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What do you notice concerning the values of \( \cos (\alpha-\beta) \) and \( \cos \alpha \) (Hint - are the values the same or different?)

Ask by Vaughan Elliott. in South Africa
Feb 27,2025

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Answer

The values of \( \cos (\alpha - \beta) \) and \( \cos \alpha \) are generally different unless \( \beta \) is 0 or a multiple of \( 2\pi \).

Solution

Recall that the cosine of a difference is given by the identity   cos(α – β) = cos α cos β + sin α sin β. Generally, this expression is different from cos α. In fact, cos(α – β) will equal cos α only in special cases—for example, when β is 0 (or any multiple of 2π), because then cos β = 1 and sin β = 0, so   cos(α – β) = cos α · 1 + sin α · 0 = cos α. For any other value of β, the term sin α sin β (and the factor cos β differing from 1) will make cos(α – β) differ from cos α. Therefore, the values of cos(α – β) and cos α are generally different unless β is chosen so that the extra terms vanish.

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Bonus Knowledge

When observing the values of \( \cos (\alpha - \beta) \) and \( \cos \alpha \), it’s important to recognize that they are generally different due to the nature of the cosine function and the angle subtraction involved. The equation \( \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \) shows that \( \cos (\alpha - \beta) \) is a combination of both sine and cosine terms, thus making it dependent on both angles, while \( \cos \alpha \) only relies on the angle \( \alpha \). In specific scenarios, like when \( \beta = 0\), the values align perfectly as \( \cos (\alpha - 0) = \cos \alpha \). However, for most other angles, particularly when \( \beta \) is not zero, we’ll see diverging behaviors which reflect the unique influences of both angles on the cosine function. So, you’re in for a rollercoaster ride of trigonometric fun!

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