Problem 39 - Algebra 2

Question

Verify that each of the following is an identity. \(\sin (\alpha+\beta) \sin (\alpha-\beta)=\sin ^{2} \alpha-\sin ^{2} \beta\)

Step-by-Step Solution

Answer

The identity is verified.

Step 1: Recall and Apply Product-to-Sum Formulas

Recall the product-to-sum identities: \(\sin(a)\sin(b) = \frac{1}{2}[\cos(a-b) - \cos(a+b)]\). Here, let \(a = \alpha + \beta\) and \(b = \alpha - \beta\) such that: \[\sin(\alpha+\beta)\sin(\alpha-\beta) = \frac{1}{2}[\cos((\alpha+\beta)-(\alpha-\beta)) - \cos((\alpha+\beta)+(\alpha-\beta))]\]. Simplify this equation to \[\frac{1}{2}[\cos(2\beta) - \cos(2\alpha)]\].

Step 2: Recall and Apply Cosine Double Angle Formulas

Using the double angle formula for cosine, \(\cos(2x) = 1 - 2\sin^2(x)\), replace \(\cos(2\beta)\) and \(\cos(2\alpha)\): \[\cos(2\beta) = 1 - 2\sin^2(\beta)\quad \text{and} \quad \cos(2\alpha) = 1 - 2\sin^2(\alpha)\].Now, substitute these into the equation: \[\sin(\alpha+\beta)\sin(\alpha-\beta) = \frac{1}{2}[(1 - 2\sin^2(\beta)) - (1 - 2\sin^2(\alpha))].\]

Step 3: Simplify the Expression

Simplify the expression obtained in Step 2:\[\sin(\alpha+\beta)\sin(\alpha-\beta) = \frac{1}{2}[(1 - 2\sin^2(\beta)) - (1 - 2\sin^2(\alpha))]\]. This results in:\[= \frac{1}{2}[0 - 2\sin^2(\beta) + 2\sin^2(\alpha)]\]. Simplifying further, we have:\[= \frac{1}{2} \cdot 2 (\sin^2(\alpha) - \sin^2(\beta))\]. The factor of 2 cancels out, leaving us with \(\sin^2(\alpha) - \sin^2(\beta)\).

Key Concepts

Product-to-Sum Formulas

Trigonometric identities can sometimes look complex, but the product-to-sum formulas help in transforming products of sine and cosine into sums or differences, which are often easier to manage. These identities are incredibly useful when simplifying trigonometric expressions. The key product-to-sum formula is: \[\sin(a)\sin(b) = \frac{1}{2}[\cos(a-b) - \cos(a+b)]\]In our exercise, we needed to verify that \(\sin(\alpha+\beta)\sin(\alpha-\beta) = \sin^2(\alpha) - \sin^2(\beta)\). We substitute \(a = \alpha + \beta\) and \(b = \alpha - \beta\) into the product-to-sum formula. This reduces the expression into a manageable form:\[\sin(\alpha+\beta)\sin(\alpha-\beta) = \frac{1}{2}[\cos(2\beta) - \cos(2\alpha)]\]. This transformation makes it easier to incorporate other identities, such as double angle formulas, to further simplify the expression.

Double Angle Formulas

Double angle formulas are another essential part of solving trigonometric identities. They help us express functions of double angles like \(2\theta\) rather than using combinations of \(\theta\), simplifying the process greatly. One common double angle formula involves cosine:\[\cos(2x) = 1 - 2\sin^2(x)\]Using these formulas, we substitute \(\cos(2\beta)\) and \(\cos(2\alpha)\) with their equivalent expressions in terms of sine into our equation:\[\cos(2\beta) = 1 - 2\sin^2(\beta)\quad \text{and} \quad \cos(2\alpha) = 1 - 2\sin^2(\alpha) \]Substituting these back, you get:\[\frac{1}{2}[(1 - 2\sin^2(\beta)) - (1 - 2\sin^2(\alpha))]\]This further simplifies the expression, and helps reveal simpler components which can be compared or verified against known identities. Double angle formulas are powerful tools when dealing with angles that are multiples of a base angle.

Simplifying Trigonometric Expressions

The primary goal in simplifying trigonometric expressions is to convert them into a more straightforward form, verifying or deriving known identities. After applying both product-to-sum and double angle formulas, the tedious-looking expression \(\sin(\alpha+\beta)\sin(\alpha-\beta)\) boils down to:\[\frac{1}{2}[0 - 2\sin^2(\beta) + 2\sin^2(\alpha)] = \frac{1}{2} \cdot 2 (\sin^2(\alpha) - \sin^2(\beta))\]Now, the factor of 2 cancels out, leaving us with:\[\sin^2(\alpha) - \sin^2(\beta)\]This simplification matches our original statement that needed to be verified. The process of using known identities to simplify to this form highlights how interconnected and useful these trigonometric concepts are. With practice, simplifying these expressions can become a straightforward and rewarding process, helping verify and understand a wide array of trigonometric identities.