MATH 116, LECTURE 6: Trigonometric
identities
1
Trigonometric identities
Consider the function f (x) = 3 sin(x)−4 cos(x) (see Figure 1). Even though
this is a combination of two trigonometric functions, it should be clear that
we should be able to state this function in terms of just one.
Figure 1: Plot of f (x) = 3 sin(x) − 4 cos(x).
Given this intuition, which kind of function should we look for? There
is some freedom in the choice, but the commonly used function is
A sin(x + φ).
In other words, we will try to relate 3 sin(x) − 4 cos(x) to the phase-offset
(φ) sine function with amplitude A. If we could do this, we could state the
combination of two trigonometric functions as one!
How do we proceed? We have the identity sin(x + y) = sin(x) cos(y) +
cos(x) sin(y), so that
A sin(x + φ) = A cos(φ) sin(x) + A sin(φ) cos(x).
(1)
Now consider the general form
f (x) = a sin(x) + b cos(x).
(2)
In order to match (1) and (2), we need to satisfy
A cos(φ)
=a
(3)
A sin(φ) = b.
(4)
1
This is two equations ((3) and (4)) in two unknowns (A and φ). The
easiest way to go about solving the equations for A is to square (3) and (4),
then add them together to get
A2 cos2 (φ) + A2 sin2 (φ) = a2 + b2
=⇒
A2 = a2 + b2
A2 (cos2 (x) + sin2 (x)) = a2 + b2
p
=⇒ A = a2 + b2 .
=⇒
In order to solve for φ, we can divide (4) by (3) to get
b
b
tan(φ) =
.
=⇒ φ = arctan
a
a
We have to be careful, however, since arctan(x) will treat a ratio in quadrant
III as if it were a ratio in quadrant I, and a ratio in quadrant II as if it were
a ratio in quadrant IV. We will have to check the signs of a and b to make
sure the signs in (3) and (4) are correct.
The general formula for turning an expression of the form a sin(x) +
b cos(x) into an expression of the form A sin(x + φ) is:
√
1. Determine A by the formula A = a2 + b2 .
2. If a > 0 and b > 0, or a > 0 and b < 0, set φ = arctan(b/a).
3. If a < 0 and b < 0, or a < 0 and b > 0, set φ = arctan(b/a) + π.
Example 1: Write f (x) = 3 sin(x) − 4 cos(x) as a single phase-shifted sine
function.
Solution: We have a = 3 and b = −4, so by the formula we have
p
p
√
√
A = a2 + b2 = (3)2 + (−4)2 = 9 + 16 = 25 = 5.
Since a > 0 and b < 0, we have
φ = arctan(b/a) = arctan(−4/3) ≈ −0.927295.
It follows that 3 sin(x) − 4 cos(x) = 5 sin(x − 0.927295) (see Figure 1).
Example 2: Write f (x) = −2 sin(x)+5 cos(x) as a single phase-shifted sine
function.
Solution: We have a = −2 and b = 5, so by the formula we have
p
√
A = (−2)2 + 52 = 29.
2
Since a < 0 and b > 0 we have
φ = arctan(−5/2) + π = −1.19028995 + 3.14159265 = 1.951302704.
√
It follows that −2 sin(x) + 5 cos(x) = 29 sin(x + 1.951302704).
Example 3: Write f (x) = −5 sin(x) − 12 cos(x) as a single phase-shifted
sine function.
Solution: We have a = −5 and b = −12, so by the formula we have
p
√
√
A = (−5)2 + (−12)2 = 25 + 144 = 169 = 13.
Since a < 0 and b < 0 we have
−12
12
φ = arctan
+ π = arctan
+ π ≈ 4.317597861.
−5
5
It follows that −5 sin(x) − 12 cos(x) = 13 sin(x + 4.317597861).
3