Finding area of a triangle given SAS

K=1/2bcsinA: b and c-given sides, A-angle between two sides

one period of sin θ

(360°)hits x axis at 0°, 180°, and 360°

one period of cos θ

(360°)hits x axis at 90° and 270°

one period of tan θ

(180°)backwards s curve(first up second down), first asymptote at 90°, hits x axis at 0° and 180°

one period of csc θ

(360°)flip arches from sine, never hits the x axis, asymptotes at 0°, 180°, and 360°

one period of sec θ

(360°)flip arches from cosine, never hits the x axis, asymptotes at 90° and 270°

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one period of cot θ

(180°) forms s curve, hits x axis at 90°, asymptotes at 0° and 180°

SSA Ambiguous Case

If a=bsinA then 1∆, if absinA then a≥b 1∆ or a

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Hero’s formula

used when given SSS and asked for area:K=√s(s-a)(s-b)(s-c) where s=1/2(a+b+c)

finding area when given SSS of a triangle

K=1/2bcsinA

equation for trig graphs

f(x)=asinb(x-h)+k

|a| in trig graph equation

amplitude:>1=vertical stretch, <1=vertical compression, if negative graph is flipped

b in trig graph equation

2π/b=period for sin, cos, csc, sec, π/b=period for tan, cot

h in trig graph equation

horizontal shift:+=left, -=right

k in trig graph equation

vertical shift:+=up, -=down

intervals when graphing trig equations

1/4period

sin 2θ

2sinθcosθ

sin(α±β)

sinαcosβ±cosαsinβ

cos(α+-β)

cosαcosβ−+sinαsinβ

cos 2θ

cos²θ-sin²θ

tan 2θ

2tanθ/1-tan²θ

opposite

-z(straight across from angle)

conjugate

z with line over it(mirror image of angle)

reciprocal

z⁻¹(mirrored and flipped r of angle)

i³

-i

i⁴

1

regular circle

r=4sinθ

rose

r=2sin2θ, first 2:length of petals, second 2:number of petals but is even so must be doubled

limacon w/ inner loop

r=1-2cosθ

limacon w/o inner loop

r=3+2cosθ

cardioid

r=2+2sinθ

determining classical curve

r=a+bcos/sineθ:|a|=|b|→cardioid, |a|<|b|→limacon w/ inner loop, |a|>|b|→limacon w/o inner loop

lemniscate

r²=4cos2θ

r²=

x²+y²

tan(α+β)

tanα+tanβ/1-tanαtanβ