P HYSICS NOTES
Using Trigonometry with Vectors
OBJECTIVE: Much of physics is expressed in the language of vectors. We will frequently need to resolve or
“decompose” a vector into its scalar components. We also often go the other way, where we “reassemble” a vector
from its scalar components. Both of these tasks can be done with trigonometry.
Let’s first define the basic trigonometric functions for a right triangle:
In a right triangle, there is one right angle (90) and two acute angles
(acute—less than 90). We often label the acute angle that we’re
interested in with the Greek letter , pronounced “theta.”
Hypotenuse

The hypotenuse is the longest side of a right triangle. It is the side
across from the right angle. It is always one of the sides making up the
angle .
Opposite
side
Adjacent side
sin 𝜃 =
opposite side
hypotenuse
We define the three trigonometric functions—sin, .cos, and tan —
in terms of these sides. These are given at right.
cos 𝜃 =
adjacent side
hypotenuse
A helpful mnemonic is: “soh cah toa”.
tan 𝜃 =
opposite side
adjacent side
The adjacent side is the other side making up the angle .
The opposite side is the side across from the angle .
Going from magnitude and direction to components
Let’s start with the task of resolving (also called “decomposing”) a vector into its scalar components, given that we
know the vector’s magnitude (length) and direction (angle).
Consider the vector 𝐴⃑ which makes an angle  with the positive x-axis. As always, we denote the magnitude of 𝐴⃑ as A.
y
Length is A
𝐴⃑
From the picture we can see the following:
Ay

Ax
x
We rearrange to find:
Ax = A cos
cos 𝜃 =
𝐴𝑥
𝐴
sin 𝜃 =
𝐴𝑦
𝐴
Ay = A sin
Note: Angles can be given in degrees or in radians. When evaluating the trigonometric functions at a particular angle,
your calculator must be in the proper mode. Hit the MODE button and scroll down to select the correct one, and hit
ENTER.
Example. The vector 𝐴⃑ has a magnitude of A = 3 meters. It makes an angle  = 30 with the positive x-axis.
Find the scalar components Ax and Ay. (Since  is in degrees, calculator must be in degree mode.)
Solution: We use the above equations. Ax = A cos = (3 m)cos30 = (3 m)(.866) = 2.59 m
Ay = A sin = (3 m)sin30 = (3 m)(.5) = 1.5 m
So the vector 𝐴⃑ points 2.59 m in the positive x-direction, and points 1.5 m in the positive y-direction.
Note: Angles are generally defined from the positive x-axis. Angles are positive in the counter-clockwise direction,
and are negative in the clockwise direction. This is the way our calculators calculate angles.
The function cos will give us negative values if  is between 90 and 270. This is consistent with Ax = A cos, since Ax
is negative for vectors that point in directions between (but not including) these angles.
The function sin will give us negative values if  is between 180 and 360. This is consistent with Ay = A sin, since Ay
is negative for vectors that point in directions between (but not including) these angles.
When dealing with vectors that point in directions with  > 90, we have two options:
(1) Use the full positive angle as defined from the positive x-axis. Then the trig functions will automatically
give the correct signs for the scalar components.
(2) Use a “reference angle” which is less than 90 (and positive). In this case the trig functions will always be
positive and so we have to let the picture tell us the signs for the scalar components. We have to put in
the correct signs “by hand.” This method is used more often than the first because it makes more sense
visually.
⃑⃑ has a magnitude of B = 2 inches. It makes an angle  = 40 above the negative x-axis.
Example. The vector 𝐵
Find the scalar components Bx and By. (Since  is in degrees, calculator must be in degree mode.)
Method 1—Use the full angle: The full angle is 180– 40 = 140.
Using this angle, the component signs will automatically be correct.
We use the above equations. Bx = B cos140 = (2 in)cos140 = (2 in)(-.766) = -1.53 in
By = B sin140= (2 in)sin140 = (2 in)(.643) = 1.28 in
Method 2—Use the reference angle: The reference angle here is 40. Both cos40 and sin40 are positive, so
we’ll have to put in the correct signs by hand. A drawing is always helpful!
y
B cos = (2 in)cos40 = (2 in)(.766) = 1.53 in
B sin = (2 in)sin40 = (2 in)(.643) = 1.28 in
⃑⃑
𝐵
So Bx = -1.53 in
So By = 1.28 in
140
40
x
Example. The vector 𝑣⃑ has a magnitude of v = 5 m/s. It makes an angle  = -15 with the positive x-axis.
Find the scalar components vx and vy. (Since  is in degrees, calculator must be in degree mode.)
Method 1—Use the full angle: The full angle is 360– 15 = 345.
Using this angle, the component signs will automatically be correct.
We use the above equations. vx = v cos345 = (5 m/s)cos345 = (5 m/s)(.969) = 4.85 m/s
vy = v sin345= (5 m/s)sin345 = (5 m/s)(-.259) = -1.30 in
Method 2—Use the reference angle: The reference angle here is +15. Both cos15 and sin15 are positive, so
we’ll have to put in the correct signs by hand. A drawing is always helpful!
y
345
x
15
𝑣⃑
v cos = (5 m/s)cos15 = (5 m/s)(.969) = 4.85 m/s
v sin = (5 m/s)sin15 = (5 m/s)(.259) = 1.30 m/s
So vx = 4.85 m/s
So vy = -1.30 m/s
Going from components to magnitude and direction
Now let’s go the other way—given the components of a vector, let’s find its magnitude and direction.
y
From the picture at left, we can see how to get the magnitude A and the direction 
if we know the scalar components Ax and Ay .
𝐴⃑
Length is A

Ay
Ax
From the Pythagorean Theorem we have:
x
We also have tan 𝜃 =
𝐴𝑦
𝐴𝑥
𝐴 = √𝐴𝑥 2 + 𝐴𝑦 2
𝐴𝑦
, or taking the inverse: 𝜃 = tan−1 (𝐴 )
𝑥
A Note on the Inverse Trig Functions.
The trig functions take in an angle , and produce a number which is the ratio of two sides.
For example, sin30 = ½ tells us that for any triangle with angle  = 30, the ratio of the opposite side and the
hypotenuse is ½. The sides don’t have to be 1 and 2, respectively, but they must be in that ratio (e.g. 3 and 6).
The inverse trig functions take in a ratio of two sides, and produce an angle. We denote these as sin-1, cos-1, and tan-1.
For example, we just saw that sin30 = ½. It thus follows that sin-1(½) = 30.
On most calculators, we calculate the inverse trig functions using the buttons SIN-1, COS-1, or TAN-1.
You may need to first hit the 2nd button to access these.
Your calculator will give the angle in whatever your MODE setting is (degrees or radians).
When dealing with vectors that point in directions with  > 90, here is the best option:
The Pythagorean Theorem always works for finding the length or magnitude of a vector because the squares
eliminate any concerns over the signs of the components.
Because of some subtleties with the definition of the inverse trig functions, it is best to first make a drawing
of the situation, and then to use the following equations to find the magnitude and angle  respectively:
𝐴 = √𝐴𝑥 2 + 𝐴𝑦 2
|𝐴𝑦 |
𝜃 = tan−1 (
|𝐴𝑥
)
|
Note: This equation for  will only give angles
between 0 and 90. You must use with a drawing
to indicate the correct orientation of the vector.
Using the Pythagorean Theorem and the right triangle trig functions all require that the vector and its components
form a right triangle. And they do form a right triangle only because we chose to resolve vectors into two
perpendicular components. We now see the usefulness of this choice.
For examples, try working the above problems in reverse. That is, take the components that we found and verify that
the above equations (together with a drawing) give the correct magnitude and direction (the answers will be slightly
off because of rounding).