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2.1.3 Hence, compare number 2.1 .1 and 2.1 .2 and write a conclusion about \( \cos (\alpha-\beta) \). \( \qquad \) \( \qquad \) (3) 2.1.4 Use 2.1.3 \( [\cos (\alpha-\beta)=\cos \alpha \cdot \cos \beta+\sin \alpha . \sin \beta] \) to derive a formula for \( \cos (\alpha+\beta) \) (Hint: use suitable reduction formula) \( \qquad \) \( \qquad \) \( \qquad \) (4) \( \qquad \) 2.1.5 Use \( \cos (\alpha-\beta) \) to derive a formula for \( \sin (\alpha-\beta) \). (Hint: use co-function) \( \qquad \) \( \qquad \) \( \qquad \) 2.1.6 Use \( \cos (\alpha-\beta) \) to derive a formula for \( \sin (\alpha+\beta) \). (Hint: use co-function) \( \qquad \) \( \qquad \) \( \qquad \) Grade 12 Mathematics SBA 2025 Page

Ask by Gardner Delgado. in South Africa
Feb 20,2025

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cos(α – β) = cos α cos β + sin α sin β cos(α + β) = cos α cos β – sin α sin β sin(α – β) = sin α cos β – cos α sin β sin(α + β) = sin α cos β + cos α sin β

Solución

Below is one way to work through the parts. ────────────────────────────── 2.1.3 Compare 2.1.1 and 2.1.2 Suppose in earlier parts you established two expressions leading to   cos(α – β) = cos α · cos β + sin α · sin β Thus, by comparing your work in 2.1.1 and 2.1.2, you conclude that   cos(α – β) = cos α · cos β + sin α · sin β. ────────────────────────────── 2.1.4 Derive a formula for cos(α + β) Hint: Replace β with –β in the formula for cos(α – β). Recall that   cos(–β) = cos β  and   sin(–β) = – sin β. Starting with   cos(α – (–β)) = cos α · cos(–β) + sin α · sin(–β) Note that the left‐side becomes cos(α + β). Thus, we have:   cos(α + β) = cos α · cos β + sin α · (– sin β)         = cos α · cos β – sin α · sin β. ────────────────────────────── 2.1.5 Derive a formula for sin(α – β) Hint: Use a co-function identity. One common approach is to recall that sine can be written in terms of cosine:   sin θ = cos(π/2 – θ). So write sin(α – β) as   sin(α – β) = cos((π/2) – (α – β))         = cos((π/2) – α + β). Now, you can derive a cosine formula for cos((π/2) – α + β) using the addition formulas. However, many choose instead to use a similar trick to that in 2.1.4. Instead, notice that by comparing known formulas or by a similar replacement process,   sin(α – β) = sin α · cos β – cos α · sin β. This is the standard difference formula for sine. ────────────────────────────── 2.1.6 Derive a formula for sin(α + β) Hint: Use a co-function identity similarly as in 2.1.5. You can use the well‐known sine addition formula directly or derive it using the method of replacing β by –β in the formula for sin(α – β). For instance, replacing β with –β in the formula sin(α – β) gives:   sin(α – (–β)) = sin α · cos(–β) – cos α · sin(–β). Recognize that   cos(–β) = cos β  and  sin(–β) = – sin β. So it follows that   sin(α + β) = sin α · cos β – cos α · (– sin β)         = sin α · cos β + cos α · sin β. ────────────────────────────── Final Answers Recap: 2.1.3 → cos(α – β) = cos α · cos β + sin α · sin β 2.1.4 → cos(α + β) = cos α · cos β – sin α · sin β 2.1.5 → sin(α – β) = sin α · cos β – cos α · sin β 2.1.6 → sin(α + β) = sin α · cos β + cos α · sin β These are the standard angle addition and subtraction formulas for sine and cosine.

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Did you know that the formulas for cosine and sine at different angles have roots in ancient Greek mathematics? Think about Pythagoras and his famous theorem! The trigonometric identities that we're discussing today were developed from these early understandings of right triangles and circles. This blend of geometry with algebra laid the groundwork for calculus and modern scientific applications. In practical terms, these trigonometric identities aren't just classroom exercises; they're crucial in various fields! For instance, engineers use them to calculate forces, architects rely on them for design angles, and even computer graphics use these functions to create realistic images and animations. Mastering these formulas opens up a world of possibilities, making abstract concepts tangible and useful in real-life scenarios!

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