Trigonometry is simply the study of triangles – more specifically, the study of the angles and dimensions of triangles. Although this sounds quite trivial, trigonometry is a necessary and important part of modern engineering, navigation, design, architecture, and other fields.
Using trigonometry skills, students can work out the precise angle of a triangle’s sides, the distance between different points on a triangle, and other information that’s important in a wide variety of settings. Trigonometry cheat sheet is an extremely helpful means to learn this skill. We provide you with a detailed trig cheat sheet as well as a concise pdf of the trig cheat sheet for your reference.
Special Right Triangles
There are two special right triangles that will continually appear throughout your study of mathematics: the 30 60 90 triangle and the 45 45 90 triangle. The specialty of these triangles is that they provide exact answers instead of decimal approximations when using trigonometric functions. This page will deal with
Special right triangles: 45°-45°-90° triangles and 30°-60°-90° triangles.
How to solve the special right triangles.
Some examples of Pythagorean Triplets: 3-4-5 triangles, 5-12-13 triangles
45 45 90 Triangle
Our first observation is that a 45-45-90 triangle is an isosceles right triangle. This tells us that if we know the length of one of the legs, we will know the length of the other leg. This will reduce our work when trying to find the sides of the triangle. Remember that an isosceles triangle has two congruent sides and congruent base angles (in this case 45º and 45º).
Congruent 45 45 90 triangles are formed when a diagonal is drawn in a square. Remember that a square contains 4 right angles and its diagonal bisects the angles. If the side of the square is set to a length of 1 unit, the Pythagorean Theorem will find the length of the diagonal to be units.
Now The lengths of the sides of a 45 45 90 triangle are in the ratio of 1:1:√2.
This implies a right triangle with two sides of equal lengths must be a 45-45-90 triangle.
You can also see a 45-45-90 triangle by the angles. A right triangle with a 45° angle must be a 45 45 90 special right triangle.
Side1 : Side2 : Hypotenuse = x : x : x√2
Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both 3 inches.
Solution: Step 1: This is a right triangle with two equal sides so it must be a 45°-45°-90° triangle.
Step 2: You are given that both sides are 3. If the first and second value of the ratio x : x: x√2 is 3 then the length of the third side is 3√2.
The length of the hypotenuse is 3√2 inches.
30 60 90 Triangle
Another kind of special right triangle is the 30 60 90 triangle. This is the right triangle whose angles are 30-60-90. The lengths of the sides of a 30 60 90 triangle are in the ratio of 1:√3:2.
You can also recognize a 30-60-90 triangle by the angles. As long as you know that one of the angles in the right-angle triangle is either 30° or 60° then it must be a 30 60 90 special right triangle.
Side1 : Side2 : Hypotenuse = x : x√3 : 2x
Do You know these 30-60-90 ratio rules are useful, but how do you memorize? Remember the 30-60-90 triangle rules are just a matter of remembering the ratio of 1: √3: 2, and knowing that the shortest side length is always opposite the shortest angle (30°) and the longest side length is always opposite the largest angle (90°).
Some people also memorize the ratio, "x, 2x, x√3," because the "1, 2, 3" succession is easy to remember. The one precaution to using this trick is to remember that the longest side is actually the 2x, not the x times √3.
Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 4 inches and 4 inches root three.
Solution: Step 1: Test the ratio of the lengths to see if it fits the n: n√2: 2n ratio.
4: 4√3: ? = x: x√3: 2x
Step 2: Yes, it is a 30°-60°-90° triangle for x = 4
Step 3: Calculate the third side. 2x = 2 × 4 = 8
The length of the hypotenuse is 8 inches.
What are Pythagorean Triples?
Any group of 3 integer values that satisfies the equation: a2 + b2 = c2 is called a Pythagorean Triple. Any triangle that has sides that form a Pythagorean Triple must be a right triangle. Here are some examples of Pythagorean Triples: 3 4 5 Triangle and 5 12 13 Triangle.
A 3-4-5 triangle is a right triangle whose lengths are in the ratio of 3: 4: 5. When you are given the lengths of two sides of a right triangle, check the ratio of the lengths to see if it fits the 3: 4: 5 ratio. The 3 4 5 triangle angles are 36.87° and 53.13° and 90 degrees.
Side1 : Side2 : Hypotenuse = 3n : 4n : 5n
Determine the length of the hypotenuse of a right triangle if the lengths of the other two sides are 6 inches and 8 inches.
Solution: Step 1: Test the ratio of the lengths to see if it fits the 3n: 4n: 5n ratio.
6 : 8 : ? = 3(2) : 4(2) : ?
Step 2: Yes, it is a 3-4-5 triangle for n = 2.
Step 3: Calculate the third side
5n = 5 × 2 = 10
The length of the hypotenuse is 10 inches.
5 12 13 Triangle
A 5 12 13 triangle is a right-angled triangle whose lengths are in the ratio of 5: 12: 13. It is another example of a special right triangle. The 5 12 13 triangle angles are 22.6°, 67.4°, 90°.
3-4-5 and 5-12-13 are examples of the Pythagorean Triple. They are usually written as (3, 4, 5) and (5, 12, 13). In general, a Pythagorean triple consists of three positive integers such that a2 + b2 = c2. Two other commonly used Pythagorean Triples are (8, 15, 17) and (7, 24, 25)
There 6 trig ratios used for finding all the various elements in Trigonometry. They are also commonly called trigonometric functions. These six trigonometric functions are sine, cosine, secant, co-secant, tangent, and cotangent. The easy way to memorize what these trig functions mean, i.e. the ratio of which sides correspond to which trigonometric function is by using different mnemonics.
Some People Have Curly Black Hair Thick Plastered Back.
S of Some ↠ Sine
P of People ↠ Perpendicular
H of Have ↠ Hypotenuse
This gives sinx (S) =Perpendicular(P)/Hypotenuse(H)
C of Curly ↠ Cosine
B of Black ↠ Base
H of Hair ↠ Hypotenuse
This gives cosx (C) = Base(B)/Hypotenuse(H)
T of Thick ↠ Tangent
P of Plastered ↠ Perpendicular
B of Back ↠ Base
This gives tanx (T) =Perpendicular(P)/Base(B)
By using a right-angled triangle as a reference, the trigonometric functions or identities are derived:
sin θ = Opposite Side/Hypotenuse
cos θ = Adjacent Side/Hypotenuse
tan θ = Opposite Side/Adjacent Side
sec θ = Hypotenuse/Adjacent Side
cosec θ = Hypotenuse/Opposite Side
cot θ = Adjacent Side/Opposite Side
The Reciprocal Identities are given as:
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
sin θ = 1/cosec θ
cos θ = 1/sec θ
tan θ = 1/cot θ
Below is the trig values chart for angles that are mostly used for solving problems. For a better understanding of Trigonometry Table and how to learn it please visit Trigonometry Table. The trigonometry table contains the values of (sine, cosine, tangent, cotangent, secant, and cosecant) of standard angles. Let us have a look at the Trig Chart.
Angles In Degrees
Angles In Radians
Trig Identities Cheat Sheet
Periodicity Identities in Radians
These formulas are used to shift the angles by π/2, π, 2π, etc. They are also called co-function identities.
sin (π/2 – A) = cos A & cos (π/2 – A) = sin A
sin (π/2 + A) = cos A & cos (π/2 + A) = – sin A
sin (3π/2 – A) = – cos A & cos (3π/2 – A) = – sin A
sin (3π/2 + A) = – cos A & cos (3π/2 + A) = sin A
sin (π – A) = sin A & cos (π – A) = – cos A
sin (π + A) = – sin A & cos (π + A) = – cos A
sin (2π – A) = – sin A & cos (2π – A) = cos A
sin (2π + A) = sin A & cos (2π + A) = cos A
All trigonometric identities are cyclic in nature. They repeat themselves after this periodicity constant. This periodicity constant is different for different trigonometric identities.
When we try to solve the problems of heights and distances, we should be consider the following:
All the objects such as towers, trees,etc. shall be considered as linear for mathematical convenience.The angle of elevation or angle of depression is considered with reference to the horizontal line.The height of the viewer is ignored, if it is not provided in the problem.
The steps below will be useful to solve Trigonometry word problems:
We should understand the question and draw the appropriate diagram. These are the two most important things to be done in solving word problems in trigonometry.
If it is possible sometimes, we have to split the given information. Because, when we break the given information into smaller steps, we can understand them easily.
It is good to draw diagrams almost for all of the word problems in trigonometry. The diagrams we draw for the given information must be correct. Drawing diagrams for any given information will result in providing us with a clear understanding of the problem statement.
Once we start understanding any particular information clearly and the appropriate diagram is drawn, solving word problems in trigonometry would not feel like a challenging job.
After the drawing part is over, we have to give a name for each position of the diagram using alphabets (it is clearly shown in the word problem given below). Giving names for the positions would be easier for us to identify the parts of the diagram.
Now we have to use one of the three trigonometric ratios (sin, cos, and tan) to find the unknown side or angle.
Once the diagram is drawn and we have translated the English Statement (information) given in the question as a mathematical equation using trigonometric ratios correctly, 90% of the work will be over. The remaining 10% is just getting the answer. That is solving for the unknown.
Look at the example below.
A person standing on the ground at a point, which is 10 m away from the pole position. He finds the top of the pole at an angle of 60°. If the height of his eye level from the ground level is 1.2 m Find the height of the pole.
Let us go through and understand the given information. From our understanding of the information given, we can draw the below picture.
In the picture above, we have to give names for each position using alphabets.
The Person's position ---> A
Pole position ---> B
The Person's eye level ---> E
Top of the pole ---> D
The point on the pole corresponding to eye level ---> C
Now, from the above diagram, the height of the pole is BD.
BD = BC + CD
In the diagram above, we have
AE = BC
AE = 1.2 m
So, BC = 1.2 m.
Now, our goal is to get the length of CD. Once CD is known to us, we can get the length of BD using BD = BC + CD.
In the right angle triangle CDE, CD is the opposite side and EC is the adjacent side and it is known to be 10m.
In the right angle triangle CDE, we have to find the length of the opposite side (CD) and the known side is the adjacent side (EC = 10m).
In this problem, we have to use the trigonometric ratio in which we have the opposite side and adjacent side. Because, the length of the adjacent side (EC = 10m) is known and we have to find the length of the opposite side (CD).
We can use the trigonometric ratio "tan" in this word problem. Because, only in "tan", we have the opposite side and adjacent side. Now, let's look at how to find the length of a CD.
tan 60° = opp/adj
√3 = CD/EC
√3 = CD/10
10√3 = CD
10 ⋅ 1.732 = CD
17.32 = CD
The length of the pole BD = BC + CD
BD = 1.2 + 17.32
BD = 18.52
So, the length of the pole is 18.52 m.
Trigonometry Cheat Sheet pdf
You can also carry with you this PDF of the Trig Cheat Sheet we have compiled for you. This should cover the basics of all kinds of trigonometry problems.
For more practice on the trigonometry problems and detailed step by step explanations, please visit Trigonometry Problems.
Trigonometric functions are used to determine properties of any angle, relationships in any triangle, and the graphs of any recurring cycle.
Learning trigonometry helps you understand, visualize, and graph these relationships and cycles.
If you are studying on your own with staying attentive in class, you get to know the basic trigonometric concepts and likely start noticing cycles in the world around you by generalizing all the concepts which you have learned so far create your own Trigonometry Cheatsheet and refer to it for better results.
Written by Gargi Shrivastava, Cuemath Teacher
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