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Trig question help

Hi everyone, there has been a trig question bothering me for a while from past papers and prelims- we were given the answer along with the question but I don't understand the working. I've gone through the trig chapter in my textbook, and gone through extra notes online relating to higher Maths and still haven't found anything helpful in explaining.


Here is an example question:

--
f(x) = ?3sinx - cosx

Express f(x) in the form ksin(x-a), where k > 0
--

And the worked answer:

--
?3sinx - cosx = ksin(x-a)

= k [sinxcosa- cosxsina]

= kcosa . sinx - ksina . cos x


k cosa = ?3
ksina = - 1

therefore k² = 3 + 1
k = 2

etc. etc.
--

What I don't understand is the point from '= kcosa . sinx - ksina . cos x' onwards.

Why are cos a and sin x flipped, and how then does this equate to k cos a = ?3 and k sin a = - 1?

Any help appreciated, cheers.
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Avatar for vc94
vc94
12
So you have,
3sinx - cosx = k [sinxcosa- cosxsina] by the compound angle identity,
= ksinxcosa - kcosxsina

Equate coefficients of sinx gives: 3 = kcosa
Equate coefficients of cosx gives: 1 = ksina
Original post by vc94
So you have,
3sinx - cosx = k [sinxcosa- cosxsina] by the compound angle identity,
= ksinxcosa - kcosxsina

Equate coefficients of sinx gives: 3 = kcosa
Equate coefficients of cosx gives: 1 = ksina


This.

All theyve done is use an addition formula (which is basically expanding trig brackets :smile:) and then compared coefficients :smile:

Hope that helps
thanks for the help

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