In an absolute value, everything with it is counted as a positive.
∣-a∣ = --a= a
∣a∣ =a
In an equation, absolute values have two possibilities when talking about equations ∣a+b∣ =x = a+b=x = a+b= -x e.g. Solve ∣x-4∣=8 x-4=8 OR x-4= -8 x=12 x=-4 Sub both answer into the equation ∣12-4∣ =8 OR ∣-4-4∣ =8 8=8 8=8 Both solution re true so x=12 or x=-4 Absolute inequalities (method 1) If ∣a+b∣ ≤x∣a+b∣ <x= -x≤a+b ≤x-x<a+b<x e.g. ∣2x-1∣<3 -3<2x-1<3 +1-3<2x<3+1 -2<2x<4 -1<x<2 Method 2 If ∣a+b∣ ≥x∣a+b∣ >x= a+b≥x ORa+b≤-x a+b>x OR a+b<x e.g. ∣2x-1∣≥3= 2x-1≥3 OR 2x-1≤-3 2x≥3+1 2x≤-3+1 2x≥4 2x≤-2 x≥2 x≤-1 Rationalizing Any fraction involving surds needs to have the denominator rationalised. Rationalise 23 23 = 23 ×1 =23 ×33 note that33 =1 =233×3 =233 The denominator is rationalised now
Beta
Trigonometry
Basic rules Hypotenuse
Adjacent
Sin = Opposite/hypotenuse Cos = Adjacent/hypotenuse Tan = Opposite/Adjacent Opposite
Cot = 1/tan Sec= 1/sin Csc = 1cos Special rules | 300 | 450 | 600 | Sin | 12 | 12 | 32 | Cos | 32 | 12 | 12 | tan | 13 | 1 | 2
2
3
300
600
1
1
3 | 2
450
1
1
Area of triangle
There is an basic formula for finding the area of a triangle A=12BH but I’m going to look at how to find the area of a triangle WITHOUT the base OR the height.
X
Y
Z
As long were are given x,y AND z we can find the area of this triangle without using the base or height. B A=12×X×Y×sinZ
A
C c a b AREA=12absinB You can’t find the area if you don’t meet these conditions AREA=12bcsinA