How do I express y=acosx+bsinx in the form y=Rcos(x-c)?

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How do I express y=acosx+bsinx in the form y=Rcos(x-c)?

From the addition formula, we know that:

Rcos(x-c) = Rcos(x)sin(c)+Rsin(x)cos(c)

Therefore:

acos(x)+bsin(x) = Rcos(x)cos(c)+Rsin(x)sin(c)

If we equate the coefficients of cos(x) and sin(x) we see that:

acos(x) = Rcos(x)sin(c); therefore a = Rcos(c)

And that:

bsin(x) = Rsin(x)cos(c); therefore b = Rsin(c)

To find c:

If we divide one of the above results by the other:

Rsin(c)/Rcos(c) = b/a

tan(c) = b/a

Therefore, c = arctan(b/a)

To find R:

a²+b² = R²cos²(c)+R²sin²(c)

a²+b² = R²(cos²(c)+sin²(c))

As cos²(c)+sin²(c) = 1,

a²+b² = R²

(a²+b²)^1/2=R

So, overall:

acos(x)+bsin(x) = (a²+b²)^1/2cos(x-arctan(b/a))

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