Mathematics, 27.10.2020 19:10 arlettg

Review the proof of Cosine (StartFraction theta Over 2 EndFraction) = plus-or-minus StartRoot StartFraction 1 + cosine theta Over 2 EndFraction EndRoot. Which step contains an error?

Review the proof of Cosine (StartFraction theta Over 2 EndFraction) = plus-or-minus StartRoot Start

Answers: 2

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Answer from: shaheedbrown06

Error in step (5)

Step-by-step explanation:

Steps Statements

1 cos(2x) = 2cosx - 1

2 Let (2x) = θ

3 Then x = $\frac{\theta }{2}$

4 cos(θ) = $2\text{cos}^2{\theta}-1$

5 cos(θ) + 1 = $2\text{cos}^2{\frac{\theta}{2}}$

6 $\frac{1+\text{cos}\theta }{2}=cos^{2}\frac{\theta }{2}$

7 $\text{cos}^{2}\frac{\theta }{2}=\frac{1+\text{cos}\theta }{2}$

8 $\text{cos}(\frac{\theta }{2})=\pm \sqrt{\frac{1+cos(\theta)}{2} }$

By comparing the solution given in the picture attached with the solution above,

Step 5 has error.

There should be [cos(θ) + 1 = $2\text{cos}^2({\frac{\theta}{2}})$ ] in place of [-1 + cos(θ) = $2\text{cos}^2({\frac{\theta}{2}})$ ]

Answer from: Quest

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

Answer from: Quest

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

$x^2-2x+y^2-6y=26$

we now create space between the x and y terms with parenthesis.

$(x^2-2x)+(y^2-6y)=26$

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

$\frac{-2}{2} =(-1)^{2} =1$

$\frac{-6}{2} =(-3)^{2} =9$

we add the squares to both sides.

$(x^2-2x+1)+(y^2-6y+9)=26+1+9$

simplify.

$(x^2-2x+1)+(y^2-6y+9)=36$

and write the quadratics in factored form.

$(x-1)^{2} +(y-3)^{2} =36$

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

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