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Basic Definitions Trigonometry (from Ancient Sanskrit "trikona" or triangle and "mati" or measure or from Greek trigōnon "triangle" + metron "measure") has its origins in geometry of right triangles. Many simple questions in trigonometry can be answered using right triangle. Basic trigonometric functions are sine ,cosine, tangent and cotangent. The argument of these functions is an angle. An angle can be designated with vertices of the triangle A , B,.. etc., or Greek letters 𝛼, 𝛽 …etc. 𝑎
𝑏
𝑎
𝑏
sin 𝐴 = ; cos 𝐴 = ; tan 𝐴 = ; cot 𝐴 = 𝑐
𝑐
𝑏
𝑎
Example 1 Question: In the right triangle below, what is the sine of angle A? Solution: 3
sin 𝐴 = 5
Comment: Sine and cosine are always less or equal to one, therefore it is always helpful to double check your answer. In this case the answer is less than one. Example 2 Question: In the right triangle below, what is the value of tan 𝛼 ∗ tan 𝛽? Solution: It is very important to write out your solution. 𝑎
𝑏
tan 𝛼 = ; tan 𝛽 = 𝑏
𝑎
𝑏 𝑎
tan 𝛼 tan 𝛽 = ∗ = 1 𝑎 𝑏
1 Special Triangles Angles with measures of 30o, 45o and 60o are very important in trigonometry. A right triangle with a hypotenuse of 1 unit in length can be used to find the values of the trigonometric functions of these angles. In such a triangle c = 1, therefore, sin A = 𝑙𝑒𝑔 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝐴 ; cos A = 𝑙𝑒𝑔 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝐴 For example, using the 30o ‐ 60o ‐ 90o triangle below and basic geometry rules: 1
3
sin 30! = ; cos 30! = 2
2
This table should be memorized. o
o
o
30 45 60 sin α 1
2
cos α 3
2
2
2
2
2
3
2
1
2
You should also know the following identities: tan 𝛼 = sin 𝛼
cos 𝛼
; cot 𝛼 = cos 𝛼
sin 𝛼
Pythagorean identity: 𝑠𝑖𝑛! 𝛼 + 𝑐𝑜𝑠 ! 𝛼 = 1 Example 3 Question: !
If A is an acute angle and cos 𝐴 = , find sin 𝐴. !
Solution: Use a Pythagorean identity: 𝑠𝑖𝑛! 𝐴 + 𝑐𝑜𝑠 ! 𝐴 = 1 sin 𝐴 = ± 1 − 𝑐𝑜𝑠 ! 𝐴 !
!
! !
!
!
!
sin 𝐴 = ± 1 − = ± = ±
The length is positive, hence 2 2
𝑠𝑖𝑛𝐴 = 3
2 , Example 4 Question: !
If A is acute angle and cos 𝐴 = , find tan 𝐴. !
Solution: tan 𝐴 = !"# !
!"# !
. First, find sin A by referring to example 3 𝑠𝑖𝑛𝐴 = tan 𝐴 = 2 2
3
2 2 1
÷ = 2 2 3
3
The Unit Circle So far we have considered right triangles to define trigonometric functions which limit us to acute angles. However, trigonometric functions are defined for all angles. To find values of sine and cosine for a number of other commonly used angles, we will use the Unit Circle in combination with a special triangle. cos α = x ‐ coordinate of a point on the Unit Circle sin α = y ‐ coordinate of a point on the Unit Circle Using the above definition and the unit circle: sin 0! = 0, cos 0! = 1, cos 180! = − 1 … 𝑒𝑡𝑐. !
Also, as seen on picture 1, cos 120! = − !
Picture 1 From picture 2, it follows that: sin 300! = sin 270! + 30! = − sin (− 60! ) = − 3
2
3
2
1
2
cos (− 60! ) = cos 300! = The Unit Circle is also very helpful in identifying whether a value of a trigonometric function is positive or negative. For example, sin 215! 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 Mark 215! on the unit circle to verify that. 3 Picture 2 Example 4 Question: !
Given that 90! < 𝐴 < 180! and sin 𝐴 = , find cos 𝐴. !
Solution: Use Pythagorean identity: 𝑠𝑖𝑛! 𝐴 + 𝑐𝑜𝑠 ! 𝐴 = 1 cos 𝐴 = ± 1 − 𝑠𝑖𝑛! 𝐴 !
!
! !
!
!
!
cos 𝐴 = ± 1 − = ± = ±
Note, that 90! < 𝐴 < 180! . As noted by the unit circle we have to use the negative value for cos A. Answer: 𝑐𝑜𝑠𝐴 = − Example 5 Question: Given that 0! < 𝐴 < 90! and sin 𝐴 = !
!
! !
!
, find sin 2 𝐴. Solution: !
First, mark on the y‐axis. Then, draw a horizontal dashed line. It will !
intersect the unit circle at two points, which will mark two angles. !
Both angles have a sine of . Since 0! < 𝐴 < 90! and by using a special !
triangle, we can determine that: A = 45! , 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 A ≠ 135! (dashed line angle) 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 2𝐴 = 90! 𝑎𝑛𝑑 sin 90! = 0 Answer: sin 2 𝐴 = 0 (This problem can also be solved using trigonometric identities) 4 Basic Identities Pythagorean identity: 𝑠𝑖𝑛! 𝑥 + 𝑐𝑜𝑠 ! 𝑥 = 1 Double angle identities: sin 2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 cos 2𝑥 = 𝑐𝑜𝑠 ! 𝑥 − 𝑠𝑖𝑛! 𝑥 Often these three identities can be used to simplify rather complicated trigonometric expressions. It is not always clear what simplify means in regard to the expression For example, sin 2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 can be used to simplify an expression, although 𝑠𝑖𝑛2𝑥 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑙𝑜𝑜𝑘 𝑎𝑛𝑦 𝑠𝑖𝑚𝑝𝑙𝑖𝑒𝑟 𝑡ℎ𝑎𝑛 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 Sometimes, you should look at the answer choices to see which way your work should be going. !"# !
!"# !
!
!
Also, recall that: tan 𝑥 = ; cot 𝑥 = ; sec 𝑥 = ; csc 𝑥 = !"# !
!"# !
!"# !
!"# !
Example 6 Question: Given that 0! < 𝐴 < 90! and sin 𝐴 = 1/3, find sin 2 𝐴. Note: This is similar to example 5, but there is no special triangle with an angle such that sin 𝐴 = 1/3. Therefore finding A is more challenging than in the example 5. It is possible to use a calculator sine inverse function to find angle A. However, sometimes it requires additional care, because a calculator function 𝑠𝑖𝑛!! 𝑥 will only return an angle between −90! and 90! . !
For example, if 90! < 𝐴 < 180! 𝑎𝑛𝑑 sin 𝐴 = 1/2 using calculator 𝑠𝑖𝑛!! = 30! , while the correct !
value of A is 150! . Solution using identities: sin 2𝐴 = 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴, you need to find cos A first. !
! !
!
!
cos 𝐴 = ± 1 − 𝑠𝑖𝑛! 𝐴 , 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 cos 𝐴 = ± 1 − = ±
Since 0! < 𝐴 < 90! choose a positive value for cos A. cos 𝐴 =
!
! !
!
!
Finally, sin 2𝐴 = 2 ∙ ∙
5 ! !
!
4 2
= 9 Example 7 Question: Simplify the following expression: 𝑡𝑎𝑛! 𝑥 + 1 !"#! !
Solution: 𝑡𝑎𝑛! 𝑥 + 1 = + 1 = !"# ! !
!"#! !! !"# ! ! !"# ! !
= ! !"# ! !
= 𝑠𝑒𝑐 ! 𝑥 The Unit Circle is very helpful in verifying connections between sine and cosine functions. The following identities can be quickly verified geometrically: sin 𝛼 + 90! = cos 𝛼 ; cos(𝛼 + 90! ) = − sin 𝛼 ; sin(−𝛼) = − sin 𝛼 ; cos −𝛼 = cos 𝛼 Example 8 Question: sin 𝛼 = − cos 𝛼 + 90! True or false? Solution: On the unit circle, sin 𝛼 is given by the leg opposite to angle 𝛼 . Adding 90! is the same as rotating the triangle by 90 degrees. Angle between the hypotenuse of the shaded triangle and the X‐axis equals 𝛼 + 90! . By comparing the two triangles you can verify that the identity is true. Alternatively, the following identities can be used in questions similar to example 8. sin( 𝛼 ± 𝛽) = sin α cos β ± cos α sin β cos( 𝛼 ± 𝛽) = cos α cos β ∓ sin α sin β Set 𝛽 = 90! to prove identity in the example above. 6 Angle and Radian Measures Angle measure of an angle A full circle is 360 degrees. For very small angles, the degree is further divided into 60 minutes. For even smaller measurements, the minute is divided again into 60 seconds. Angle measures are commonly used in geometry. Radian measure of an angle Radian measure represent the arc length of an angle inscribed in the unit circle. We can think of this as a number line wound around a circle. See picture 1. Then, each number on the number line corresponds to an angle. For example: The whole circle with a radius of 1 has circumference of 2π, therefore2 π = 360o. Consequently half of the unit circle is π, therefore π = 180o.. !
!
The number = 0.52 … is one sixth of π, and π = 180o. , hence marks a 30o angle. See picture 2. !
!
Winding of the number line Picture 2 Picture 1 One radian is often defined as an angle which cuts an arc with the arc length equal to the radius. R = 1 yields the Unit Circle interpretation of radians. 7 How to Convert Radians to Degrees and Back (use that π = 180o) Degrees to radians Radians to degrees !
Convert radians to degrees. !
π = 180o , therefore !
!"# !
!
= Convert 135! degrees to radians π = 180o divide both sides by 180! to get: 𝜋
1=
180!
π 3 π
135! = 135! ×1 = 135!
=
!
180
4
= 30! Many calculators can convert the angle measures from degrees to radians and vice versa. Warning! Always check whether your calculator is set up for degrees or radians. Change the setting if necessary Angle measure in radians is often designated with x. Note: you will not be allowed to use the unit circle like the one on the right. 𝜋
𝜋
𝜋
Can’t use on the test 4
6
3
1
3
2
sin x 2
2
2
1
3
2
cos x 2
2
2
Suppose you need to find sin
!"
!
and cos
!"
!"
!
. Graph your own unit circle, think about as if you go around the !
unit circle. !"
!
= 𝑡𝑟𝑎𝑣𝑒𝑙 2𝜋 𝑢𝑛𝑖𝑡𝑠 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑡ℎ𝑒𝑛 𝑏𝑎𝑐𝑘 𝑢𝑝 units. !
!
Inscribe the special triangle, 5π
π
3
5π
1
sin
= sin 2𝜋 − = − ; cos
= 3
3
2
3
2
8 !"
!"
!
!
You may convert to degrees: = 5 ∙
!
!
= five times 60 degrees = 300! Equations Most trigonometric equations lead to finding 𝑥 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑠𝑖𝑛 𝑥 = 𝑐, 𝑜𝑟 𝑐𝑜𝑠 𝑥 = 𝑐 where c is a number. Let us consider the following example: Example 9 Question: !
!
Find all 𝑥, < 𝑥 < 𝜋, such that cos 𝑥 = − !
!
Comment: This is an example of a very simple trigonometric equation. Solution: 1. Cosine is X –coordinate of a 2. The line marks two angles‐
3. To obtain the angle, go along !
point on the unit circle. Mark the solid line and the dashed the unit circle a distance of !
!
!
line. The solid line corresponds − on the X‐axis. Draw a radians and then additional !
!
!
to an angle between and 𝜋. vertical line. !
radians. π π
2π
𝑥 = + = 2 6
3
!"
Answer: 𝑥 = !
Note: it was very important to pick the correct angle from the two choices marked on the unit circle. !
!"
!"
!"
If we change the condition < 𝑥 < 𝜋 to 𝜋 < 𝑥 < the answer will be 𝑥 = , NOT 𝑥 = !
!
!
!
9 Example 10 Question: !
Solve the following equation 2𝑐𝑜𝑠 𝑥 = 4𝑐𝑜𝑠 𝑥 + 1 , for 𝑥 from the interval 0 < 𝑥 < . !
Solution: The equation has only one trigonometric function ‐ cos 𝑥 . We can solve the equation for cos 𝑥. Some people like to replace cos 𝑥 with another variable. Let’s substitute u = cos 𝑥, then 1
2𝑢 = 4𝑢 + 1, 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑢, 𝑢 = − 2
!
cos 𝑥 = − (This equation is solved in the example 7.) !
!"
Answer: 𝑥 = !
Some equations require the use of identities first. It becomes necessary when equation contains several trigonometric functions. Example 11 Question: !
Solve the following equation: 𝑠𝑖𝑛! 𝑥 = 𝑐𝑜𝑠 ! 𝑥 , for 𝑥 from the interval 0 < 𝑥 < . !
Solution: The equation has two trigonometric functions: cos 𝑥 𝑎𝑛𝑑 sin 𝑥. We can use a Pythagorean identity 𝑠𝑖𝑛! 𝑥 + 𝑐𝑜𝑠 ! 𝑥 = 1 to “replace” cos 𝑥 𝑤𝑖𝑡ℎ sin 𝑥. 𝑠𝑖𝑛! 𝑥 = 𝑐𝑜𝑠 ! 𝑥 𝑠𝑖𝑛! 𝑥 = 1 − 𝑠𝑖𝑛! 𝑥 2𝑠𝑖𝑛! 𝑥 = 1 !
𝑠𝑖𝑛! 𝑥 = , therefore sin 𝑥 = ±
sin 𝑥 = !
!
!
, since 0 < 𝑥 < , sin 𝑥 is positive !
Graph the unit circle (see example 5) 𝐴𝑛𝑠𝑤𝑒𝑟: 𝑥 = !
10 !
!
!
!
Graphs of Trigonometric Functions Graphs of the sine and cosine functions will all look similar to each other and resemble a “wave”. For example, the picture to the right shows the graphs of the following functions: 𝑦 = sin 𝑥 𝑦 = cos 𝑥 𝑦 = sin 2𝑥 π
y = 1.5cos ( x + ) 4
One may notice that the graphs have different heights and widths. Also, some of the graphs go through the origin while others are shifted sideways. In mathematics the “height” of the graphs for sine and cosine function is called amplitude and the “width” is called the period. For the graph of 𝑦 = sin 𝑥 , below, the amplitude A = 1 and the period P = 2π = 6.28… As seen on the unit circle to the right, the amplitude of the function 𝑦 = sin 𝑥 equals to one, because −1 ≤ sin 𝑥 ≤ 1 and the period of the function 𝑦 = sin 𝑥 equals 2 𝜋 because 2 𝜋 constitutes one full turn. Hence, the graph repeats itself every 2 𝜋 units. This is why the graphs of trigonometric functions are usually given over one period, assuming that one can extend the picture as far to the right or to the left as necessary. Having these properties in mind about the sine and cosine functions and the graphing techniques we use to perform transformations are two approaches that can be used to graph an arbitrary sine or cosine function. These two approaches are outlined in the example below: 11 Example 12 Question: Graph 𝑦 = 1.5sin 2𝑥 Approach 1: Use transformation of simple functions. First graph 𝑦 = sin 𝑥, then shrink it 3 times along the x‐axis and stretch 1.5 times along the y‐axis Approach 2: Find the amplitude and period using the formulas below: 𝐹𝑜𝑟 𝑦 = Asin 𝜔𝑥 2𝜋
𝜔
!!
For example, to graph 𝑦 = 1.5sin 2𝑥 find the amplitude A =1.5, period P = = 𝜋. Therefore you !
should fit a sine wave in the box with the dimensions shown below. 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 = A , 𝑝𝑒𝑟𝑖𝑜𝑑 𝑃 = 12