All of TRIGONOMETRY in 36 minutes! (top 10 must knows)

Summary

Highlights

• Delving into the properties of similar triangles and how they simplify solving triangle problems. 📐
• Exploring the foundational acronym SOHCAHTOA for right triangle calculations. 🧠
• Unpacking the sine and cosine laws for tackling non-right angle triangles. 📏
• Introduction to the unit circle for understanding angles beyond standard measures. 🔄
• Seeing sine and cosine as wave functions in their respective graphs. 🌊
• Understanding radians as a measure and their application in circle-related problems. 🔢
• Applying special triangle knowledge to find precise trigonometric values. 🔍
• Exploring the core trig identities that aid in simplifying complex expressions. 🔑
• Utilizing the CAST rule to identify positivity of trigonometric functions in different quadrants. 🎯
• Illustrating methods for solving various trigonometric equations effectively. 🧩

Key Takeaways

• 1. Similar Triangles: Understand how similar triangles maintain angle and side ratios. 📐
• 2. SOHCAHTOA: Essential for mastering basic trigonometry with right triangles. 🧠
• 3. Sine and Cosine Laws: Crucial for solving non-right triangles. 📏
• 4. Unit Circle: Broadens trig understanding beyond 0-90°. 🔄
• 5. Trig Graphs: Discover sine and cosine as wave-like functions. 🌊
• 6. Radians: Alternative angle measure to degrees, key for advanced trig. 🔢
• 7. Special Triangles: Use them to find exact trig values easily. 🔎
• 8. Trig Identities: Learn foundational identities to simplify expressions. 🔑
• 9. CAST Rule: Quickly determine which trig functions are positive. 🎯
• 10. Solving Equations: Techniques to find solutions across cycles. 🧩

Overview

Chapters

• 00:00 - 03:50: Similar Triangles The chapter discusses the concept of similar triangles. Similar triangles are defined as triangles that have the exact same shape but may differ in size. The idea is illustrated by drawing two triangles that start off being identical in size. If one triangle is resized while maintaining its angle measurements, it becomes a similar triangle to the original.
• 03:50 - 07:00: SOHCAHTOA In this chapter titled 'SOHCAHTOA,' the concept of similar triangles is discussed, focusing on their properties. Similar triangles maintain the equality of corresponding angles, meaning angles A and D, B and E, and C and F are all equal. Additionally, the chapter emphasizes that the ratios of corresponding sides in similar triangles are also equal, as illustrated by comparing side AB to side DE.
• 07:00 - 10:00: Sine Law and Cosine Law The chapter focuses on the Sine Law and Cosine Law, specifically on solving problems involving triangles by using properties of similar triangles. It highlights the method to solve for missing sides and angles. The example presented refers to using proportionality in similar triangles to find the length of a side labeled 'X.' It begins with a necessary step of proving the similarity between the two triangles—the larger outer triangle and the smaller triangle contained within it.
• 10:00 - 13:30: Special Triangles This chapter on 'Special Triangles' discusses the three principal ways to prove the similarity of triangles: Angle-Angle (AA) Similarity, Side-Side-Side (SSS) Similarity, and Side-Angle-Side (SAS) Similarity. It emphasizes the Angle-Angle similarity where having two pairs of equal angles is sufficient to establish triangle similarity. The transcript also mentions the use of angle theorems to prove similarity by identifying parallel lines and specific patterns formed by the intersecting lines.
• 13:30 - 19:00: CAST Rule and the Unit Circle The chapter delves into the CAST rule and its application in exploring the relationship between angles and their corresponding sides within triangles, specifically in the context of the unit circle. It explains that when three pairs of corresponding angles are equal in two triangles, the triangles are similar, meaning their corresponding sides have proportional lengths. The explanation is grounded in the geometric principle that similarity of triangles arises from angle congruence, using triangles DEA and BCA as examples to illustrate this concept.
• 19:00 - 23:00: Exact Values and Reference Angles In the chapter titled 'Exact Values and Reference Angles,' the focus is on understanding the relationships between different line segments in geometric figures and solving equations using algebra. The content includes a problem-solving example where two line segments, BC and De, are compared with another pair of segments, B and Da. The speaker demonstrates the process of solving for an unknown variable (X) by setting up a proportion based on the lengths of these segments, using algebraic techniques like cross-multiplication to solve the equation. The equation given is 10 * 6 = 60 and 4 * X + 6 = 4X + 24, which simplifies to 36 = 4X, and dividing by 4 gives X = 9.
• 23:00 - 26:00: Sine and Cosine as Functions The chapter discusses the use of properties of similar triangles to solve unknowns in trigonometry. It introduces the acronym 'SOAA' as a mnemonic to remember the three primary trigonometric ratios. The explanation involves drawing a right triangle, marking a 90° angle, and labeling one of the acute angles as Theta, designated as the reference angle.
• 26:00 - 29:00: Radians The chapter "Radians" discusses the concept of similar triangles, emphasizing that if the angle Θ remains the same, triangles are similar regardless of their size. It explains that in similar triangles, the ratios of corresponding sides remain constant. This is illustrated using a right triangle where the hypotenuse is labeled as the longest side.
• 29:00 - 32:00: Trigonometric Identities The chapter begins with establishing the basic elements of trigonometric functions, focusing on a reference angle Theta. It defines the sides of a right triangle in relation to this angle: the 'opposite' side (labeled 'o'), the 'adjacent' side (labeled 'a'), and the hypotenuse. The text explains that these three sides can form three different pairings, which lead to the three primary trigonometric ratios. Specifically, the ratio comparing the side opposite to Theta with the hypotenuse is identified as the sine ratio. The chapter sets the framework for exploring these ratios further within the context of the mnemonic "SOHCAHTOA" used to remember them.
• 32:00 - 36:00: Solving Trigonometric Equations The chapter focuses on solving trigonometric equations, specifically using reference angles and basic trigonometric functions like sine (sin), cosine (cos), and tangent (tan). The transcript mentions understanding these ratios: sin as the ratio of the opposite side to the hypotenuse, and tan as the ratio of the opposite side to the adjacent side. It emphasizes using these primary trigonometric ratios to find missing sides or angles in right triangles, demonstrated with examples. The transcript particularly points out solving for a missing side in a right triangle with a reference angle of 40 degrees.

All of TRIGONOMETRY in 36 minutes! (top 10 must knows) Transcription

• 00:00 - 00:30 let me show you the top 10 most important things for you to know about trigonometry must know number one similar triangles similar triangles are triangles that have the exact same shape but might have different sizes let me draw two triangles so you can see what I mean now I've started by drawing these two triangles the exact same size and shape so they're congruent if I were to take triangle DF and make it smaller without changing any of its angle so
• 00:30 - 01:00 keep the shape the exact same it is still similar to triangle ABC and similar triangles have two very important properties their pairs of corresponding angles are equal so in these two triangles you can see that angle A and D are equal to each other B and E are equal and C and F are equal the other important property is that ratios of corresponding sides are also equal so the ratio of side AB to side de
• 01:00 - 01:30 would be equal to the ratio of side AC to DF which would be equal to the ratio of side BC to EF and we can use these properties to solve for missing sides and angles in triangles that are similar to each other let me show you an example in this question we're asked to solve for the length of X in order to do that I'll have to use these two similar triangles that I see the big triangle and the smaller triangle within it we'll have to start by proving they're similar
• 01:30 - 02:00 which there are three ways to prove similarity you can do angle angle similarity just find two pairs of equal angles side side side similarity where you show that the ratios of all three pairs of sides are equal or side angle side similarity where you find two equal ratios of pairs of sides and one matching pair of angles for this question we can do angle angle similarity if you know your angle theorems we have two parallel lines here connected by this side I know that the angle is inside of that f pattern called
• 02:00 - 02:30 corresponding angles are equal to each other so I could say that angle D is equal to angle B same thing on the other side of the triangle I know that these angles in here are equal to each other so angle E equals angle C and then of course the two triangles both share angle a So based on all three pairs of angles being equal I know the triangles are the same shape so I could say that triangle DEA is similar to Triangle BCA and because I know that they're similar I know that the ratios of course corresonding sides have to be equal to
• 02:30 - 03:00 each other so I know that the length of BC / the length of De is going to be equal to the length of B / the length of Da and then if I sub into this equation what I know I can then solve this equation using Algebra I can cross multiply 10 * 6 is 60 and 4 * x + 6 is 4x + 24 solving this for X I'd have 36 = 4X divide the 4 I have x = 9
• 03:00 - 03:30 there you go now you can see how to use properties of similar triangles to solve for unknowns must know number two soaa this is an acronym that helps you remember the three primary trig ratios let me draw a right triangle in this right triangle I've indicated the 90° angle and then I've labeled one of the acute angles angle Theta we'll call that the reference angle all other right triangles with this same reference angle
• 03:30 - 04:00 Theta would be similar to each other right no matter how much I change the size of this triangle as long as I keep that angle Theta the same it has the exact same shape meaning the triangles are similar which means based on what we know about similar triangles they would have the same ratios of corresponding sides in any right triangle there are three sides let's name them the longest side of a right triangle is always called the hypotenuse I'll label that H the side across from our reference angle
• 04:00 - 04:30 Theta we'll call that the opposite side I'll call it o and the side touching our reference angle Theta we'll call that the adjacent side because it's right beside our reference angle I'll label that a because there are three sides there are three different pairs we could make of these three sides that's why you see three ratios in soaa if we find the ratio of the side opposite from the reference angle divid by the hypotenuse we call that the sign ratio cosine of
• 04:30 - 05:00 the reference angle is the adjacent side divid by the hypotenuse and tan is a function that means we're looking at the ratio of the opposite side divided by the adjacent side now these primary ratios can be used to find a missing side or a missing angle of a right triangle let me show you both examples let's look at this first example where we're going to solve for a missing side x I notice it's a right triangle with a reference angle of 40 all right triangles with a reference angle of 40
• 05:00 - 05:30 are similar to each other they all have the same ratios of sides so whatever x / 8 is will be the same ratio of all other right triangles with a reference angle of 40 so let's communicate that from 40 the side we're looking for is adjacent to it and 8 is the longest side of the right triangle so that's the hypotenuse the function that involves adjacent and hypotenuse is cosine so I know coine of 40° equals the adjacent side x / by the hypotenuse eight your calculator has
• 05:30 - 06:00 stored within it what the ratio of adjacent over hypotenuse is for all right triangles with a reference angle of 40 if you typed cosine of 40° on your calculator it'll give you that ratio and then we can just multiply the 8 to the other side of that ratio to isolate X SOA can also be used to find a missing angle inside of a right triangle if we're looking for this angle here notice that 8 is opposite from it and 17 is across from the right angle so that's the hypotenuse the ratio that involves
• 06:00 - 06:30 opposite and hypotenuse is the sign ratio so I could say s of the angle equals the opposite side 8 ID the hypotenuse 17 now all right triangles that have an opposite over hypotenuse ratio of 8 17 are similar to each other which means they've all got the same angles so we can solve for angle Theta your calculator will be able to tell you if you use the inverse sign function which is sometimes called ark sign of 8 17 by using the inverse sign function
• 06:30 - 07:00 you're telling your calculator that you are inputting the ratio and looking for the angle and your calculator would tell you that the angle is about 28.07 de so that's how you use soaa to find a missing side or a missing angle must know number three sign law and cosine law now not all triangles you work with are going to have a 90° angle if they don't the the only way you'll be
• 07:00 - 07:30 able to solve for a missing side or angle is by using S law or cosine law cosine law has two versions one version if we're solving for a missing side and one version that is just a rearranged version of it where we're solving for a missing angle and then there's sign law which just tells you the ratio of any side divided by sign of its opposite angle is equal cosine law has two scenarios where it's used and S law has two scenarios where it's used so I'll show you each scenario of both of these the scenario where we use cosine law to find a side length is when you know two
• 07:30 - 08:00 sides and the angle contained by those two sides in a triangle so notice in this triangle we know two sides and the angle contained by those two sides so we could solve for the opposite angle using cosine law if I follow the pattern of cosine law it tells me that the unknown side squared equals the sum of the squares of the other two sides minus 2 * the product of the known sides time cosine of the angle contained by the two sides that will tell you what x^2 equals to get X make sure you square root
• 08:00 - 08:30 whatever this value is and in this question it's about 1996 let me shrink this and let's do another question this rearranged version of cosine law is used to solve for an angle if you know all three sides of an oblique triangle if I'm looking for this angle Theta when I use this cosine law the important thing to notice is that the side that starts in the numerator is the one opposite from the angle you're looking for so if I'm finding cosine of theta the side opposite from Theta is 13
• 08:30 - 09:00 so I must start with 13 s and then subtract the squares of the other two sides and then divide by -2 times the second two sides 14 and 16 this will give us what cosine Theta is equal to and when we know the ratio and want the angle we would do inverse cosine of that ratio and it'll give us the angle and we would get an angle of about .82 de sign lock can be used when you know two sides and an angle opposite from one of the two sides based on S law I know that the
• 09:00 - 09:30 ratio of 64 / s of its opposite angle would be equal to the ratio of 42 / by S of its opposite angle and then algebraically we could rearrange this to isolate sin Theta and then do inverse sign of that ratio and you would get the angle which is about 6093 de the other scenario where you would use sign law is if you know two of the angles and one of the sides in this triangle I know that x / s of its opposite angle 47 would be equal to 5 /
• 09:30 - 10:00 s of its opposite angle 51 if you rearrange to isolate X you would find out that the value of x is about 4.71 units and there you go now you know how to use S law and cosine law to solve for missing sides and angles of oblique triangles must know number four the special triangles there are two special triangles that are going to allow you to find exact values of s Co and tan for a
• 10:00 - 10:30 few special angles let me construct those triangles for you the first one is an isoceles right triangle if this triangle is isoceles I know that the two legs of the right triangle are equal in length I could assign them both a length of one and I know these two angles would have to be equal to each other and since they both add to 90 I know they're both 45° if I use Pythagorean theorem to solve for the length of the hypotenuse I would get the square root of two and
• 10:30 - 11:00 then I could use these Dimensions to figure out the S Co and tan ratios for 45 now not all right triangles with a reference angle of 45 are going to have lengths of 1 1 and < tk2 but their ratios of lengths would all be the same because all the triangles no matter how big or small would have the same shape meaning the same ratios let's assume that this 45 is our reference angle that would make this opposite this adjacent this hypotenuse let me write the three primary ratios for 45° s of 45° is
• 11:00 - 11:30 opposite over hypotenuse so 1/ < tk2 cosine of 45 is adjacent over hypotenuse which is also 1 over < tk2 and tan of 45 is opposite over adjacent which is 1 over 1 which is just one now the sign and cosine ratios have a square root in their denominators you should rationalize those denominators and you can do that by multiplying top and bottom of both of those fractions by < tk2 so those are the exact values for
• 11:30 - 12:00 sin Co tan of 45° the other special triangle is a half equilateral triangle if this equilateral triangle started off with all side lengths being two and of course all the angles being equal so they would all be 60 if I cut this equilateral triangle in half that 60° angle that was at the top of this triangle gets bisected into 2 30° angles and the side at the bottom also gets bisected so this length is one now I don't actually need the left half of
• 12:00 - 12:30 this triangle so I'm going to erase it and then I could use Pythagorean theorem to solve for that side and I would find out its length is equal to the < TK of 3 and now I notice there are two angles in here that I could find the exact values of their s Co and tan ratios if I start by finding the ratios for the 30° reference angle so s of 30 is opposite over hypotenuse which is a half COS of 30 is adjacent over hypotenuse which is < tk32 and tan of 30 is opposite over adjacent which is 1 < tk3 but if we
• 12:30 - 13:00 rationalize that by multiplying top and bottom byun3 you get < tk3 over3 now if we make 60 the reference angle s of 60 would be opposite over hypotenuse So < tk3 over2 Co of 60 is adjacent over hypotenuse so 1 over 2 and tan of 60 is opposite over adjacent which is < tk3 over 1 which is just Ro < tk3 so there you go if you ever want a primary ratio for either 30 45 or 60 you can use your special triangles to find the exact values of those
• 13:00 - 13:30 ratios must know number five cast Rule and the unit circle these tools are going to broaden your understanding of trigonometry and allow you to start using it for angles bigger than angles you would find inside of a triangle let's start with the unit circle I've started by drawing a cartisian grid and on this cartisian grid I'm going to label the positive xais 0 de and then moving in a counterclockwise Direction
• 13:30 - 14:00 I'm going to label 90 180 270 and then co-terminal with 0° would be 360° the reason why I label them like that is because when we think of angles now we're not going to think of them as angles in a triangle but we're going to think of an initial arm which starts on the positive xais and then rotate some angle counterclockwise in a circle away from that positive xais so if you rotated 90°
• 14:00 - 14:30 you would be up here 180 here 270 here 360 back to where we started which is why we say it's co-terminal with 0 degrees now wherever that initial arm finishes we call that its terminal arm and if we know where that terminal arm intersects a circle that has a radius of one that actually tells us a lot about its s Co and tan ratios Let Me Show You by drawing a circle that has a rad of one and I should also mention that this
• 14:30 - 15:00 unit circle whose radius is one it's centered at the origin now this yellow terminal arm has rotated some angle Theta so I'm going to label that angle Theta and then I'm going to construct a right angle triangle inside of this circle this point that it intersects the unit circle at is some XY point if this is point XY then I know to get to that point I would go X units that way and then y units that way so I could label
• 15:00 - 15:30 this triangle side lengths of X and Y and because this is a unit circle I know the radius of that circle is one now if you remember soaa you would be able to write the S cosine and tan ratios for angle Theta based on the side lengths of this triangle that we see inside of the unit circle s of theta equals the opposite side y / the hypotenuse 1 and y/ 1 is just y cosine of theta equals
• 15:30 - 16:00 the adjacent side / the hypotenuse so x / 1 which is just X and tan of theta is opposite over adjacent so y / X so the important thing you need to notice is the S ratio of angle Theta is equal to the y-coordinate of where you intersect the unit circle and cosine of theta is equal to the x coordinate of where the terminal arm intersects the unit circle so if we know the point where the
• 16:00 - 16:30 terminal arm intersects the unit circle then we in fact know the cosine and S ratios for angle Theta so I could say that point XY is equal to cos Theta comma sin Theta and what points on this unit circle would we know the coordinates of well because the radius is one I know moving one unit in all those four directions I could label four points on this unit circle this point would be the point 1 0 up here is the 01 here's -1 0 and here's
• 16:30 - 17:00 01 so if I asked you to find s of 90° you wouldn't have to ask a calculator if you rotated 90° your terminal arm would intersect the unit circle at that point right there and the sign ratio is equal to the y-coordinate of where you intersect the unit circle so s of 90 is equal to one let me test you with one more what if I asked you to find cosine of 27 ° if you rotated
• 17:00 - 17:30 270° your terminal arm would intersect the unit circle at this point right here and the cosine ratio is equal to the x coordinate so cosine of 270 is equal to 0 Let's look at another tool which is very directly related to this unit circle this tool is called the cast Rule and it's just an acronym that helps you remember what ratios are positive in each quadrant and for the cast rule you write c a s t in each of those
• 17:30 - 18:00 quadrants I've also heard people say all students take calculus to help you remember that acronym but what these letters tell you is in each of these quadrants which ratios are positive a stands for all and then of course s is for S T is for tan and C is for cosine and then why does this make sense well if we think back to the unit circle I know that the y-coordinate of where you intersect the unit circle is the sign
• 18:00 - 18:30 ratio where are y-coordinates positive above the xaxis in quadrant 1 and 2 that's why the cast rule tells us in Quadrant One S is positive and in quadrant 2 s is also positive and where Y coordinates negative that's below the x-axis in quadrant 3 and four that's why in those quadrants sign would be negative only tan is positive in quadrant 3 and only cosine is positive in Quadrant 4 and cosine is equal to the x coordinate of where you intersect the unit circle and where x coordinates
• 18:30 - 19:00 positive in quadrant 1 and four and castal tells us in quadrant 1 Co is positive in Quadrant 4 Co is positive but in quadrant 2 and three only s and tan are positive meaning cosine would be negative in both those quadrants hopefully you understand how the unit circle and cast rule can be useful and in the next couple sections I'll show you how we can actually use them must know number six you you need to be able to find the exact value of
• 19:00 - 19:30 ratios for angles that are bigger than 90° to do that you're going to need to remember your special triangles and the unit circle let me draw those quickly as a reminder let's say you were asked to find the exact value of the sign ratio of 150° the first thing you want to do is figure out where that principal angle of 150° falls on a cartisian grid a principal angle of 150° means from the positive x-axis rotate counterclockwise 150° and that would land you about right here
• 19:30 - 20:00 the angle from positive X counterclockwise to that terminal arm is 150 now each principal angle has a reference angle the reference angle is the angle between the terminal arm and the closest xais the reference angle is the angle between the terminal arm and the closest X AIS the angle between 150 and 180 would be 30° so I could say that the reference angle is 30° this reference angle and the cast rule is going to help me find the sign ratio for
• 20:00 - 20:30 150 and as a reminder the cast rule tells me that in quadrant number two where the terminal arm is the sign ratio is positive for that reason I know that s of 150 is going to be equal to the positive sign ratio of the reference angle so it'll be equal to S of 30° and from my special triangle I know that s of 30 well s means opposite over hypotenuse so s of 30 is 1/ 2 so my
• 20:30 - 21:00 final answer for S of 150 is that it's equal to a half let me take a second and explain to you why s of 150 is equal to S of 30 if I were to draw a principal angle of 30° that would mean rotate 30° counterclockwise from the positive X and that will land you in quadrant number one the reference angle is 30° and if we look at where both of those terminal arms intersect the unit circle if we look at the points where the those terminal arms intersect the unit circle
• 21:00 - 21:30 notice that they have the exact same y-coordinate and if you remember if we look over here the y-coordinate of where we intersect the unit circle is equal to the sign ratio of the angle so of course s of 150 and S of 30 have the same value because they intersect the unit circle at a at two points that have the same yv value and the y- value of those points would of course be a half let me do one more example quickly for you I'll shrink this to make room let's find cosine of
• 21:30 - 22:00 225° I'll start by drawing my cartisian git and figuring out where 225° Falls if I were to rotate 225 I would go past 180 into quadrant number three so the principal angle is 225 and the reference angle which is the angle between the terminal arm and the closest x axis so the angle between 225 and 180 is 45° and don't forget based on cast rule in this quadrant only tan ratios are posi POS which means any ratio of an
• 22:00 - 22:30 angle in that quadrant will have to be negative so I can change the principal angle of 225 to the reference angle of 45 as long as I make the cosine ratio negative so I can say this equals the negative cosine of the reference angle 45° and cosine of 45 from your special triangle you would notice cosine is adjacent over hypotenuse so I would have -1 / < tk2 which if I rationalize that would be Nega < tk2 / 2 that would be
• 22:30 - 23:00 the exact value of cosine of 225 there you go now you know how to use your special triangles the unit circle and the cast rule to find exact values for trig ratios must know number seven s and cosine as functions let's start with s if we were to think of s as a function so yal sinx shows the relationship between X which is our angle and y which
• 23:00 - 23:30 is equal to the whole sign ratio of the angle we could create a bunch of XY Pairs and graph and see what the function looks like so that's what I'll do I'll make a table of values where my independent variable X are going to be angles and the Y values are the sign ratios of those angles the angles that I'm going to pick are at 90° intervals starting at zero so 0 90 180 270 360 in
• 23:30 - 24:00 order to figure out the sign ratios for each of these angles it'll be helpful if you remember the unit circle based on the unit circle hopefully you remember that the point where we intersect the unit circle the x coordinate tells us the coine ratio for the angle and the y-coordinate tells us the sign ratio for the angle so if I want s of zero I look if the terminal arm intersected the unit circle at 0° the y-coordinate would be 0 so s of 0 is 0 and then just continue looking at the y-coordinates as you rotate around the unit circle and you
• 24:00 - 24:30 would figure out the sign ratios go 0 1 010 and if this terminal arm kept rotating around we're going to have that same pattern of yv values that just keeps repeating over and over again that's why this is called a periodic function it has a pattern of yv values which we see here that just repeat over and over again let me graph that for you if I plot the Five Points that I have in my table I get one cycle of this sign function and it looks like a wave that
• 24:30 - 25:00 oscillates up and down between y values of 1 and 1 if I were to continue this pattern I would get another cycle of this function the domain of this function is infinite this pattern just keeps continuing forever in both directions let me now show you what a cosine function looks like so I'm just going to shrink this so I have room let's graph the function y equals cosine X where the x value is the angle and the Y value is the cosine ratio of that angle I'll choose the same X values and
• 25:00 - 25:30 to get the Y values I can just look at the x coordinates as I rotate around the unit circle since that's what the cosine ratio of the angle is equal to so as I rotate around my X values are 1 0 1 0 and one and that pattern of Y values would continue this will just give me one cycle of the graph and let me show you that if I wanted another cycle I would just continue the pattern let's talk about a couple Key properties of these graphs the two things we need to to look at are things called the amplitude and period of these functions
• 25:30 - 26:00 the definition of amplitude is that it's half the distance between the Max and Min y values of a periodic function well for both s and Co the maximum yvalue is 1 and the minimum yvalue is NE 1 so the full distance between Max and Min is 2 half of that is one therefore the amplitude is 1 the period of a periodic function is the horizontal length of one cycle if I map off one cycle of sign and one cycle of cosine notice the
• 26:00 - 26:30 horizontal length of both of those Cycles is 360° there you go hopefully you now understand how we could think of s and cosine as functions and you have a general idea of what the graphs look like must know number eight radians up until now all the angles we've been working with were measured in degrees but there's another unit of measurement we could think about for angles and that is radians let me show you what a radi is if I draw a circle it'll help you
• 26:30 - 27:00 visualize it let me draw an angle that subtends an AR length of this circle in this circle that I've drawn I have the radius here and this right here is what we call the Arc of the circle if we were to measure this angle in degrees we would just figure out how many 360th of this circle this section occupies but if we're going to measure in radians we're going to figure out how many radiuses have we moved around the Arc of this circle and how we figure out
• 27:00 - 27:30 how many radiuses we've moved around the arc is we use the formula The Arc Length divid by the radius so whatever the ratio of this Arc Length divided by this radius is that's the measure of that angle in radians so what does one radian look like well if this AR length was exactly equal to the length of this radius that would be one radian right in the angle calculation when we do Arc
• 27:30 - 28:00 over radius if the arc and radius are equal really what I'm doing is the radius divid the radius which is of course one so how many radians are in a full circle well do you remember the relationship between the circumference of a circle and its radius the relationship is that the circumference equals 2 pi * the radius so that tells us that the radius fits around the circumference of the circle 2 pi times therefore there are 2 pi radians in a circle so in a circle we could say
• 28:00 - 28:30 there's 360° or 2 pi radians those are equivalent to each other so let's pick a common degree to convert to radians how about 30° if I want to convert it to radians well how many radians are in each degree in this equation I made here let's change this 360° to 1 Dee by dividing both sides by 360 if I do that you'll see that 1° equal < 180 Radian to change 30° to radians I would have to
• 28:30 - 29:00 multip byk / 180 30 / 180 reduces to 1 6 so this would equal pi/ 6 radians so all of the trigonometry that we did up till now in degrees we could easily convert all of it to radians by just multiplying any degree measure by pi/ 180 must know number nine trig identities first of all what is a trig identity a trig identity is an equation
• 29:00 - 29:30 that is true for all values of the variable and there are three fundamental identities that you should be familiar with the reciprocal identities the quotient identities and of course the Pythagorean identity the reciprocal identities are that cosecant of any angle is equal to the reciprocal of so one over s of that same angle secant is the reciprocal of cosine and cotangent is the reciprocal of tangent the
• 29:30 - 30:00 quotient identities tell us that Tan x is equal to the quotient of sinx and cos x and since coent is the reciprocal of tangent that must mean that coent is the reciprocal of sin over Co which is cos over s and the Pythagorean identity is that sin x plus cos x equal 1 and there are common ways to rearrange that now any of these equations are true no matter what value of x you Sub in so Sub in anything you want for these X's into
• 30:00 - 30:30 sin^2 X Plus cos s x and it's always going to equal 1 now what we can do with these identities is prove other identities so let me make some room and then we'll prove an identity let's see if we can prove that tan^ 2 x - sin^2 x is equal to sin^2 x * tan^ 2 x if this is an identity it'll be true for all values of the variable so how we prove that is separating into left side and right side of the equation and work with both of these sides completely
• 30:30 - 31:00 separately from each other so I'll draw a line down the middle and start rewriting these expressions in equivalent ways to reveal that they are in fact the exact same expression so I'm not allowed to change the value of either of them what I could do on the left I know that based on the quotient identity tan of any angle equals the quotient of s over Co so I can change tan^ 2 x to sin^2 X over cos 2 x and then I'm subtracting sin s x but remember that sin^2 X is technically
• 31:00 - 31:30 over one I can multiply top and bottom of that by co^ 2 x to get a common denominator I haven't changed its value I've just multiplied it by one so now I have a common denominator of cos^2 X with both of those terms so I can combine those into a single fraction in the numerator I have two terms and both of those terms have a s^ squ x so I could common and factor that out and now I have a quotient of sin^2 x and cos s x based on the quotient I identity I know that that's equal to tan^ 2 x and that's
• 31:30 - 32:00 being multiplied by a 1 - cos^2 x which based on the Pythagorean identity I know is equal to sin^2 X and now notice that's the exact same expression as I have on the right side of the equation so the two sides of this equation are actually the exact same expression they're exactly equivalent to each other for all values of X so we've proven the identity we can say left side equals right side or QED d must know number 10 solving trig
• 32:00 - 32:30 equations I'm going to give you two trig equations to solve the first one is sinx = -1 < tk2 and for this one I only want the solutions between 0 and 2 pi which means only the Solutions in the first cycle of this function since s is a periodic function there's going to be an infinite number of times it's equal to1 < tk2 so by restricting the domain I'm only asking you for two of the answers in this case so let's go ahead and find
• 32:30 - 33:00 those two answers you're going to have to remember your special triangles so let me draw those off to the side when solving a trig equation I like to draw a cartisian grid and based on cast rule I like to figure out where my answers are going to be since the sign ratio is negative I know s ratios are negative in two quadrants in quadrant 3 and in Quadrant 4 so I'll will place a terminal arm in both of those quadrants in order for the angles that get to these terminal arms to have the the same sign ratio I know that those terminal arms
• 33:00 - 33:30 must have the same reference angle and what is the reference angle that creates a ratio of one/ < tk2 well s of pi/ 4 equals opposite over hypotenuse 1/ < tk2 so I know the reference angle for both of these terminal arms is pi over 4 and now I just have to calculate the two angles to get to those terminal arms so my first angle I'll call this X1 I'm looking for the angle that goes from positive X counterclockwise all the way to that terminal arm well notice it went past Pi by < over 4 so I could say that
• 33:30 - 34:00 X1 is equal to < + < over 4 which is 5 pi over 4 and to get to X2 I would have to go not a full 2 pi I would have to go pi over 4 less than that so I can do 2 pi minus piun over 4 to get my second principal angle and 2 piun - piun over 4 is 7 piun over 4 so there are two answers to this equation between 0 and 2 pi the answers are 5 < over 4 and 7 piun over 4 let's do a second equation where I don't restrict the domain let's say we
• 34:00 - 34:30 have 2 sin^2 x - 3 sin x + 1 = 0 I would start by factoring this quadratic expression by finding numbers that have a product of 2 * 1 so a product of two and a sum of -3 the numbers that satisfy that product and sum are -2 And1 there's lots of factoring methods you could use I'll just split the middle term of this quadratic into -2 sinx - sin x and now I would Factor by grouping I would take a 2 sin x from the first two terms and take a Nega 1 from the last two terms I
• 34:30 - 35:00 can take out this common binomial of sin x -1 and after I take that out I'm left with 2 sinx - 1 as my second Factor now this product would be zero if either of these factors were zero the first Factor would be Zer if sin x was equal to 1 and the second Factor would be zero if sin x was equal to a half now sinx equaling one if you know your unit circle that answer will be easy if you rotate an angle of Pi / 2 it intersects the unit circle at a y-coordinate of one and I
• 35:00 - 35:30 know the sign ratio is equal to the y-coordinate so the angle that would make the sign ratio be one is Pi / 2 so the first answer this to this equation is Pi / 2 now because the domain of this equation has not been restricted I know that I could continue rotating around this circle and every two Pi radians I rotate around I'm going to be back intersecting at that same point so how I could communicate that is I could say that I can add 2 pi K times where K has
• 35:30 - 36:00 to be an integer so there's a set of answers that come from the first Factor being zero now how can this second Factor be zero let me draw a quick cartisian grid I have a positive sign ratio while sign ratios are positive in quadrant one and two so there will be a terminal arm in both of those quadrants both of those terminal arms have the same reference angle and their reference angle is going to be Pi / 6 and I know that because s of Pi / 6 is 1 / 2 opposite over hypotenuse so I'll place the reference angle of pi/ 6 in both of those quadrants and now I have to
• 36:00 - 36:30 calculate the principal angle to get to each of those terminal arms in Quadrant One the principal angle is just equal to the reference angle so it's pi/ 6 and my second terminal arm which is my third answer for for X overall to get to there I would go Pi 6 less than Pi piun minus < 6 is 5 piun 6 and from either of those terminal arms you could rotate 2 pi radians and be back to the same spot so to communicate that I would say to either of these answers I could add 2 pi K where K has to be an integer so these
• 36:30 - 37:00 three sets of X values give me all the solutions to this original equation hopefully this video helped you gain more confidence in your understanding of trigonometry let me know about what you want to top 10 of next