A square has an area of 561.69m2. Work out the perimeter of the square.
1 answer:
You might be interested in
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A
Step 1) Draw a dashed line through the points (0,6) and (4,7). These two points are on the line y = (1/4)x+6. To find those points, you plug in x = 0 to get y = 6. Similarly, plug in x = 4 to get y = 7. The dashed line indicates that none of the points on this line are part of the solution set.
Step 2) Draw a dashed line through (0,-1) and (1,1). These two points are on the line y = 2x-1. They are found in a similar fashion as done in step 1.
Step 3) Shade the region that is above both dashed lines. We shade above because of the "greater than" sign. This is shown in the attached image I am providing below. The red shaded region represents all of the possible points that are the solution set. Once again, any point on the dashed line is not in the solution set.
Step 2) Draw a dashed line through (0,-1) and (1,1). These two points are on the line y = 2x-1. They are found in a similar fashion as done in step 1.
Step 3) Shade the region that is above both dashed lines. We shade above because of the "greater than" sign. This is shown in the attached image I am providing below. The red shaded region represents all of the possible points that are the solution set. Once again, any point on the dashed line is not in the solution set.