Capabilities, Relations and Trigonometry | The Final NESA Maths Reference Sheet Information

(textual content{For} ax^3 + bx^2 +cx + d = 0 😉
Sum of roots: (α+β+γ = textual content{  } – frac{b}{a})
Sum in Pairs:  (αβ+αγ+βγ = frac{c}{a})
Product of roots: ( αβγ = -frac{d}{a})

(textual content{log}_a x = frac{textual content{log}_b x}{textual content{log}_b a})

(a^x = e^{xtext{In}a})

(A= frac{1}{2}abtext{ sin}C)

(frac{a}{textual content{sin}A} = frac{b}{textual content{sin}B} = frac{c}{textual content{sin}C})

(c^2=a^2+b^2-2abtext{ cos}C)

(textual content{cos}C = frac{a^2+b^2-c^2}{2ab})

(l=r theta)

(A=frac{1}{2}r^2theta)

Trigonometric identities

(textual content{sec}A = frac{1}{textual content{cos}A}, textual content{cos}A ≠0)

(textual content{cosec}A = frac{1}{textual content{sin}A}, textual content{sin}A ≠0)

(textual content{cot}A = frac{textual content{cos}A}{textual content{sin}A}, textual content{sin}A ≠0)

(textual content{cos}^2x+textual content{sin}^2x=1)

Compound angles

(textual content{sin}(A+B)=textual content{sin}Atext{cos}B+textual content{cos}Atext{sin}B)

(textual content{cos}(A+B)=textual content{cos}Atext{cos}B-text{sin}Atext{sin}B)

(textual content{tan}(A+B)=frac{textual content{tan}A+textual content{tan}B}{1-text{tan}Atext{tan}B})

(textual content{If } t=textual content{tan} frac{A}{2} textual content{ then } )
(textual content{sin}A = frac{2t}{1+t^2} )
(textual content{cos}A = frac{1-t^2}{1+t^2} )
(textual content{tan}A = frac{2t}{1-t^2} )

(textual content{cos}Atext{cos}B=frac{1}{2}[text{cos}(A-B)+text{cos}(A+B)])

(textual content{sin}Atext{sin}B=frac{1}{2} [text{cos}(A-B) -text{cos}(A+B)])

(textual content{sin}Atext{cos}B = frac{1}{2} [text{sin}(A+B)+text{sin}(A-B)])

(textual content{cos}Atext{sin}B=frac{1}{2}[text{sin}(A+B)-text{sin}(A-B)])

(textual content{sin}^2nx=frac{1}{2}(1-text{cos}2nx))

(textual content{cos}^2nx=frac{1}{2}(1+textual content{cos}2nx))

start{align*}
y &= frac{-b± sqrt{b^2-4ac}}{2a}
& textual content{the place a, b and c are the coefficients of a quadratic equation:} -2y^2+3y+1=0
finish{align*}

Therefore, substitute (a=-2), (b=3) and (c=1) into (y=frac{-b± sqrt{b^2-4ac}}{2a}):

start{align*}
y &= frac{-3± sqrt{3^2-4 occasions (-2) occasions 1}}{2 occasions (-2)}
&= frac{-3± sqrt{9+8}}{-4}
&= frac{-3± sqrt{17}}{-4}
&= frac{3± sqrt{17}}{4}
∴ y &= frac{3+ sqrt{17}}{4} textual content{  or  } y = frac{3- sqrt{17}}{4}
finish{align*}

start{align*}
textual content{Let } α, β, γ textual content{ be roots of the given cubic equation}

αβγ &= -frac{d}{a}
&= -d textual content{  (since a=1)}
&= -6 textual content{  (product of the roots is -6)}
⇒ d &=6

α+β+γ &= – frac{b}{a}
⇒ textual content{Equation 1: } α+β+γ &= -b textual content{  (since a=1)}

αβ+αγ+βγ &= frac{c}{a}
⇒ textual content{Equation 2: } αβ+αγ+βγ &= c textual content{  (since a=1)}
finish{align*}

It’s on condition that (x=1) and (x=3) are options to the cubic equation, (x=1) and (x=3) are two roots of the equation.

Therefore, substitute (α=1) and (β=3) into (αβγ = -6):

start{align*}
(1)(3)γ &= -6
∴ γ &= -2
finish{align*}

To search out the coefficients b and c (we’ve already discovered (a=1) and (d=6)):

Substitute (α=1), (β=3) and (γ=-2) into equation 1: (α+β+γ = -b) and equation 2: (αβ+αγ+βγ = c):

start{align*}
textual content{Equation 1: } 1+3+-2 &= -b
∴b &= -2

textual content{Equation 2: } (1)(3)+(1)(-2)+(3)(2) &= c
3+(-2)+6 &= c
∴ c &= 7
finish{align*}

Due to this fact, the coefficients of the equation (ax^3 + bx^2 +cx + d = 0) are (a=1), (b=-2),(c=7) and (d=6).

start{align*}
textual content{Substitute } y &= x textual content{ into } y = 2x – 4 :
x &= 2x-4
∴ x &= 4

textual content{Substitute } x &= 4 textual content{ into } y = 2x – 4 :
y &= 2(4) – 4
∴ y &= 4

⇒ textual content{Centre of circle (h, ok) } & = (4, 4)
finish{align*}

To search out the radius r of the circle, we are able to use the data that the realm of this circle is (16π ) items². The system for the realm of a circle is: ( A = pi r^2 )

start{align*}
textual content{Substitute } A & = 16 pi textual content{ into } A = pi r^2 :
16 pi &= pi r^2
16 &= r^2
∴r &= 4 (r > 0)
finish{align*}

The equation of the circle is: ((x-h)^2+(y-k)^2=r^2) the place ((h, ok)) = ((4, 4)) and (r = 4)

start{align*}
(x-4)^2+(y-4)^2 & =4^2
⇒ textual content{Cartesian equation of the circle: } (x-4)^2+(y-4)^2 & =16
finish{align*}

e.g.

(textual content{log}_8 9 = frac{textual content{log}_{10} 9}{textual content{log}_{10} 8} = frac{ textual content{ln}9}{textual content{ln}8})

(textual content{log}_2 5 = frac{textual content{log}_{10} 5}{textual content{log}_{10} 2}= frac{ textual content{ln}5}{textual content{ln}2})

start{align*}
& int a^x dx
&= int e^{xtext{ln}a} dx
&= frac{1}{textual content{ln}a}  e^{xtext{ln}a} + c
&= frac{1}{textual content{ln}a}  a^x + c
finish{align*}

Evaluation your information of integration by substitution right here.

b) cos(90°-A) = sinA = (frac{4}{5})

Alternatively, recognise that (90°-A) is an angle within the triangle.

Therefore, cos(90°-A) = (frac{textual content{adj}}{textual content{hyp}}) = (frac{4}{5})

c) tan90° = undefined

As we are able to see on this proper angle triangle, the size reverse 90° is the hypotenuse. Therefore, the expression ( textual content{tan}90° = frac{textual content{opp}}{textual content{adj}}) is undefined.

start{align*}
4 sqrt{3} &= frac{1}{2} (3)(x)textual content{sin}60°
8 sqrt{3} &= (3)(x)textual content{sin}60°
8 sqrt{3} &= (3)(x)frac{sqrt{3}}{2}
x &= frac{ 8 sqrt{3} occasions 2}{3sqrt{3}}
∴x &= frac{16}{3} textual content{ cm}
finish{align*}

This equation will also be rearranged as:

(textual content{cos}C = frac{a^2+b^2-c^2}{2ab})

start{align*}
θ + 45° + 30° &= 180°  textual content{  (angle sum of a triangle)}
∴θ &= 105°
finish{align*}

Thus, we are able to use the sine rule:

Substitute (a=4), (A=30°), (b=x), (B=θ=105°) into (frac{a}{textual content{sin}A} = frac{b}{textual content{sin}B}):

start{align*}
frac{4}{textual content{sin}30°} &= frac{x}{textual content{sin}105°} textual content{ (sine rule)}
x &= frac{4 occasions textual content{sin}105°}{textual content{sin}30°}
∴ x &= 7.727 textual content{ cm     (to the closest 3 decimal locations)}
finish{align*}

(utilizing θ in radians: ( pi ^c = 180°))

(utilizing θ in radians: ( pi ^c = 180°))

So, to seek out the most important space of the sector, substitute (r=6) and (theta=frac{π}{3}^c) into (A= frac{1}{2} r^2 theta):

start{align*}
A &= frac{1}{2} occasions 6^2 occasions frac{π}{3}
A &= 6 pi
∴ textual content{Most space of the sector} &= 6 pi textual content{ m}^2
finish{align*}

Angle (A) and (B) can take any worth (they are often equal or unequal) and the identification will maintain true.

Substitute sinA, cosA or tanA with these ‘t-values’. That approach you gained’t should cope with trigonometric features, solely algebraic expressions involving t. Upon getting simplified your expression involving t, typically you will have to substitute (t=textual content{tan} frac{A}{2}) to seek out the values of (A).

Word: When A = 180°, (t=textual content{tan} frac{A}{2} = textual content{tan}90) is undefined. Meaning that you’ll want to individually verify whether or not A = 180° it’s a resolution if you use the t-formula. See Instance 21 beneath.

textual content{On condition that }2text{cos}2B – textual content{sinB}2B  &= 1
⇒ 2 occasions frac{1-t^2}{1+t^2} – frac{2t}{1+t^2}  &= 1
(1+t^2) occasions (2 occasions frac{1-t^2}{1+t^2} – frac{2t}{1+t^2}) &= (1+t^2) occasions 1
2 (1-t^2) – 2t &= 1+t^2
2 – 2t^2 – 2t &= 1+t^2
-3t^2 – 2t + 1 &= 0
3t^2 + 2t – 1 &= 0
(3t-1)(t+1) &= 0
3t-1 &= 0 textual content{   or   } t+1=0
t &= frac{1}{3} textual content{   or   } t=-1
⇒ textual content{tan}B &= frac{1}{3} textual content{   or   } textual content{tan}B=-1
∴ B &= textual content{tan}^{-1} frac{1}{3} textual content{   or   } B = 180 + {tan}^{-1}(-1)   textual content{   (since 0<B<180°)}
∴ B &= 18.43° textual content{   or   } 135°
finish{align*}