Hey guys! I know a lot of you out there are having troubles with trig, I mean, I myself had troubles initially. And so, I decided to share with you guys all I know about Trig in this thread. I guess you could say it covers the entire Trig course in Spesh. I'm trying to keep this tutorial short and sweet so it would be easier to understand. Of course it's not going to be perfect so if I miss anything or did anything incorrectly, please post and I will fix it asap
I will regularly be updating this tutorial as I can't do the whole thing at once.
Hope you guys learn a thing or two from this.
TRIGONOMETRYContents:- Addition Formulae
- Double angle Formulae
- Trigonometric Equations
- Trig Identities
- Re-expression of acosx + bsinx
- Sine Rules
- Cos Rules
ADDITION FORMULAEI'm going to skip the derivations of the compound angle formulas and double angle formulas. If you're really interested in the derivation however, here's a pdf which shows the derivation
http://www2.warwick.ac.uk/services/elearning/mathsfit/trigonometry/5/xa2gt1pcaf.pdf (Credits to the person who made that
Compound Angle FormulaeYou will need to remember these formulas so note them down.
- sin(u+v) = sinucosv + sinvcosu
- sin(u-v) = sinucosv - sinvcosu[An easy way of remembering the two sine rules is to remember that if you are finding sin(u+v), you are finding the sin of the two angles ADDED, and so, it is sinucosv PLUS sinvcosu. Vice versa sin(u-v).]
- cos(u+v) = cosucosv - sinusinv
- cos(u-v) = cosucosv + sinusinv
[An easy way of remembering the two cosine rules is to remember that the sign between cosucosv and sinusinv is opposite to the sign inside the cos brackets. For example, if you are finding cos(u+v), it would be cosucosv
MINUS sinusinv, because opposite of + is -. Vice versa for cos(u-v).]
- tan(u+v) = (tanu + tanv) / (1 - tanutanv)
- tan(u-v) = (tanu - tanv) / (1 + tanutanv)
[An easy way of remembering the two tan rules is to remember that the sign inside the tan brackets determines the sign between the tanu and tanv in the numerator, and to remember that the sign between the 1 and tanutanv, is always opposite to the sign between tanu and tanv in the numerator. For example, tan(u+v). There is an addition between the angles u and v, hence there is an addition between tanu and tanv in the numerator. The sign of between 1 and tanutanv is opposite to the sign between tanu and tav in the numerator, hence it is 1
MINUS tanutanv. Therefore tan(u+v) = (tanu
PLUS tanv) / (1
MINUS tanutanv)]
Okay, now that we know the formulas, lets try out some problems.
1. Find an exact expression for:a. sin(15°)
sin(15°) = sin(45° - 30°)
Because there is a
MINUS between 45° and 30°, we know that there must also be a
MINUS between sinucosv and sinvcosu. Therefore we use the sin(u-v) compound angle formula.
= sin(45°)cos(30°) - sin(30°)cos(45°)
Using the exact values table, we get:
= (√2/2)(√3/2) - (1/2)(√2/2)
= (√2/2) [ √3/2 - 1/2 ]
= (1/√2) [ (√3 - 1)/2]
=
(√3 - 1) / 2√2
b. cos( 11π/12) [sorry this π is pi]
= cos( (π/4) + (2π/3) )
= cos(π/4)cos(2π/3) - sin(π/4)sin(2π/3)
= (√2 /2)( - 1/2) - (√2 /2)(√3 /2)
= (√2 /2) [ -1/2 - √3/2 ]
= (1/ √2) [ (-1 -√3) /2]
=
- (√3 + 1) / 2√2
2. Simplify (1/ √2).sin(α + β) + (1/ √2).cos(α + β)From our exact values table, we know that sin(π/4) = cos(π/4) = √2/2 = 1/√2
So, let the first 1/√2 = sin(π/4) and the second 1/√2 = cos(π/4)
We then get:
sin(π/4).sin(α + β) + cos(π/4).cos(α + β)
= cos(π/4).cos(α + β) + sin(π/4).sin(α + β)
What does this resemble? Let u = π/4 and v = α + β
= cosu.cosv + sinu.sinv
= cos(u-v)
Subbing the angles back into u and v gives
=
cos( π/4 - (α + β)) Magic isn't it? Sometimes you just have to think outside the box, or use different ways to approach a problem when you get stuck.
This question also has an alternative answer, that being
sin(( π/4 + (α + β)). [That was derived from using complementary angles, but you could also work it out by subbing the first 1/√2 as cos(π/4) and the second 1/√2 as sin(π/4)]
3. Show that 1/sin(10°) - √3/cos(10°) = 4Firstly cross multiply the LHS so that they share a common denominator.
LHS = [ cos(10°) - √3sin(10°) ] / sin(10°)cos(10°)
Multiply numerator and denominator by 2
= 2[ cos(10°) - √3sin(10°) ] / 2sin(10°)cos(10°)
√3 = tan(60°)
= 2[ cos(10°) - tan(60°)sin(10°) ] / 2sin(10°)cos(10°)
= 2[ cos(10°) - [sin(60°)/cos(60°)]sin(10°) ] / 2sin(10°)cos(10°)
Now at this part, completely ignore the denominator or else you will get confused.
Ignoring the denominator, we have:
2[ cos(10°) - [sin(60°)/cos(60°)]sin(10°) ]
= 2[ cos(10°) - sin(10°)sin(60°)/cos(60°) ]
= 2[ [cos(10°)cos(60°) - sin(10°)sin(60°)] /cos(60°) ]
= 2[ cos(10° + 60°) / (1/2) ]
= 4cos(70°)
Now we look at the denominator we ignored before, alone:
2sin(10°)cos(10°) [I know there is a double angle formula, but we'll stick to compound angle formulas first]
= sin(10°)cos(10°) + cos(10°)sin(10°)
= sin(10° + 10°)
= sin(20°)
Putting the numerator and denominator back together, we have:
= 4cos(70°) / sin(20°)
What do we notice about cos(70°) and sin(20°)? They are in fact complementary angles! cos(70°) = sin( 90° - 70°) = sin(20°)
= 4sin(20°) / sin(20°)
The sin(20°) cancels out
=
4Therefore 1/sin(10°) - √3/cos(10°) = 4, as required.Just a tip. It's up to you whether or not you put 'as required', though I heard that it was safer to do this when doing show/prove questions.
4. If α+β+γ = π, prove that tanα + tanβ + tanγ = tanα.tanβ.tanγUsing α+β+γ = π,
α + β = π - γ Tan both sides
tan(α + β) = tan(π - γ)
(tanα + tanβ) / 1 - tanαtanβ = (tanπ - tanγ) / 1 + tanπtanγ tanπ = 0
= 0 - tanγ / 1 + (0)tanγ
= - tanγ multiply (1 - tanαtanβ) over to RHS
tanα + tanβ = - tanγ(1 - tanαtanβ)
tanα + tanβ = - tanγ + tanαtanβtanγ add tanγ to both sides
tanα + tanβ + tanγ = tanα.tanβ.tanγ
Therefore tanα + tanβ + tanγ = tanα.tanβ.tanγ, as required. I will continue with the Double Angle Formula part another time and should finish it in a day or two.
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