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| Mathematics Magazine for Grades 1-12 |
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Theory:
A quadratic function is any function equivalent to one of the form f(x) = ax2 + bx + c Here are some examples of quadratic functions Ø f(x) = x2 – 9 Ø
f(x) = x2 + 2x +1 A quadratic equation is any equation equivalent to one of the form ax2 + bx + c = 0 Here are some examples of quadratic equations Ø x2 – 9 = 0 Ø
x2 + 2x +1 = 0 Solutions from the Previous Issue:
1.
Prove that
Solution:
=
= 2.
a + b + c= π. Prove that Solution: a+b = pi - c => cos(a + b) = cos(π - c) => cos(a + b) = -cos(c) => cos(a) ∙cos(b) - sin(a) ∙sin(b) = - cos(c) => cos(a) ∙cos(b) + cos(c) = sin(a) ∙sin(b) => (cos(a) ∙ cos(b) + cos(c))2 = (sin(a) ∙sin(b))2 => [cos(a) ∙cos(b)]2 +2.cos(a) ∙cos(b) ∙cos(c) + cos2(c) => [1 - cos2(a)] ∙ [1 - cos2(b)] => cos(a) ∙cos(a) + cos(b) ∙cos(b) + cos(c) ∙cos(c) + 2cos(a) ∙cos(b) ∙cos(c) =1 3. Prove that: sin(a) + sin(b) - sin(c) = sin(a
+ b + c) + 4sin
Solution:
sin(a) + sin(b) =
sin (a + b + c) – sin (c) = 2
Then sin(a)+ sin(b) + sin(c) - sin(a + b + c)
=
=2
sin(a)+ sin(b) + sin(c) =
= sin(a + b + c) +2
4. Solve: 2∙tan(x) = sin(4x) - 2.sin(2x) ∙tan(x) ∙cos(2x) ∙tan(x) Solution: 2∙tan(x) = sin(4x) - sin(4x) ∙tan2 (x) <=> 2∙tan(x) = sin(4x) ∙ (1 - tan2(x))
<=> 2∙sin(x) ∙ cos(x) = sin(4x) ∙ (cos2 (x) - sin2 (x)) and cos(x) not 0 <=> sin(2x) - 2∙sin(2x). cos(2x) ∙ cos(2x) = 0 and cos(x) not 0 <=> sin(2x) ∙ [1 - 2∙cos2 (2x)] = 0 and cos(x) not 0 <=> 2∙sin(x) ∙ cos(x) ∙ [1 - 2∙ cos2 (2x)] = 0 and cos(x) not 0 <=>
sin(x) = 0 or cos(2x) =
<=>
x = k∙ π
or 2x =
<=>
x = k∙ π
or x =
5.
Prove that: sin(2∙arctan(x)) =
Solution: Let arctan(x) = a => tan(a) = x sin(2 arctan(x)) = sin(2a) = 2sin(a)cos(a) = 2 tan(a).cos(a).cos(a)
Because: Proposed Exercises: Plot the inequalities: 1. (x - 1)2 + (y+2)2 ≥ 9 2. 2x + 3y – 5 ≤ 0 3. 1 < x2 + y2 ≤ 4 4. y ≤ x3 +2x – 1 |
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