Trig Identity

[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?

[berkeman]

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?

Does this help?

http://en.wikipedia.org/wiki/Trig_identities

.

[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!

[mathman]

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

[tiny-tim]

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

Hi HasuChObe! :smile:

(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(?) :wink: