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    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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    I was struggling with the same thing up until the weekend when I found these notes they helped a lot
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    3sinx-cosx Ksin(x-a):ksinxcosa-kcosxsina

    Kcosasinx-ksinacosx.
    What comes infront of sinx it's 3 so kcosa;3

    same applies for what comes infront of cosx -1 so -ksina:-1 hope tht sort of helped for each type of question just go through same process
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    thanks peeps
 
 
 

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