Further Trig question (1 Viewer)

[goldenmusic]

I have been spending hours trying to figure out these questions:
Solve for 0<θ<360

5cosθ - 12sinθ = -3
4sinθ - cosθ + 3 = 0
sinθ - cosθ = 1
√ 2cosθ + sinθ = 1

I know we're supposed to use the formula asinθ + bcosθ = rsin(θ+α), where r=√a²+b² and tanα = b/a.

Please help by solving them and showing working
Thanks!

 

[mirakon]

I see you're in prelim.

btw, do you know double angle formula? It is required in auxiliary angles e.g.

rsin(a+b)=rsinacosb+rsinbcosa

If you don't know double-angle formulae and have not yet learnt auxiliary angles to a full extent, there is really no point in doing these questions as yet.

 

[adomad]

use the auxiliary angle method

 

[goldenmusic]

Hmm, i do know those d-angles,
eg. sin2x = 2sinxcosx

are those what you need?

 

[hscishard]

I see you're in prelim.

btw, do you know double angle formula? It is required in auxiliary angles e.g.

rsin(a+b)=rsinacosb=rsinbcosa

If you don't know double-angle formulae and have not yet learnt auxiliary angles to a full extent, there is really no point in doing these questions as yet.

You can use the t-formula as well, without the need of learning the double angle formulae.

Interestingly, you can let sinx = root (1-cos^2x) and similary for cox.

 

[mirakon]

Yes and no. They are vital to auxiliary angles theory-wise but auxiliary angles is

asinθ + bcosθ = rsin(θ+α)

as in your question. Essentially what you do is solve for r (you were correct that it's the quare root of a^2+b^2.

Subsequently expand the RHS using double-angle formulae.

Equate the coeffients of sin theta or cos theta on both sides (whichever one) and then solve for sin alpha or cos alpha, depending on what you equated.

I hope that makes sense.

 

[b00m]

jesus... i'm doing the hsc and i cant even remmeber how to do those..

kekeke.. 3 replies in 10 seconds

 

[cutemouse]

Or use t results.

 

[mirakon]

You can use the t-formula as well, without the need of learning the double angle formulae.

Interestingly, you can let sinx = root (1-cos^2x) and similary for cox.

Hmmmmmmmm................ nevertheless in extension maths, you're looking for the most elegant solutions. t-formulae in this case is far too lenghty in comparison to the standard procedure of auxiliary angles as outlined above. Remember you're working to a tight time-limit, there may be multiple ways to get to an answer, but be sure to choose the fastest, simplest one.

 

[hscishard]

Hmmmmmmmm................ nevertheless in extension maths, you're looking for the most elegant solutions. t-formulae in this case is far too lenghty in comparison to the standard procedure of auxiliary angles as outlined above. Remember you're working to a tight time-limit, there may be multiple ways to get to an answer, but be sure to choose the fastest, simplest one.

That is true. But lets face it. You finish the exams well before the time limit. Lol
Quickest,simplest = best

 

[mirakon]

It really depends on the standard of your school as well. In terms of hardness:

catholic>independent>selective>public

I don't mean to brag, but I'm quite adept in maths and so are some other students in my cohort, and even then we found it a challenge to finish all our prelim papers on time, but maybe that's just because our exams were very hard.

 

[cutemouse]

t-formulae in this case is far too lenghty in comparison to the standard procedure of auxiliary angles as outlined above.

Yes but with the auxiliary angle method you sometimes need to deal with issues relating to the domain. With the t method you don't need to do this and all you need to do is to check if Pi is a solution to the original equation.

 

[hscishard]

It really depends on the standard of your school as well. In terms of hardness:

catholic>independent>selective>public

What?
Selective should be the first.
Btw, I go to a Catholic, private school...easy as ^&*(.
Your school must be awsome.

No I'm not going to reveal my schools name.

 

[iSplicer]

It really depends on the standard of your school as well. In terms of hardness:

catholic>independent>selective>public

I lol'ed

 

[adomad]

I don't mean to brag, but I'm quite adept in maths and so are some other students in my cohort, and even then we found it a challenge to finish all our prelim papers on time, but maybe that's just because our exams were very hard.

or you all suck

 

[_Lone Wolf_]

Auxialiary Angle is the quickest and easiest method to solve these types of problems.

--------------------------------------------------------------------------------------------------------------------------------------
e.g.
5cosθ - 12sinθ = rcos(θ+a)
= r(cosθcosa - sinθsina)
= rcosθcosa - rcosθsina

rcosa = 5
cosa = 5/r (adj/hyp)

rsina = 12
sina = 12/r (opp/hyp)

r^2 = 5^2 + 12^2 (by pythagoras principles and trigonometry)
r^2 = 169
r = 13

a = tan^-1(opp/adj)
= tan^-1(12/5)
= 67 degrees

Therefore:

5cosθ - 12sinθ = 13cos(θ+67)

13cos(θ+67) = -3
cos(θ+67) = -3/13 0 => θ => 360
67 => θ+67 => 427
θ+67 = 103
θ = 36

Hopefully no miscalculation on my part.

Btw GoldenMusic for:

[pSinθ + qCosθ] USE TRASFORMATION --> rSin(θ+a)
[pSinθ - qCosθ] USE TRANSFORMATION --> rSin(θ-a)
[pCosθ + qSinθ] USE TRANSFORMATION --> rCos(θ-a)
[pCosθ - qSinθ] USE TRANSFROMATION --> rCos(θ+a)

--------------------------------------------------------------------------------------------------------------------------------------

By the way, all you guys, you're real smart, eh. You didn't even read his question properly, "Please help by solving them and showing working".

He wants help, not people bragging about how smart they are.

 

[cutemouse]

Auxialiary Angle is the quickest and easiest method to solve these types of problems.

Yes but sometimes you need to deal with getting solutions within the required domain. With t-results you don't need to deal with this (as I stated above).

By the way, all you guys, you're real smart, eh.

Same goes to you