How do i prove algebraicly that a sin(Bx)+b cos(Bx) is the same as Asin(Bx+C)?

Question

Whomever answers this, will you provide a step-by-step guide on how to do it.

Answer 1

I have to go teach my class in about 4 minutes, but when I return, if no one has done it, I can show you how. It requires factoring out a specific value that makes the coefficients "a" and "b" turn into two values that satisfy x^2 + y^2 = 1. In other words, I know it starts out by factoring out sqrt(a^2 + b^2) from

a sin(Bx)+b cos(Bx)


to get


sqrt(a^2 + b^2) [ a/sqrt(a^2 + b^2) * sin (Bx) + b/sqrt(a^2 + b^2) * cos(Bx)]


EDIT: Okay here we go. And the "orthogonal rotating vectors" proof below is usually over most people's heads, or at least, most people who would ask this question on Yahoo answers.


First, you factor out SQRT(a^2+b^2) as I mentioned above. For the sake of simplicity, I will let A = SQRT(a^2+b^2). So factor A out of

a sin(Bx)+b cos(Bx)

to get

A [a/A sin(Bx) + b/A cos(Bx)]




Since (a/A)^2 + (b/A)^2 = 1 (you might need to actually verify this for your purposes), there exists some angle C such that

cos (C) = a/A and sin (C) = b/A

(also, you should point out that both a/A and b/A have absolute values less than one, so that C is real and not complex with a nonzero imaginary part).

So make the substitutions to get

A [cos(C) sin(Bx) + sin(C) cos(Bx)].





According to the formula

sin(y+z) = sin y cos z + cos y sin z,

the expression in the brackets must equal sin(Bx + C). So now the whole expression becomes

A sin (Bx + C).





So in conclusion,

a sin(Bx)+b cos(Bx) = Asin(Bx+C),

where A = SQRT(a^2+b^2), and C = arccos(a/A).




There is a better representation for C, however.

Since cos C = a/A and sin C = b/A, you have

tan C = (sin C)/(cos C) = (b/A)/(a/A) = b/a.

Thus, tan C = b/a, and C = arctan (b/a). This definition of C does not rely on the creation of a third variable (A).





Hope this helps!

Answer 2

This is the "well known trigonometric identity" which is harder to find as hen's teeth, so you end up having to derive it yourself!
Sinθ and cosθ can be represented as orthogonal rotating vectors, e^jθ and e^j(θ + π/2)
Viewing the problem as one in vector addition, with cos(Bx) orthogonal to sin(Bx), the resultant vector will have a magnitude of A^2 = a^2 + b^2, and an angle with respect to asin(Bx) of C = arctan(b/a). therefore,
if A = √(a^2 + b^2),
and C = arctan(b/a),
then a sin(Bx) + b cos(Bx) = Asin(Bx + C)

Answer 3

are A,B and C the angles of a triangle?