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forsale [732]
3 years ago
13

Q9: Identify the graph of the equation and write an equation of the translated or rotated graph in general form. (Picture Provid

ed Below)

Mathematics
1 answer:
MArishka [77]3 years ago
6 0

Answer:

D. Circle; (x²) + (y²)-4=0

Step-by-step explanation

Simplify the equation by dividing both sides by 2

2x² + 2y² =8

x² + y² =4

this gives a circle form where we can apply (x-h) ² +(y-k)² =r² to find center and radius of the circle.

Thus r=2, h=0, k=0

Center of circle is at (h, k), (0,0)

Radius of circle is 2

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<u>Step-by-step explanation:</u>

Use the Double Angle Identity:  cos 2A = 2cos²A - 1

Use the Difference Identity:  cos (A - B) = cosA · cosB + sinA · sinB

Use the Unit Circle to evaluate:  cos (π/2) = 0   &     sin (π/2) = 1

<u>Proof  LHS → RHS</u>

\text{Given:}\qquad \qquad \qquad \qquad 2\cos^2\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)-1\\\\\\\text{Double Angle Identity:}\quad \cos2\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)\\\\\\\text{Simplify:}\qquad \qquad \qquad \quad \cos\bigg(\dfrac{\pi}{2}-A\bigg)\\\\\\\text{Difference Identity:}\qquad \cos\dfrac{\pi}{2}\cdot \cos A+\sin \dfrac{\pi}{2}\cdot \sin A\\\\\\\text{Unit Circle:}\qquad \qquad \qquad 0\cdot \cos A+1\cdot \sin A\\\\\\\text{Simplify:}\qquad \qquad \qquad \qquad \sin A

LHS = RHS:   sin A = sin A   \checkmark

 

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