Precalculus: Notes 1

1.Be careful when using factoring by grouping
method and its particular case:

factoring by grouping

f(x) = 2 x² - 6x + 4 = 0

of the form :

a x² + b x + c = 0

Somme x1 + x2 = - 6
Produit x1 . x2 = 2 . 4 = 8


Hence:

x1 = - 4
x2 = - 2

f(x) = (1/a)(ax + m)(ax + n)
2 x² - 6x + 4 = (1/2)(2x - 4)(2x - 2) =
(x - 2)(2x - 2) = 2(x - 2)(x -1)

Always simplify the expressions.

Its particular case

f(x) = x² - 9x + 14 = 0

of the form:
x² - Sx + P = 0

Somme x1 + x2 = 9
Produit x1 . x2 = 14
x1 = 2
x2 = 7

f(x) = (x - x1)(x - x2)

f(x) = (x - 2)(x - 7)

2. sinus is an even function

sin²x + cos²x = 1     (1)
cos(x + y) = cosx cosy - sinx siny     (2)
sin(x + y) = sinx cosy + cosx siny     (3)




5π/4 = π + π/4
sin (5π/4) = sin (π + π/4)
Applying the relationship (3) yields:
sin(π + π/4) = sinπ cos(π/4) + cosπ sin(π/4)

we have
sinπ = 0
cosπ = - 1

Therefore

sin (5π/4) = 0. cos(π/4) -1 . sin(π/4) =
- sin(π/4)

Since sin is an even function

- sin(π/4) = sin(- π/4)

Therefore:

sin (5π/4) = sin(- π/4)