Trigonometry: Auxillary Rule (1 Viewer)

[munkaii]

Just a question about the expression:

asinx+bsinx=c=rsin(x+alpha)

for this eqn.

r= (aa + bb)^1/2

tan-1 x = b/a <<<<<<< is this the same expression for both sine and cos rule? or is it different?

 

[Slidey]

Why not just derive it each time? It's quick, and there's less chance of error that way:

Given:

asinx+bcosx=c

You could go:
asinx+bcosx=R(sinxcos@+cosxsin@)=Rsin(x+@)
Equate:
Rcos@=a
Rsin@=b
R^2(cos^2(@)+sin^2(@))=a^2+b^2
.'. R=sqrt(a^2+b^2)
But also:
Rsin@/Rcos@=b/a
tan@=b/a
So tan^-1(b/a)=@ (QUITE different to what you had!)

Suppose you instead went:
asinx+bcosx=R(cosxcos@+sinxsin@)=Rcos(x-@)
Equate:
Rcos@=a
Rsin@=b
tan@=b/a (Still)

That is: yes, arctan(b/a)=@ works for any of the 4 identities you use.

 

[Will Hunting]

The above method is called the subsidiary angle method, and is sometimes called the auxiliary angle method because the angle introduced ( i.e. &alpha; ) is secondary to the angle you're solving for (i.e. x).

is this the same expression for both sine and cos rule? or is it different?

It's different, although memorising the methods for each situation is not at all difficult. Here's how it's done:

When the LHS begins with an expression in sin&Theta;, signs are the same for both equations i.e.

asin&Theta; + bcos&Theta; = c = rsin( &Theta; + &alpha; )
(There is a plus sign in both equations)

asin&Theta; - bcos&Theta; = c = rsin( &Theta; - &alpha; )
(There is a minus sign in both equations)

Conversely, when the LHS begins with an expression in cos&Theta;, signs are contrary, or they alternate from one side to the other i.e.

acos&Theta; + bsin&Theta; = c = rcos( &Theta; - &alpha; )
(There is a plus sign on the left and a minus sign on the right)

acos&Theta; - bsin&Theta; = c = rcos( &Theta; + &alpha; )
(There is a minus sign on the left and a plus sign on the right)

You've probably recognised that these results exhibit the same features as results for sums and differences of angles. In actual fact, the processes involved are exactly the same, as Slide Rule has demonstrated above, so if you're up to speed with sums and differences, you're sure to shred this up too Good luck!

 

[munkaii]

ah. i knew the sums and differences connection.
Just wasnt sure about the tan alpha part.

ive got a nother q while im at it.
Express sin theta in terms of sin theta/2 and cos theta/2.< basic enoguh i can do that.

Given t= tan theta/2, show that sin theta = 2t /1+t^2

 

[JamiL]

yer u could derive it each tym or u could just fukn learn it
lol... it isn that hard.

 

[rama_v]

ive got a nother q while im at it.
Express sin theta in terms of sin theta/2 and cos theta/2.< basic enoguh i can do that.

Given t= tan theta/2, show that sin theta = 2t /1+t^2

Its in the Fitzpatrik book if you want to see it. The way to do it is draw up a right angled triangle with tan@/2 as the acute angle. Then from there the opposite side is t and the adjacent side is 1. so tan @/2 = t , sin@/2 = t/sqrt(1+t^2) , cos @/2 = 1/sqrt(1+t^2) .

You know that sin2@ = 2sin@cos@ so therefore sin@ = 2sin(@/2)cos(@/2)
just plug in the stuff you've derived above and you get 2 x t/sqrt(1+t^2) x 1/sqrt(1+t^2)

which equals 2t/1+t^2

 

[m_isk]

ah. i knew the sums and differences connection.
Just wasnt sure about the tan alpha part.

ive got a nother q while im at it.
Express sin theta in terms of sin theta/2 and cos theta/2.< basic enoguh i can do that.

Given t= tan theta/2, show that sin theta = 2t /1+t^2

(i) sin@=2sin(@/2)cos(@/2)

(ii) given t =tan(@/2),
well, tan(@) = tan(@/2+@/2) =2tan(@/2) / 1-tan(@/2)^2
=2t/1+t^2 given tan(@/2)=t
then, you should draw a right angled triangle where tan(@) = is the crap we said..then using pythagoras you will find that the third side, which is the hypotenuse, is 1+t^2..
sin(@) equals opposite/hypotenuse =2t/1+t^2
Note: you should be able to find this in any 3u book under the "t" method

 

[m_isk]

great timing rama!! (i cant believe this..that's the second time in 20 mins

 

[rama_v]

lol! looks like you were unlucky hehe

 

[velox]

yer u could derive it each tym or u could just fukn learn it
lol... it isn that hard.

sometimes in exams they get you to derive it. So i would know how to derive it if i were u.

 

[Trebla]

The way I was taught to derive the t results of sinθ was without the reference of a right-angled triangle:

Since:
sin2A = 2sinA.cosA
Let A = θ/2
.: sinθ = 2sin(θ/2).cos(θ/2)

Make sin²(θ/2) + cos²(θ/2) as the denominator of 2sin(θ/2).cos(θ/2)
(since sin²(θ/2) + cos²(θ/2) = 1)

.: sinθ = ____2sin(θ/2).cos(θ/2)____
-------------> sin²(θ/2) + cos²(θ/2)

Divide the numerator and denominator by cos²(θ/2)
(since the numerator and denominator have a common denominator, this denominator would normally cancel out)

.: sinθ = ____2sin(θ/2).cos(θ/2)____ ÷ ____sin²(θ/2) + cos²(θ/2)___
...............................cos²(θ/2)...................................cos²(θ/2)........................


The first term should simplify to 2tan(θ/2):

.: sinθ = 2tan(θ/2) ÷ ____sin²(θ/2) + cos²(θ/2)___
...................................................cos²(θ/2).......................................................

The second term should be split into separate fractions and simplified:
.:sinθ = 2tan(θ/2) ÷ [____sin²(θ/2)____ + ____cos²(θ/2)___]
................................[........cos²(θ/2)....................cos²(θ/2)......].......................


Simplify the two fractions and you get:

sin θ = 2tan(θ/2) ÷ [tan²(θ/2) + 1]

.: sin θ = ____2tan(θ/2)____
...................[tan²(θ/2) + 1].....................................................................................

Given that t = tan(θ/2)

.: sinθ = ___2t___
.................t² + 1......................................................................................................

How's that?

[IGNORE the dotted lines or elipses. They're there because I can't do more than 2 space bars that will show on the post and I didn't want to use the slash as a fraction line because that will make it confusing especially with complicated fractions]

 

[JamiL]

sometimes in exams they get you to derive it. So i would know how to derive it if i were u.

yes u HAVE 2 no how 2 derive it, as u said they can ask u. but noing how 2 derive it should make it easyer 2 remember. in a case when they say scetch
cos@+root2sin@.
in this case u could derive it but it would be a waste of tym, if u new the formula u can do it in a second. its not a big deal thou
lol

 

[munkaii]

The way I was taught to derive the t results of sinθ was without the reference of a right-angled triangle:

Since:
sin2θ = 2sinθ.cosθ
Let θ = θ/2
.: sinθ = 2sin(θ/2).cos(θ/2)

Make sin²(θ/2) + cos²(θ/2) as the denominator of 2sin(θ/2).cos(θ/2)
(since sin²(θ/2) + cos²(θ/2) = 1)

.: sinθ = ____2sin(θ/2).cos(θ/2)____
-------------> sin²(θ/2) + cos²(θ/2)

Divide the numerator and denominator by cos²(θ/2)
(since the numerator and denominator have a common denominator, this denominator would normally cancel out)

.: sinθ = ____2sin(θ/2).cos(θ/2)____ ÷ ____sin²(θ/2) + cos²(θ/2)___
...............................cos²(θ/2)...................................cos²(θ/2)........................


The first term should simplify to 2tan(θ/2):

.: sinθ = 2tan(θ/2) ÷ ____sin²(θ/2) + cos²(θ/2)___
...................................................cos²(θ/2).......................................................

The second term should be split into separate fractions and simplified:
.:sinθ = 2tan(θ/2) ÷ [____sin²(θ/2)____ + ____cos²(θ/2)___]
................................[........cos²(θ/2)....................cos²(θ/2)......].......................


Simplify the two fractions and you get:

sin θ = 2tan(θ/2) ÷ [tan²(θ/2) + 1]

.: sin θ = ____2tan(θ/2)____
...................[tan²(θ/2) + 1].....................................................................................

Given that t = tan(θ/2)

.: sinθ = ___2t___
.................t² + 1......................................................................................................

How's that?

[IGNORE the dotted lines or elipses. They're there because I can't do more than 2 space bars that will show on the post and I didn't want to use the slash as a fraction line because that will make it confusing especially with complicated fractions]

cheers! i was looking for a way besides triangle method. ta

 

[Slidey]

in this case u could derive it but it would be a waste of tym, if u new the formula u can do it in a second. its not a big deal thou
lol

If you derive it, "you can do it in a second".

 

[Will Hunting]

Since:
sin2θ = 2sinθ.cosθ
Let θ = θ/2
.: sinθ = 2sin(θ/2).cos(θ/2)

This should be rephrased,

sin2A = 2sinAcosA, for some A &ne; &Theta;
Let A = &Theta;/2
Then sin&Theta; = 2sin(&Theta;/2)cos(&Theta;/2)

&Theta; = &Theta;/2 implies &Theta; = 0, and only 0

But from the triangle proof, the t results can be seen to hold for &Theta; &ne; 0 aswell.

this denominator would normally cancel out

It will always cancel out, provided it's added to both the numerator and the denominator of the starting expression.

 

[JamiL]

If you derive it, "you can do it in a second".

lol... if ur that quick slide... u have 2 admitt if u dont have 2 derive it, it is much quicker.

 

[munkaii]

But!

yes but somehow pull a wrong answer and you lose a majority of your marks

 

[amandagirgenti]

The above method is called the subsidiary angle method, and is sometimes called the auxiliary angle method because the angle introduced ( i.e. α ) is secondary to the angle you're solving for (i.e. x).




It's different, although memorising the methods for each situation is not at all difficult. Here's how it's done:

When the LHS begins with an expression in sinΘ, signs are the same for both equations i.e.

asinΘ + bcosΘ = c = rsin( Θ + α )
(There is a plus sign in both equations)

asinΘ - bcosΘ = c = rsin( Θ - α )
(There is a minus sign in both equations)

Conversely, when the LHS begins with an expression in cosΘ, signs are contrary, or they alternate from one side to the other i.e.

acosΘ + bsinΘ = c = rcos( Θ - α )
(There is a plus sign on the left and a minus sign on the right)

acosΘ - bsinΘ = c = rcos( Θ + α )
(There is a minus sign on the left and a plus sign on the right)

You've probably recognised that these results exhibit the same features as results for sums and differences of angles. In actual fact, the processes involved are exactly the same, as Slide Rule has demonstrated above, so if you're up to speed with sums and differences, you're sure to shred this up too Good luck!

thank u soo much 4 this explanation... i was wondering why wen the expression begins with cos, the signs on the left and right are opposite

thanx agenn