The interior angles formed by the sides of a quadrilateral have measures that sum to 360°.
2 answers:
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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
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<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
Answer:
Approximately 17.88x or
Step-by-step explanation:
Use pythagorean formula. In a rhombus the diagonals bisect each other and they are perpendicular, so you could have a right triangle with legs of 2x and 4x, the hypoteneuse would then be which is approximately 4.47x. In a rhombus all 4 sides are the same, so multiply that by 4 and you get the perimeter. 4(4.47x) = 17.88x or if you simplify the radical instead it's
Answer:
f(x) = -3x - 21
Step-by-step explanation:
Graphed this equation and it passes through (-6, -3), also known as Point A.
