Express cosx - √3sinx= -1 in the form of Acos(x+a)?
Hey guys, can you please help me solve:
Express cosx - √3sinx= -1 in the form of Acos(x+a)
I can't work it out and i've tried several times... can you please show the working out and how you get it.
And here's the answer ^_^
2 cos〖(x/1 + π/3)〗
Thank you!!! SO MUCH!!! :D
Update:Oops sorry i wrote it wrong, the question is, Express cosx - √3sinx in the form of Acos(x+a)
Please ignore the negative one.
Thanks!!!!
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Acos(x + a) ≡ Acosxcosa - Asinxsina
=> cosx - √3sinx ≡ Acosxcosa - Asinxsina
so, 1 = Acosa
and √3 = Asina.....then dividing gives:
--> tana = √3
i.e. a = π/3
squaring gives A² = 1² + (√3)² = 4 => A = 2
so, cosx - √3sinx ≡ 2cos(x + π/3)
:)>
cos(x) - sqrt(3) * sin(x) =>
2 * ((1/2) * cos(x) - (sqrt(3)/2) * sin(x)) =>
2 * (cos(pi/3) * cos(x) - sin(pi/3) * sin(x)) =>
2 * cos(x + pi/3)