(Note that the -1 exponent means arctan.) The phase shift is the quantity tan-1(a/b), it has the effect of shifting the graph of the sine function to the left or right.
This trig identity shows that a combination of sine and cosine functions can be written as a single sine function with a phase shift.
(Note that the -1 exponent means arctan.) The phase shift is the quantity tan-1(a/b), it has the effect of shifting the graph of the sine function to the left or right.
To derive this trig identity, we presume that the combination a cos(t) + b sin(t) can be written in the form c sin(K+t) for unknown constants c,K.
a cos(t) + b sin(t) = c sin(K+t)
a cos(t) + b sin(t) = c sin(K) cos(t) + c cos(K) sin(t)
We used the formula for sine of a sum of angles to expand the right hand side above. To have equality for any value of t, the coefficients of cos(t) and sin(t) must be equal on the left and right sides of the equation.
a = c sin(K)
b = c cos(K)
Solving this system of simultaneous equations leads us to
c = ± √(a2 + b2)
K = tan-1(a/b)
So the trig identity for b<>0 is
a cos(t) + b sin(t) = ± √(a2+b2) sin(t + tan-1(a/b))
The ambiguity of whether to choose + or - is related to the ambiguity of the value of
cos(t) + sin(t) = √2 sin(t + π/4) = -√2 sin(t + 5π/4)
If we limit the range of tan-1(a/b) to within (π/2, -π/2) then we can always choose the + solution. The trig identity is then as stated at the top of this page.