) radians to the result: \(\alpha = \arctan(-2) + \pi\)

Write the final expression

Now we have all the values of A, ω, and α. Thus, the final expression is: \(6 \sin 2t - 3 \cos 2t = 3\sqrt{5} \cos(2t - (\arctan(-2) + \pi))\)

Key concepts

Sum-to-Product Formulas

Understanding the sum-to-product formulas is essential for simplifying and manipulating expressions in trigonometry. These formulas show how to write the sum or difference of trigonometric functions as the product of functions. They are particularly useful for integrating trigonometric functions and for solving trigonometric equations.

Here's the general form for converting a sum into a product:

• For sine: \(\text{sin}(x) + \text{sin}(y) = 2 \text{sin}\left(\frac{x + y}{2}\right)\text{cos}\left(\frac{x - y}{2}\right)\)
• For cosine:\(\text{cos}(x) + \text{cos}(y) = 2 \text{cos}\left(\frac{x + y}{2}\right)\text{cos}\left(\frac{x - y}{2}\right)\)

These formulas are particularly useful in transforming expressions into a format that can be easily compared or combined with other terms, as seen in the original exercise, leading to a more straightforward solution.

Angle Subtraction Formula

The angle subtraction formula is a fundamental concept in trigonometry. It relates the cosine or sine of the difference of two angles to the products of sines and cosines of the individual angles:

• \(\text{cos}(A - B) = \text{cos}(A)\text{cos}(B) + \text{sin}(A)\text{sin}(B)\)
• \(\text{sin}(A - B) = \text{sin}(A)\text{cos}(B) - \text{cos}(A)\text{sin}(B)\)

These formulas are invaluable in simplifying complex trigonometric expressions and are used to rewrite the given expression in the exercise in the desired form of \(A \text{cos}(\omega t - \alpha)\).

Pythagorean Sum

The Pythagorean sum follows from the Pythagorean Theorem applied to trigonometry. In trigonometry, the theorem says that for any angle \( \theta \), the square of the sine plus the square of the cosine of \( \theta \) equals one: \(\text{sin}^{2}(\theta) + \text{cos}^{2}(\theta) = 1\).

When converting trigonometric expressions, the Pythagorean sum helps us determine the magnitude of a resultant vector using its components. In the context of the solved problem, it was used to calculate the amplitude \(A\) by finding the square root of the sum of the squares of the coefficients from the original sine and cosine terms. This amplifies our understanding of how to combine two trigonometric functions into one with a specified phase and amplitude.

Inverse Tangent Function

The inverse tangent function, commonly denoted as \(\arctan(x)\) or \(\tan^{-1}(x)\), is used to find an angle when the tangent of the angle is known. Since the tangent function is periodic, the inverse tangent function will provide an angle typically between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

This function is key in determining the phase shift or angle \(\alpha\) in trigonometric expressions. To maintain the principle of \(\alpha \geq 0\), we may need to adjust the resulting angle from the inverse tangent function, as it can produce a negative result. In the original exercise, since \(\tan \alpha\) was negative, we found the principal value using \(\arctan(-2)\) and then adjusted by adding \(\pi\) radians to ensure \(\alpha\) was non-negative, as required by the problem.