Question

Prove the trigonometric identity below. Show all steps algebraically

(𝑥sinθ−𝑦cosθ)2 +(𝑥cosθ+𝑦sinθ)2 =𝑥2 +𝑦2

Answer 1

Not the same question, so here it goes

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

Let x = sinA and let y = cosA

Left Side = (𝑥sinθ−𝑦cosθ)^2 +(𝑥cosθ+𝑦sinθ)^2
= (sinAsinθ - cosAcosθ) + (sinAcosθ + cosAsinθ)^2
= ( cos(A+θ))^2 + (sin(A+θ))^2
= 1

RS = (sinA)^2 + (cosA)^2
= 1
= LS

Answer 2

the 2's are exponents

Answer 3

Someone else posted that earlier:

www.jiskha.com/questions/1866365/solve-the-following-equation-and-state-the-general-solution-for-all-values-of-x-in-exact

Answer 4

clever solution, but if |x| or |y| >1 then they cannot be sin and cos of A. In that case, some more algebra will be needed, but it still works out to x^2 + y^2.

Answer 5

[AI-generated answer — Explain Bot]

To prove the given trigonometric identity algebraically, we can expand both sides of the equation and simplify to demonstrate they are equal. Let's start with the left side:

(𝑥sinθ - 𝑦cosθ)^2 + (𝑥cosθ + 𝑦sinθ)^2

Expanding the first term using the binomial formula:

= (𝑥^2 sin^2θ - 2𝑥𝑦sinθcosθ + 𝑦^2 cos^2θ) + (𝑥^2 cos^2θ + 2𝑥𝑦sinθcosθ + 𝑦^2 sin^2θ)

Now, combining like terms, we get:

= 𝑥^2 sin^2θ + 𝑦^2 cos^2θ + 𝑥^2 cos^2θ + 𝑦^2 sin^2θ - 2𝑥𝑦sinθcosθ + 2𝑥𝑦sinθcosθ

Notice that the middle terms cancel each other out (𝑥^2 cos^2θ - 2𝑥𝑦sinθcosθ + 𝑦^2 sin^2θ + 2𝑥𝑦sinθcosθ), leaving:

= 𝑥^2 sin^2θ + 𝑦^2 cos^2θ + 𝑥^2 cos^2θ + 𝑦^2 sin^2θ

Now, combining like terms once more:

= 𝑥^2 (sin^2θ + cos^2θ) + 𝑦^2 (cos^2θ + sin^2θ)

Using the fundamental trigonometric identity sin^2θ + cos^2θ = 1, we can simplify the equation further:

= 𝑥^2 (1) + 𝑦^2 (1)

= 𝑥^2 + 𝑦^2

Hence, we have proven that (𝑥sinθ - 𝑦cosθ)^2 + (𝑥cosθ + 𝑦sinθ)^2 equals 𝑥^2 + 𝑦^2 using algebraic steps.