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Given x^2+y^2=r^2 and the figure of the right triangle with legs x and y and hypotenuse r, prove cos^2θ+sin^2θ=1.

I need assistance filling out the blanks on the attached document.

Given x2y2r2 and the figure of the right triangle with legs x and y and hypotenuse r prove cos2θsin2θ1 I need assistance filling out the blanks on the attached class=
Given x2y2r2 and the figure of the right triangle with legs x and y and hypotenuse r prove cos2θsin2θ1 I need assistance filling out the blanks on the attached class=
Given x2y2r2 and the figure of the right triangle with legs x and y and hypotenuse r prove cos2θsin2θ1 I need assistance filling out the blanks on the attached class=

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LammettHash

By definition of cosine and sine,

cos(θ) = x/r

sin(θ) = y/r

so that

cos²(θ) + sin²(θ) = (x/r)² + (y/r)²

… = x²/r² + y²/r²

… = (x² + y²)/r²

… = r²/r²

… = 1

To that end, I would say

• [blank1] = "Division property of equality"

That is, we divide both sides of an equation by the same number and equality still holds since r ≠ 0

• [blank2] = "Definition of cos"

• [blank3] = "cos²(θ) = x²/r²"

• [blank4] = "Defintion of sin"

• [blank5] = "sin²(θ) = y²/r²"

• [blank6] = "Simplify"

More specifically, x² + y² = r² is given, so

x²/r² + y²/r² = (x² + y²)/r² = r²/r² = 1

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