Question

Prove the trigonometric identity below. Show all steps algebraically

(𝑥sinθ − 𝑦cosθ)^2 + (𝑥cosθ + 𝑦sinθ)^2 = 𝑥^2 + 𝑦^2

Answer 1

not again

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Answer 2

Sorry for reposting this problem but I couldn't understand the process from the previous post. However, i do get it now, thanks anyways

Answer 3

[AI-generated answer — Explain Bot]

To prove the given trigonometric identity, we start with the left-hand side (LHS) and simplify it using algebraic steps.

LHS: (𝑥sinθ − 𝑦cosθ)^2 + (𝑥cosθ + 𝑦sinθ)^2

Expanding the squares, we get:
LHS: 𝑥^2sin^2θ - 2𝑥𝑦sinθcosθ + 𝑦^2cos^2θ + 𝑥^2cos^2θ + 2𝑥𝑦sinθcosθ + 𝑦^2sin^2θ

Combining like terms, we see that the middle two terms cancel out:
LHS: 𝑥^2sin^2θ + 𝑦^2cos^2θ + 𝑥^2cos^2θ + 𝑦^2sin^2θ

Rearranging the terms, we have:
LHS: 𝑥^2(sin^2θ + cos^2θ) + 𝑦^2(sin^2θ + cos^2θ)

The trigonometric identity sin^2θ + cos^2θ = 1 holds for any angle θ. Therefore, we can substitute 1 in place of sin^2θ + cos^2θ:

LHS: 𝑥^2(1) + 𝑦^2(1) = 𝑥^2 + 𝑦^2

Hence, the left-hand side (LHS) is equal to the right-hand side (RHS), proving the trigonometric identity (𝑥sinθ − 𝑦cosθ)^2 + (𝑥cosθ + 𝑦sinθ)^2 = 𝑥^2 + 𝑦^2.