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Converting Expression `acos(t)+bsin(t)` into the Form `Asin(t+alpha)`

Any expression of the form `acos(t)+bsin(t)` can be written in the form `Asin(t+alpha)`.

To do this let's factor out `sqrt(a^2+b^2)`:

`acos(t)+bsin(t)=sqrt(a^2+b^2)(a/(sqrt(a^2+b^2)) cos(t)+b/(sqrt(a^2+b^2))sin(t))`.

But `(a/sqrt(a^2+b^2))^2+(b/sqrt(a^2+b^2))^2=(a^2)/(a^2+b^2)+(b^2)/(a^2+b^2)=(a^2+b^2)/(a^2+b^2)=1`. Therefore, from identity `sin^2(alpha)+cos^2(alpha)=1` it follows that exists such angle `alpha` that `sin(alpha)=a/(sqrt(a^2+b^2))` and `cos(alpha)=b/(sqrt(a^2+b^2))`.

If we denote `sqrt(a^2+b^2)` through `A` then we can write that

`color(red)(acos(t)+bsin(t)=sqrt(a^2+b^2)(a/(sqrt(a^2+b^2)) cos(t)+b/(sqrt(a^2+b^2))sin(t)))color(red)(=)`


Numbers `a,b,A,alpha` are connected by following relations:

  • `a=Asin(alpha)`,
  • `b=Acos(alpha)`,
  • `A=sqrt(a^2+b^2)`,
  • `sin(alpha)=a/(sqrt(a^2+b^2))=a/A`,
  • `cos(alpha)=b/sqrt(a^2+b^2)=b/A`.

Example. Convert `3sin(2t)+4cos(2t)` into the form `Asin(2t+alpha)`.

First of all note that coefficient `a` is coefficient near cosine and coeffcient `b` is coeffcient near sine.

That's why a=4, b=3, so `A=sqrt(a^2+b^2)=sqrt(4^2+3^2)=sqrt(25)=5`, `sin(alpha)=a/A=4/5`, `cos(alpha)=b/A=3/5`.

Therefore, `3sin(2t)+4cos(2t)=5sin(2t+alpha)` where `sin(alpha)=4/5,cos(alpha)=3/5`.