This guide is intended to provide you with the trigonometry that you need to know in order to do physics in 2 dimensions.
Trigonometric or trig functions are simply ratios involving the sides of right triangles.
Consider the right triangle shown below. We're interested in relationships involving the angle, θ. Greek letters are frequently used to denote angles, and θ, pronounced theta, is the most common. People have found it convenient to define the ratios (see Table 1) and to give them the names sine, cosine, and tangent.
Less commonly used functions are the reciprocal functions given in Table 2.
Table 1: Commonlyused functions 
Name 
Function 
Ratio 
sine 
sinθ 
L_{o}/H 
cosine 
cosθ 
L_{a}/H 
tangent 
tanθ 
L_{o}/L_{a} 


Table 2. Reciprocal functions 
Name 
Function 
Ratio 
cosecant 
cscθ = 1/sinθ 
H/L_{o} 
secant 
secθ = 1/cosθ 
H/L_{a} 
cotangent 
cotθ = 1/tanθ 
L_{a}/L_{o} 

Whenever you see one of these trig functions in your reading, remember the definitions. They're just ratios of two sides of a right triangle. They're handy, because the values depend only on the angle. All similar, right triangles with the same angle, θ, have the same values for the trig functions. Here's a mnemonic that people use to recall the ratios: SOHCAHTOA. Say it out loud and you won't forget it. It stands for "Sine = Opposite over Hypotenuse; Cosine = Adjacent over Hypotenuse; Tangent = Opposite over Adjacent.
Table 2 gives multiplicative inverses (reciprocals) of the trig functions. Another way to write a multiplicative inverse is the following: (tanθ)^{1} = cotθ. Note that the entire function, tanθ in this case, goes in the parentheses.
The notation, tan^{1}( ), means something entirely different than a multiplicative inverse. "tan^{1}(y/x)" means to find the angle whose tangent is (y/x). That is, θ = tan^{1}(y/x). Sometimes, this function is called the arctangent to avoid confusion with the term inverse. The tan^{1} function on a calculator is the arctangent function. Note that there are several equivalent notations:
 θ = tan^{1}(y/x)
 θ = arctan(y/x)
 θ = atan(y/x)
Similar considerations apply to the arcsine and arccosine functions. 

In physics problems, you'll see angles like 30°, 45°, and 60° used frequently. That's because the values of trig functions for these angles are easy to retrieve mentally without resorting to using a calculator (or a set of tables in an earlier time). The common ones are given in Table 3.
Table 3. Values of the trigonometric functions for special angles 
Angle 
Triangle 
sin 
cos 
tan 
Explanation 
45° 

1/√2 
1/√2 
1/1 
Given that the legs are each 1 unit, the Pythagorean Theorem gives the hypotenuse. Note that sin45° = cos45° as a consequence of the fact that the legs of the triangle are equal. 
60° 

√3/2 
1/2 
√3/1 
An equilateral triangle with sides of 2 units is convenient to use. The altitude is the square root of 3 by the Pythagorean Theorem.
The geometrical relationships are evident from viewing the triangle.
Values of the trig functions for both 30° and 60° angles can be determined. 
30° 
1/2 
√3/2 
1/√3 
0° 

0 
1 
0 
To get values for an angle of 0°, which of course is not a triangle at all, just imagine that the hypotenuse is swept down to give smaller and smaller values of θ.
In the process, the hypotenuse approaches a length equal to that of the side adjacent to θ.
At 0°, the hypotenuse coincides with the adjacent side, and the side opposite has shrunk to 0. 
90° 

1 
0 
undefined 
This is the opposite process as for 0°. Swing the hypotenuse up so that θ increases toward 90°.
Note that the tangent approaches infinity. 
37° 

3/5 
4/5 
3/4 
37° and 53° may seem like strange angles to use, but they happen to beto 2 significant figuresthe angles of a 345 triangle. 
53° 
4/5 
3/5 
4/3 
For now, the most important application of the trig functions is in determining the components of vectors. Given the direction angle and magnitude of a vector, the x and ycomponents can be determined using the sine and cosine functions. Here's how the process works. 
A. Vectors and Coordinate Systems
Suppose the vector A has a magnitude of A and an angle (also called direction angle) of θ. Note the following.

A vector may be represented either with an arrow over the top, as in the diagram, or in boldface, like this: A. You must use one of the two notations in order to represent a vector without ambiguity.

The magnitude, A, of the vector is represented in normal type without emphasis.

In order to be able to state the direction of a vector, it must be placed in a coordinate system. The vector shown to the right is placed in an xy coordinate system. Such Cartesian coordinates are standard. Directions on the surface of the Earth are also common.
 In an xy coordinate system, the direction angle of a vector is measured counterclockwise from the +x axis. Direction angles therefore range from 0° to 360°.


B. Components of a Vector
The components of a vector are found by dropping perpendiculars from the tip of the vector to the x and yaxes. Notice in the diagram that the components A_{x} and A_{y} are the legs of the right triangle adjacent and opposite to θ. Therefore,
cosθ = A_{x}/A and sinθ = A_{y}/A.
Rearranging, the components are given by the following.
A_{x} = Acosθ and A_{y} = Asinθ.
In case you're wondering why one would want to know the components, they're useful for adding vectors. This method is described in the Ch. 3 reading. 



C. Signs of the Components
The components of a vector can be positive or negative depending on which quadrant the vector is in. Table 4 gives the signs; however, you can always figure them out just by looking at the vector in an xy coordinate system and seeing whether the perpendicular dropped from the tip of the vector intersects the positive or negative extension of the axis.
You can also see that the signs of the trig functions depend on the quadrant into which the vector points. These are also included in the table. Again, it's not necessary to memorize these, as you can figure them out when you need them. 
Table 4 
Sign of component 
Sign of trig function 
Quadrant 
x 
y 
sin 
cos 
tan 
I 
+ 
+ 
+ 
+ 
+ 
II 
 
+ 
+ 
 
 
III 
 
 
 
 
+ 
IV 
+ 
 
 
+ 
 



D. Finding Magnitude and Direction Angle from Components
Let's suppose that the components of the vector are known and one wants to find the magnitude and direction angle of a vector. By the Pythagorean Theorem, the hypotenuse is square root of the sum of the squares of the two sides. Hence,
A = (A_{x}^{2} + A_{y}^{2})^{0.5}.
The direction angle is obtained using the tangent function.
tanθ = A_{y}/A_{x}_{}
The direction angle is then given by: θ = tan^{1}(A_{y}/A_{x}). In order to find the angle on a calculator, first calculate the ratio A_{y}/A_{x}. Then apply the tan^{1} function. 
Note that the inverse trig functions on a calculator don't tell you which quadrant the angle is in. This is something you determine from the context of the problem. Let's say, for example, that the vector A shown to the right has components of A_{x} = 2.5 and A_{y} = 4.5. Both components are negative, since the vector is in the third quadrant. The magnitude of the vector is
A = (A_{x}^{2} + A_{y}^{2})^{0.5} = ((2.5)^{2} + (4.5)^{2})^{0.5} = 5.1
The direction angle is found from
θ = tan^{1}((4.5)/(2.5)) = tan^{1}(1.8)
Now if you put 1.8 into your calculator and take the inverse tangent, the result is 61° to 2 significant figures.What the calculator has given you is the angle φ (pronounced fee) shown in the diagram. It's up to you to perform the following operation to determine the angle θ measured counterclockwise from the +x axis.
θ = 180° + φ = 241°


Let's look at another example. We rotate the vector into the fourth quadrant. Let's say the vector has the same magnitude as before, 5.1. It's not the same vector, though, because it has a different direction angle. So we'll give it a new name, B. Suppose the components are now B_{x} = 4.7 and B_{y} = 1.9. [You can verify that ((4.7)^{2} + (1.9)^{2})^{0.5} = 5.1.]
The direction angle is found from
θ = tan^{1}((1.9)/(4.7)) = tan^{1}(0.40)
The calculator returns 22°, which is the angle φ as before. In order to determine the angle θ measured counterclockwise from the +x axis, you do the following.
θ = 360° + φ = 338°


There are some useful identities involving trig functions. Here are the most common ones. 
From the diagram to the right, you can see that sinθ = Ay/A. Now consider angle φ, which is the complement of θ. From an examination of the right triangle that includes φ, cosφ = Ay/A. That's because Ay is the leg adjacent to φ. Thus, we have
sinθ = cosφ = cos(90 °  θ)
Here are three more identities that we give without proof:
sin(θ) = sinθ
cos(θ) = cosθ
sinθ / cosθ = tanθ


Now consider the right triangle to the right. We have:
sinθ = a/c and cosθ = b/c
Square each trig function and add the squares to obtain:
(sinθ)^{2} + (cosθ)^{2} = (a/c)^{2} + (b/c)^{2} = (a^{2} + b)^{2} / c^{2}
The numerator (a^{2} + b)^{2} is equal to c^{2} by the Pythagorean Theorem. Therefore, we have:
(sinθ)^{2} + (cosθ)^{2} = 1


Other useful identities involving the trig functions are found in Appendix 1 of your text. 

