This trig identity shows that a combination of sine and cosine functions can be written as a single sine function with a phase shift.
$$a \cos{t} + b \sin{t} = \sqrt{a^2 + b^2} \; \sin(t + \tan^{-1} \frac{a}{b}) $$
for b ≠ 0 and − π⁄2 < tan−1 a⁄b < π⁄2
(Note that tan−1 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)
If we limit the arctan to be within
− π⁄2 < tan−1 a⁄b < π⁄2
then we can always choose the + in front of the square root.
