auxilliary angle question (1 Viewer)

[243_robbo]

when completing these transformation questions, my textbook gives me four formulas to learn for different sin + cos, sin - cos, cos + sin, cos - sin combinations. I dont really want to learn the long winded way as some do, but is it correct to rearrange to question so i can just use the one formula:

a.sin(x) + b.cos(x) = r.sin(x + α)
________
where r √a^2 + b^2 , tan α = b/a


nd just use negative a's and b's


for example

5.cos(x) - 3.sin (x) = (-3)sin(x) + (5)cos(x)


thanks for any help you can offer

 

[Trev]

Yeah that's fine, unless they state which formula they want you to use.

 

[243_robbo]

and by telling you which formula they want you to use then you can just sub into that

 

[243_robbo]

ok so idid that question above and i got

sqrt(34).sin (x - 59) ---roughly


and the answer is sqrt(34).cos(x + 31)


and because sin(x) = cos(90 -x)

my answer = sqrt(34).cos (90 -(x - 59))
= sqrt(34).cos (149 - x)

but cos is symetrical about y axis so

= sqrt(34).cos (x -149)

so does that equal

sqrt(34).cos(x + 31) from the answers?

 

[Riviet]

sqrt34.sin(x-59)=sqrt34.cos[90-(x-59)]
=sqrt34.cos(90-x+59)
=sqrt34.cos(149-x)
=sqrt34.cos(180-31-x)
=sqrt34.cos(180-(31+x))
=-sqrt34.cos(x+31). The negative comes from the fact that cos is negative in second quadrant. Check your working.

 

[Ror bones]

i try to memorise as little as possible

with the the transformation method as long as you know the sin (a+b) and
cos(a+b) expansions you are right

 

[243_robbo]

even then i only think you need to know the sin one, ror.


o yeah and thanks riviet, ive seen some of your posts your a maths god



do u reckon itd be ok if i left my answer as the sine one cos its the same thing really?

 

[Riviet]

If the question didn't specify what you need to transform the equation into, then either one's fine.