so this video is for those of you who are about to take trig either in high school or in college and so one of the concepts that i'm going to cover is an expression called sohcahtoa and this expression it helps us to remember the formulas dealing with the three most common trig ratios sine cosine and tangent so the first one so tells you that sine of some angle theta is equal to the opposite side divided by the hypotenuse cosine is equal to the adjacent side divided by the hypotenuse and for the last part toa it tells us that the tangent ratio is equal to the opposite side divided by the adjacent side so let's say if we have a right triangle and so this is a 90 degree angle and we're going to say this is the angle theta so opposite to theta would represent this side on the right adjacent to it is the side below it or next to the angle and across the 90 degree angle is the longest side of the triangle and so that is the hypotenuse of the triangle so let's say we have a right triangle that looks like this and let's say it's a 3 4 5 right triangle and here's the angle theta so what is sine theta based on that triangle before we get the answer let's identify which side is opposite adjacent and the hypotenuse with respect to the angle so 3 is adjacent to the angle 4 is opposite to it and 5 is the hypotenuse so we know that sine theta based on the so part of socatopa is equal to the opposite side divided by the hypotenuse of the right triangle so opposite to theta is four and the hypotenuse is five so sine theta is going to be four over five now let's apply the same thing to find the cosine ratio cosine is adjacent over hypotenuse so the adjacent side has a measure of three and the hypotenuse is five so cosine theta is three over five now to find the tangent ratio it's equal to the length of the opposite side divided by the adjacent side so in this case it's four over three now you can also find tangent by taking the sine ratio and divided by the cosine ratio so if you take four over five and divide it by three over five the fives will cancel giving you four over three now let's try another example so here's another right triangle and let's say we're given two of the three sides of this right triangle we're going to say the angle theta is right here go ahead and find the value of sine theta cosine theta and tangent theta based on this example so feel free to take a minute and pause the video now the first thing we need to do is find the missing side so whenever you have a right triangle you could use this formula this is going to be a b and the hypotenuse is c so we could use the pythagorean theorem a squared plus b squared is equal to c squared in this case we're looking for the hypotenuse which is c so we can write c squared is equal to let's say that a is 5 and b is 12. so it's going to be 5 squared plus 12 squared now 5 squared or 5 times 5 that's 25 12 times 12 is 144 and 25 plus 144 is 169. now we need to take the square root of both sides so the square root of 169 is 13. so the hypotenuse of the right triangle is 13. so now that we have that let's determine the sine and the cosine ratios sine theta is going to be equal to the opposite value which is going to be 5 in this case so opposite to theta is 5 adjacent to it is 12 and the hypotenuse is always the longest side so sine theta is going to be the opposite side divided by the hypotenuse so that's 5 over 13. now cosine theta is equal to the adjacent side divided by the length of the hypotenuse so it's 12 over 13. and tangent based on the total part of socatoa it's going to equal the opposite side divided by the adjacent side so in this case it's 5 over 12. and so that's a quick and simple way to evaluate the trig ratios given a right triangle now i want to mention that there are certain right triangles special right triangles that you should commit to memory the first one is the three four five right triangle which you've seen already the second one you've also seen the 5 12 13 triangle there's also the 8 15 17 triangle and then you have the 7 24 25 right triangle i've also seen this one the 9 40 41 right triangle and also the 11 60 61 right triangle however in the typical trig course these four are the most common that you'll see and so make sure you commit these to memory if you don't want to use the pythagorean theorem to find the missing side so let's work on an example so let's say this side is 14 and this side is 50 and here is the angle theta so go ahead and find the values of sine cosine and tangent now we need to find a missing side and this is a special right triangle so the ones i've given you were these which one of these do you think corresponds to the triangle that we see on the right notice that 50 is twice the value of 25. and 14 is twice the value of seven so any of these numbers can work or any ratios of those numbers can work as well so if we take 7 24 and 25 and multiply each number by 2 this will give us the 14 48 50 right triangle so the missing side is 48. now sine theta is going to be equal to the opposite side which has the length of 14 divided by the length of the hypotenuse which is 50. so if we divide both numbers by 2 this reduces to 7 over 25. now cosine theta is equal to the adjacent side which is 48 divided by the hypotenuse and so that reduces to the special triangle that we had before the 7 24 25 triangle tangent is going to be equal to the opposite side which is 14 divided by the adjacent side that's 48 and so that reduces to 7 over 24. now what about c get cosecant and cotangent cosecant is the reciprocal of sine it's 1 over sine so if sine is equal to 7 over 25 cosecant is going to be 25 over 7. now to calculate the secant ratio this is going to be the reciprocal of cosine so if cosine is 24 over 25 secant is going to be 25 over 24. and finally we have cotangent which is the reciprocal of the tangent ratio so tangent is 7 over 24 which means that cotangent is going to be 24 over 7. now sometimes you might be given a right triangle trigonometry problem that looks something like this so let's say you're told that sine theta is equal to 8 divided by 17 and that theta is somewhere between 0 degrees and 90 degrees what is the value of cosine theta and tangent theta but let's start with cosine how can we find the value of cosine now the first thing you need to do is you need to draw a right triangle but in the appropriate quadrant so let's draw the x-axis and the y-axis that line doesn't look straight so this is quadrant one here we have quadrant two quadrant three and quadrant four so the angle is typically measured from the positive x-axis so this is going to be 0 degrees here we have 90 180 270 and 0 is the same as 360. so if theta is between 0 and 90 that tells us that we have a right triangle in quadrant 1. so let's draw our right triangle in quadrant one now we're going to place our angle theta between the x-axis which is here and the hypotenuse of the right triangle now we know that sine is equal to the opposite side divided by the hypotenuse so the hypotenuse has a length of 17 and opposite to theta is the opposite side which is going to be eight now this is a special right triangle it's the 8 15 17 right triangle so the missing side has to be 15. so now that we have everything filled in we can now calculate the value of cosine theta and tangent theta so cosine is going to be equal to the length of the adjacent side divided by the length of the hypotenuse so it's 15 over 17 and tangent is going to be equal to the opposite side divided by the adjacent side so it's 8 over 15. and so that's it for this example but now what's going to happen if we have another problem where let's say it's not in quadrant one so let's say that sine theta is equal to two over five and the angle is somewhere between pi over two and pi well the first thing we need to know is what is pi and what is pi over two pi and pi over two those are angles measured in radians you need to know that pi is equal to 180 degrees and pi over two is basically 180 divided by two so that's 90 degrees so since the angle is between 90 and 180 based on the quadrants that we drew we know that we're dealing with quadrant two so let's begin by drawing the triangle in quadrant two now let's consider the value of sine theta which is equal to the opposite side divided by the hypotenuse so the hypotenuse has a length of five and opposite to theta is going to be two now what is the missing side here now we don't have a special right triangle so we have no choice but to use the pythagorean theorem c squared is equal to a squared plus b squared now c in this example is five let's say that b is two and so we're looking for a so this should be 2 squared 5 squared is 25 2 squared is 4. 25 minus 4 is 21 and so the missing side is the square root of 21 which we cannot simplify so now what is cosine theta cosine we know it's equal to the adjacent side divided by the hypotenuse so it's going to be the square root of 21 over 5. now you need to be careful the y value is positive in quadrant two because you're going up you're going in the positive y direction relative to the origin the x value is negative it's on the negative x axis so therefore this should be negative square root 21. so in quadrant two cosine is equal to a negative value but sine is positive now let's calculate tangent so tangent theta is going to be equal to the opposite side divided by the adjacent side so it's going to be 2 over negative square root 21. now in a typical trigonometry course you need to rationalize any fractions with a square root in the bottom so to do that you need to multiply the top and the bottom by the square root of 21. and so the final answer is going to be negative 2 square root 21 divided by 21 the square root of 21 times the square root of 21 is going to be just 21. now let's calculate the other ratios starting with cosecant theta and secant so cosecant is the reciprocal of the sine function so all you need to do is flip this fraction so if sine theta is two over five cosecant is five over two now let's move on to secant secant is the reciprocal of cosine so we gotta flip that fraction and so initially it's going to be negative 5 divided by the square root of 21 but we need to rationalize it so let's multiply the top and the bottom by the square root of 21. so it's going to be negative 5 square root 21 over 21. now the last one that we need to find is cotangent theta now for this one we know that cotangent is the reciprocal of tangent however you don't want to flip this answer because it's going to give you a square root on the bottom and then you'll have to rationalize it again rather you want to flip this part of the answer because these two are the same but if you flipped this part the square root will be on top and you do not need to rationalize it again so cotangent theta is equal to negative square root 21 over 2 and that's it so that's how you could find the other three trig functions when you're dealing with triangles that don't when you're dealing with triangles that are not special by triangles i almost lost my train of thought there now sometimes you might be asked to find the exact value of this expression what is the exact value of sine of 30 degrees now there are some calculators that will give you the exact value however you may need to do this without a calculator which is typically what you'll see on an exam so how can we evaluate sine 30 without a calculator one you could memorize the answer two you could use the unit circle by the way for those of you who want more info on the unit circle or more examples on problems like these i recommend that you check out my new trigonometry playlist i'm going to post a link in the description section of this video so you can take a look at that or if you subscribe to my channel you'll get access to new videos that i'm going to upload in the future as well now in that trig playlist you could find answers to almost any topic that you're going to encounter in a typical trade course so i highly recommend you take a look at that which can help you in your class now let's get back to this problem so what can we do to evaluate this trig function at 30 degrees one way is to know two special right triangles i'm going to give you the first one the first one is called the 30 60 90 triangle so this angle is 30 this is 60 and the box is basically a 90 degree right angle across the 30 is 1 across the 90 is 2 and across the 60 is the square root of 3. so if you commit this triangle to memory you can evaluate things like sine 30 cosine 60 tangent 30 and so forth using sohcahtoa so we know that sine is equal to the opposite side that is opposite to 30 not 60 because we're evaluating sine 30 so opposite to 30 is 1 and the hypotenuse is always across the 90 degree angle so the hypotenuse is 2. therefore sine 30 is one-half now let's say if we wish to evaluate cosine of 30 degrees so cosine is equal to the adjacent side divided by the hypotenuse so that's going to be the square root of 3 over 2 and so that's how you could find the exact value of things like sine 30 cosine 60 and so forth now let's say if we wish to evaluate cosine pi over four how can we do so so we're given an angle in radians and we need to convert it to an angle in degrees to convert radians to degrees take the angle that you're given and multiply it by 180 over pi so that's the conversion factor because pi is equal to 180. so you want to set it up in such a way that pi cancels and this tells you that you need to divide so we're going to divide 180 over four now 180 divided by four half of 180 is 90 and half of 90 is 45 so whenever you wish to divide something by four you could just half it two times so this is equivalent to cosine of 45 degrees now there's another special reference triangle that you need to know and it's the 45 45 90 right triangle so across the 45 angle the side is equal to one across the 90 angle is the square root of two so cosine 45 we can pick any of these two angles let's use this one it's equal to the adjacent side divided by the hypotenuse based on sohcahtoa so it's one divided by the square root of two and of course we need to rationalize this answer and so the square root of 2 times the square root of 2 is the square root of 4 which is 2. therefore cosine 45 is equal to the square root of two over two now let's calculate tangent of pi over four or a tangent of 45 degrees now let's use this angle so tangent is going to be equal to the length of the opposite side divided by the adjacent side so one divided by one is one so tangent of pi over four or tangent of 45 degrees is one now what about this problem evaluate sine of 240 degrees go ahead and try that now if you need to evaluate a trig function with an angle that is greater than 90 or not between 0 and 90 you need to find a reference angle one way to do that visually is to draw the angle so positive angles you need to measure them traveling in a counterclockwise direction and for negative angles you need to travel in a clockwise direction or in the direction of a clock so we know this is 0 90 180 270. so 240 is somewhere in this region so this would be an angle of 240 measured from the positive x-axis so what i'm going to do is turn this into a right triangle and so if this angle is 240 and this angle is 180 the difference between 240 and 180 is 60. so this angle is known as the reference angle the angle between the x-axis and your terminal ray which is where your ray should stop that difference is the reference angle now we can complete the triangle we have a 30 60 90 triangle the three angles of a right triangle must always add up to 180 so across the 60 has to be the square root of three the side length across the 30 degree angle has to be one and across the 90 degree angle is 2. now the next thing we need to do is include the appropriate signs so the triangle is in quadrant 3 where it's on the left side or in the negative x-axis so this should be negative one it's also below the x-axis towards the negative y-direction so this should be negative square root three but the hypotenuse is always positive so keep that in mind so now we can evaluate sine of 240 using the triangle that we set up here so it's going to be associated with this reference angle opposite to the reference angle is negative square root 3 and the hypotenuse is 2. so the answer is going to be negative square root 3 divided by 2. and you could check that with a calculator make sure it's in degree mode if you type in sine 240 you should get negative square root 3 over 2 which as a decimal is equal to negative 0.866 and so forth so you should get this answer now let's try another example cosine of let's say five pi over six feel free to pause the video and evaluate that expression without the use of a calculator now the first thing that i would recommend doing is converting the angle from radians to degrees so let's multiply by 180 over pi and so pi will cancel now 180 divided by 6 well if we divide 18 by 6 we'll get 3 so all we need to do is add a zero so 180 divided by six is thirty so we have five times thirty now five times three is fifteen so five times thirty is one fifty so this is the same as cosine of 150 degrees so 150 is in quadrant two so let's draw the triangle in quadrant two so from here to here is an angle of 150 degrees now the reference angle is going to be basically this acute angle between the x-axis and the terminal ray which is where this line ends so we know this is 180 and the terminal rate is at an angle of 150 so the difference between the two is 30 degrees and so that's the reference angle of 150 the reference angle is the same as the angle in quadrant one to calculate it in quadrant two it's always going to be the difference between the angle is going to be the difference between 180 and the angle in quadrant two to calculate the reference angle if it's in quadrant three is the difference between the angle in quadrant three and one eighty and if it's in quadrant four it's going to be 360 minus the angle in quadrant four now let's go ahead and complete the triangle so this is going to be 60 degrees and the side length across the 30 degree angle is 1 across the 60 degree angle it's the square root of 3 and across the the right angle or the hypotenuse is going to be 2. now this side is above the x-axis going in a positive y-direction so this is going to be positive one so whenever you're drawing a triangle that is not in quadrant one make sure to add the appropriate signs the hypotenuse will always be positive but in quadrant two the triangle is located on the negative x-axis so this is going to be negative square root 3. so now we can evaluate cosine 150 focusing on the reference angle of 30. so cosine is equal to the adjacent side over the hypotenuse so cosine 150 is going to be negative square root 3 over 2. cosine 30 if you type this in your calculator it's positive square root 3 over 2. so notice that an angle and its reference angle have the same magnitude however based on where the triangle is located based on the quadrant the sine will change so cosine 150 it's negative square root three over two even though cosine thirty is positive square root three over two have you ever heard of the expression all students take calculus it's a trig expression and it's useful for helping students to remember which trig ratio is positive or negative in a certain quadrant so the first letter a is for quadrant one this is for quadrant two for quadrant three and so forth so the all part tells us that sine cosine and tangent they're all positive in quadrant one so s c and t sine cosine and tangent they're all positive in the first quadrant and then students or s that tells us that only sine is positive in quadrant two everything else cosine and tangent they're both negative in quadrant two and then take t tells us that only tangent is positive in quadrant three everything else sine and cosine they're both negative and quadrant three and then the last one calculus is for quadrant four so only cosine c is positive in quadrant four everything else is negative and so that's what that expression helps you to keep in mind all students take calculus so the three trig functions they're all positive in quadrant one in quadrant two only sine is positive take in quadrant three tangent is positive calculus in quadrant four cosine is positive it also helps to realize that sine is related to the y value so whenever y is positive sine will be positive and so y is positive in quadrants one and two y is negative in quadrants three and four because that's below the x-axis now cosine is associated with the x value x is positive on the right side so in quadrants one and four cosine is positive x is negative on the left side so in quadrants two and three cosine is negative tangent is basically related to y over x so tangent is always negative when sine and cosine do not have the same value as you can see one is positive and the other is negative tangent is positive whenever sine and cosine have the same value because a positive divided by a positive will give you a positive result and a negative divided by another negative number will also give you a positive result so you can think of tangent as y over x now let's try this example tangent of negative 120 degrees go ahead and work on that example now first let's draw a picture now we have a negative angle so we can't travel in the counterclockwise direction rather we need to go clockwise so this is zero this is negative 90 negative 180 negative 270. so here's zero negative 90 and we're going to stop at negative 120. now the difference between negative 180 and negative 120 is 60. so 60 is the reference angle it's the angle between the x-axis and the terminal ray so once again we have a 30 60 90 right triangle so across the 30 is one across the 60 will be the square root of 3 across the 90 is 2. now in quadrant three x and y are both negative we could see that y is below the x-axis and x is on the left side where it's negative so now we can evaluate tangent so tangent of the reference angle is going to be equal to the opposite side divided by the hypotenuse i mean not the hypotenuse but the adjacent side so it's negative square root 3 divided by negative one which is just positive square root three so as we could see in quadrant three the result is that tangent is equal to a positive number now let's try this one secant of 225 degrees so let's draw a triangle so we have a positive angle this is 90 180 225 so the difference between 225 and 180 is 45 so that's going to be the reference angle for this particular problem so what we have is a 45 45 90 reference triangle so across the 45 angle will be equal to 1 and across the 90 is going to be the square root of 2. now in quadrant 3 both x and y are negative now it's better if we find cosine before we evaluate secant secant is one over cosine now we know that cosine is equal to the adjacent side let's use this angle so adjacent to that angle is negative one divided by the hypotenuse and so cosine is negative one over the square root of two so if this is cosine then secant is the reciprocal of that fraction which means it's negative square root two over one or simply negative square root 2. to show your work from this step you can multiply the top and the bottom by negative square root 2. so here these will cancel and the negative signs will cancel at the bottom so you get a final answer of negative square root 2 for this example now let's try another example so let's go with cosecant negative 13 pi over six go ahead and try that example now just like before let's convert the angle from radians to degrees so let's multiply by 180 over pi now 180 divided by six we know it's thirty and thirty times thirteen you can think of thirteen as being ten plus three and so 10 times 30 is 300 3 times 30 is 90. 300 plus 90 is 390. so this is equal to cosecant of negative 390. now if you have a large positive or negative angle that's not between 0 or 360 what you can do is add or subtract by 360 because if you do that whatever position you start from you're going to end at that position so if i start here and i add 360 i'm going to make one full rotation around the circle and it's going to lead me to the same starting point these are known as coterminal angles and anytime you evaluate a coterminal angle with a trig ratio the answer will always be the same so if i do negative 390 and add 360 to it this is going to be negative 30. it does cosecant of negative 390 is the same as cosecant of negative 30. and if you want to get rid of the negative angle you could add 360 again to negative 30. and so that's going to be positive 330 for those of you who do not wish to deal with any negative angles now let's draw this so let's say if we wanted to draw this angle 330 this would be 90 180 270 and 330 will be somewhere in that region so that's an angle of 330. now we can also get the same ray by going in this direction which will be negative 30. or if you want to get this angle this would be negative 90 negative 180 negative 270 negative 360 negative 390. all three angles lead us to this ray so therefore because they have the same terminal position they are called coterminal angles co means equal or the same at least that's the basic idea behind that word either case we could see that the reference angle that is the angle between the x-axis and the terminal ray is 30 because this angle is negative 30. and the reference angle is really what we need to get the answer so once again we have a 30 60 90 triangle so across the 30 is going to be 1 across the 60 square root 3 and across the 90 2. so in quadrant four x is positive but y is negative so focusing on the reference angle we need to evaluate cosecant and cosecant is one over sine and sine is going to be opposite over hypotenuse so opposite to the reference angle is negative 1 and the hypotenuse is 2. so one divided by negative one-half if we multiply the top and the bottom by negative two negative one-half and negative two will cancel to positive one and so the answer is one times negative two or simply negative two so this is the value of our original problem cosecant negative thirteen pi over six that's the answer well that's it for this video now for those of you who want more topics and trig or even more example problems like the ones you see on the board take a look at the description section of this video i'm going to post a link to my new trigonometry video playlist which you can get more info and more problems now if you haven't already done so feel free to subscribe to this channel all you need to do is click the red button at the bottom right of the video and you could do so and whatever you do don't forget to click the notification bell otherwise you may 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