# Trig Identity

1. Jun 18, 2009

### HasuChObe

I forget how this one goes.

A cos(x) + B sin (x) = C sin (x + invtan(?))

How do you go about condensing both these terms into 1 like the above?

Last edited: Jun 18, 2009
2. Jun 18, 2009

### Staff: Mentor

3. Jun 18, 2009

### jgens

I'm not sure how much help the wikipedia page will be so I'll provide a relatively simple derivation here:

Given,

Asin(x) + Bcos(x)

We can define the sine and cosine of an angle y by considering a right triangle with side lengths A and B. The hypotenuse is then given by,

C = sqrt(A2 + B2)

Consequently, the sine and cosine of y are given by the following formulas,

sin(y) = B/sqrt(A2 + B2)

cos(y) = A/sqrt(A2 + B2)

Substituting these values into the equation produces,

C[sin(x)cos(y) + cos(x)sin(y)] = Asin(x) + Bcos(x)

Therefore,

Csin(x + y) = Asin(x) + Bcos(x)

Now, we only need determing an expression for y. Using our expressions for sin(y) and cos(y), we know that,

tan(y) = B/A

y = arctan(B/A) = tan-1(B/A)

and consequently,

Asin(x) + Bcos(x) = Csin(x + tan-1(B/A))

Hope this helps!

4. Jun 18, 2009

### mathman

Let A=Ccos(y) and B=Csin(y). So you see immediatedly that:
C2=A2+B2
and B/A=tany.

5. Jun 19, 2009

### tiny-tim

Hi HasuChObe!

(I think I'm saying the same as other people, but let's just isolate the principle …)

The object is to get the LHS to look like cos(x)sin(?) + sin(x)cos(?)