in this video we're going to focus on the unit circle so in the unit circle you need to know that this is the first quadrant this is quadrant 2 quadrant 3 and Quadrant 4 perhaps you heard of the expression all students take calculus in quadrant one all s cosine Tang functions are positive so that's the all part and all students take calculus in quadrant two s is positive in quadrant three tangent is positive in Quadrant 4 cosine is positive those are some things to know cosine is associated with the x value sign is associated with the yalue so this is 0 deg 90 18 80 270 this is 30 45 and 60 and then there's 120 150 well actually before 150 we have 135 and then 150 after 150 the next angle in degrees is 210 then 225 and 240 after which we have 300 315 and 330 now you need to know the Angles and radians that correspond to these values so 30 is the same as pi over 6 150 is 5 pi over 6 210 is 7 Pi 6 and 330 is 11 Pi 6 all of these share the same reference angle which is 30° next you have the pi of fours so 45 is 1 Pi 4 135 is 3 pi over 4 225 is 5 piun over 4 and 315 that's 7 pi over 4 Now 60 is pi over 3 or 1 pi over 3 120 is double that amount so that's 2 pi over 3 240 is 4 pi over 3 and 300 is 5 pi over 3 90 is pi/ 2 180 is pi and 270 is 3 pi over 2 0 and 360 are co-terminal angles 360 is also equal to 2 pi by the way so now what we need to do is go over the values that correspond to these angles you need to know the XY coordinates so for zero the coordinates are 1 comma 0 30 it's X is going to be < tk3 over 2 and Y is 12 at 45 it's < tk2 over2 and at 60 it's going to be 12 comma < tk3 / 2 and at 90 it's uh 0 comma 1 now if you know the values for the first quadrant it's going to be the same for the other quadrants but the signs may change so in quadrant 2 x is negative but Y is positive so notice that you have one Z on the right side 180 is going to be negative 1 now at 30 we have root3 over 2 comma 12 at 150 it's going to be very similar only X is going to be negative so it's going to be negative < tk3 / 2 comma positive 12 135 shares the same reference angle as 45 so therefore they're going to have the same values so it's going to bek2 over2 2 over2 and 120 and 60 they share the same reference angle so it's going to be -2 and positive3 over2 so as you can see the values on the left is the same as the values on the right the only difference is the X values are negative so now let's look at quadrant 3 we're going to have the same values but X and Y are both negative so at 210 210 150 and 30 they share the same reference angle so the values is going to be the same but it's going to be < tk3 / 2 comma half 225 is similar to 45 but negative so < tk2 over2 comma < tk2 over2 and 240 similar to 60 so we're going to have -2 and3 over 2 now 01 270 is opposite to 90 so instead of being 01 we're going to have 01 okay now for the quadrant four angles 330 is similar to 30 they share the same reference angle but Y is negative in Quadrant 4 x is positive so it's going to be posi3 positive < tk3 / 2 comma half and 315 is similar to 45 but X I mean but Y is still negative so it's < tk2 over2 comma < tk22 and 300 has a reference angle of 60 so it's going to be 12 comma < tk3 / 2 so X and Y are positive in quadrant one in quadrant 2 x is negative Y is positive in quadrant 3 X and Y are both negative but in Quadrant 4 x is positive Y is negative so that that's a unit circle but sometimes it may be difficult to memorize the unit circle so there's another technique that can help you if you need to evaluate a s cosine or tangent function but before we get into that let's talk about how to use the unit circle to evaluate s and cosine so let's say if we want to figure out what s of 60 is s is associated with the yvalue S of 60 is < tk3 over 2 cosine is associated with the x value cosine 60 is 12 now if you want to find tangent 60 tangent 60 is y/x it's going to beunk 3 over 2 / 12 the twos are going to cancel so tan 60 is simply root3 so here's another example let's say if we want to evaluate s 180 s is going to be the Y value sin 180 is zero cosine 180 is the x value value it's 1 tangent 180 is y/x it's 0 over1 which is 0 now let's say if we want to evaluate 300 S I mean not 300 but 330 sin 330 is negative a half is the Y value cosine 330 that's positive < tk3 /2 tangent 330 is half / < tk3 over2 it's YX the twos will cancel so it's going to be -1 over < tk3 and if you rationalize it it's going to bek3 over3 so that's how you can use the unit circle to evaluate s cosine and tangent uh functions if you have a memorized but now let's talk about how to evaluate trig functions if we don't have the unit circle or if we don't have it memorized so let's say if we want to evaluate s of 30° how can we do so without the help of the unit circle so you need to know the 30 6090 triangle and this I would commit it to memory it's going to be very helpful across the 30 is 1 across the 60 is root3 across the 90 or the hypotenuse it's two now the something called sooa and we're going to focus on the so part because s stands for sign s s of an angle let's say Theta is equal to the opposite side divided by the hypotenuse side o stands for opposite H stands for hypotenuse a stands for adjacent so s Theta is opposite ID hypotenuse so we're looking for S 30 opposite to 30 is one so one is opposite to it two is the hypotenuse < tk3 is adjacent to 30 so sin 30 is opposite which is 1 / the hypotenuse which is across the box so that's two so sin 30 is a half and if you go to the unit circle if you locate 30 and if you look at the y-coordinate it should be2 so now let's say if we wish to evaluate cosine of 30 degrees so now we need to look at the C part of soaa so cosine is equal to adjacent / hypotenuse adjacent is < tk3 so it's going to be Roo tk3 / the hypotenuse which is 2 so cosine 30 is < tk3 over2 now the next thing we're going to look at is tangent of 30 so if we focus on the tolla part tan 30 it's opposite over adjacent opposite is one adjacent is root3 so it's going to be 1 / Roo tk3 now you don't want to have a radical on a bottom in the denominator of a fraction so what you want to do is you want to rationalize it you want to multiply top and bottom by root3 so 1 * Ro tk3 well that's justo tk3 < tk3 * tk3 is the < TK of 9 and it's < TK of 9 is 3 So Tan 30 is < tk3 over 3 now the next thing you need to be familiar with is the 4545 90° triangle across the 45 is a one and across the the 90° angle or the hypotenuse is < tk2 so if you want to evaluate let's say s of 45 s of 45 you can use any of the 45 angles it's opposite which is 1 / the hypotenuse of < tk2 and you do have to rationalize it but when you do you're going to get < tk2 over 2 sin 45 that's what it is < tk2 over2 cosine 45 is going to be the same thing it's going to be adjacent divid hypotenuse which is 1/ < tk2 which rationalized to < tk2 over < tk2 over two now tangent 45 what do you think tangent 45 is so tangent is equal to the opposite side over the adjacent side so opposite is one and adjacent is one so it's one over one tan 45 is simply one well that's great what if we have an angle that's not in quadrant one so for example let's say if we wish to evaluate s of 120 120 is not part of the 30 60 90 triangle so how can we do this now there's something called the reference angle and that's what you want to use in this situation the reference angle in quadrant one is the same as the angle in quadrant one so you don't need it if your angle's in quadrant one in quadrant 2 the reference angle is 180 minus the angle in quadrant 2 in quadrant 3 the reference angle is the angle minus 180 and in Quadrant 4 the reference angle is 360 minus the angle in Quadrant 4 so let's draw it so 120 that's going to be somewhere in quadrant 2 so this angle is 120 notice that it forms a reference angle of 60° the reference angle is the angle between the hypotenuse of the triangle and the x-axis so we could turn this into a triangle if you want so notice that we get the 30 6090 triangle so across the 30 we said is a one across the 60 is a root3 and across the 90 is a two now because the triangle is in quadrant 2 we know X is negative Y is positive so we're going to put negative 1 here and leave this as positive3 the hypotenuse is always going to be positive so s of 120 is equal to S of 60 the only thing you have to worry about is if it's going to be positive or negative now s cosine or tangent of any reference angle which is always between 0 and 90 it's going to be positive because everything is positive in quadrant 1 so sin 60 is positive now s 120 s is associated with the yalue so a sign positive or negative in quadrant 2 notice that in quadrant 2 the Y value goes up so therefore Y is positive which means that s 120 is positive so sign is positive in quadrant 2 remember all students take calculus the S part is for Quadrant 2 sign is positive so sin 120 is equal to positive sin 60 so now using so we can evaluate sin 60 so it's going to be opposite opposite to 60 is the posi root3 / the hypotenuse which is 2 so sin 60 and therefore sin 120 is < tk3 / 2 now it's your turn evaluate cosign of 240 so 240 is in quadrant three so that's 240° and this is 180 so keep in mind the reference angle in quadrant 3 it's equal to the quadrant 3 angle minus 180 so it's 240 - 180 which is 60 so this angle inside the triangle is 60° so if we turn it to a right triangle this is going to be 90 and the other angle is 30 so we have the 30 6090 triangle again now across the 30 uh we have a one across the 60 is root3 across the 90 is 2 now 240 is in quadrant 3 so X and Y are both negative so we can see that cosine 240 is equal to is associated with cosine 60 because 60 is a reference angle but cosine is negative in quadrant 3 so let's put a negative sign so now let's evaluate it so cosine is adjacent ID hypotenuse adjacent to the 60 is-1 and hypotenuse is two so it's going to be -1/2 keep in mind cosine 60 is POS 1 over2 but we need to add the negative sign so overall it should be 1 over2 because cosine 240 240 is in quadrant 3 and cosine is negative overall in quadrant 3 so that's cosine 240 negative half and you can confirm it with the unit circle so try this one tangent of 330 what's tangent of 330 now tangent is y overx or it's also opposite over hypotenuse but let's draw it so 330 is going to be somewhere over here so this entire angle is 330° now the reference angle in Quadrant 4 is 360 minus the angle in Quadrant 4 and the angle is 330 so therefore we have a reference angle of 30° so now we can turn this into a triangle the 30 6090 triangle across the 30 is 1 across the 60 is < tk3 and hypotenuse is always two now in Quadrant 4 x is positive but Y is negative as you go to the right you can see X is positive but Y is going down towards the negative Zone so tangent 330 is going to be a negative value because tangent is negative in quadrants 2 and four but tangent is positive in 1 and three so it's going to be tan 30 so now let's focus on tan 30 tangent is opposite opposite to 30 is1 divid adjacent adjacent or next to 30 is root3 so so it's going to be -1/ < tk3 overall tangent is negative in Quadrant 4 now let's rationalize it let's multiply top and bottom byun3 so therefore tan 330 is < tk3 over 3 try this one cosine 225 so 225 is in quadrant 3 and S and cosine are Nega in quadrant 3 so notice the difference between 225 and the xaxis 225 - 180 that's a 45 difference so what we have is the 45 45 90 triangle so across the 45s is one and across the 90 is root two because this is in quadrant 3 X and Y are both negative so cosine 225 is the same as negative cosine 45 and cosine 225 is going to be adjacent doesn't matter which 45 you pick let's pick this one so adjacent it's 1 / the hypotenuse < tk2 and don't forget to rationalize it so cosine 225 is < tk22 / 2 so now let's say if you get an angle that is not part of the special triangles either the 30 60 90 triangle or the the 45 4590 triangle so for example let's say if you need to evaluate tangent 90 how would you do it so you can't really draw a triangle for this you simply need to know that at 90° the XY coordinate on a unit circle is 0 comma 1 and tangent is y / X tangent is s / cosine so it's sin 90 over cosine 90 now turns out that sin 990 is the y-coordinate it's one cosine 90 is zero whenever you get a zero in the denominator of a fraction it's undefined so tangent 90 is undefined if you were to type in tan 90 in your calculator it's going to tell you what are you doing it's going to you're going to get an error So Tan 90 is undefined tan 270 is also undefined coent of 0 and 180 are undefined as well so just make sure you're aware of that you never know you might see that on the test now let's say if you want to evaluate s of 5 pi/ 6 how would you do it so here we have the angle in radians and if you're going to use a unit circle you can simply take the paper and just simply find out what the y-coordinate is and get the answer so keep in mind you can go to Google Images and type in unit circle and just print it out and memorize it but if you want to use the 30 6090 triangle here's what you need to do let's convert radians into degrees in order to do this you need to multiply the angle in radians by 180 over Pi so notice that the pies will cancel so 180 is basically 18 * 10 and 18 is 6 * 3 so we can cancel a six so what we have left over is 5 * 3 which is 15 and 15 * 10 that's 150 so therefore 5 piun / 6 is equal to 150° so we're looking for S of 150 so let's draw it 150 is in quadrant 2 and it's 30° away from 180 so the reference angle is 30 so we can draw the triangle so we have another 30 60 90 triangle across the 30 is 1 across the 60 is < tk3 across the 90 is 2 now in quadrant 2 we know that X is negative but Y is positive so now sin 150 is associated with sin 30 so let's let's focus on the 30° angle the reference angle so s is opposite over hypotenuse based on SOA TOA so opposite to the 30 is 1 and the hypotenuse is two so therefore s of 150 which is the same as s 5i 6 it's 12 and you could check that with the UN Circle so now it's your turn try this example what is cosine of of 11 Pi / 6 so let's convert radians into degrees so let's multiply by 180 over Pi so we can cancel the pies now 18 / 6 is 3 so 180 / 6 is 30 so what we have is 11 * 30 11 is 10 + 1 so let's distribute to 30 30 * 10 is 300 and 30 * 1 is 30 so this is 330 so we have here is cosine 330 now let's go ahead and plot it on a graph so 330 is in Quadrant 4 and this is 360 or 0 degrees so we have a 30° reference angle so it's a 30 6090 triangle again across the 30 is one across the 60 is root3 across the 90 is 2 in Quadrant 4 x is positive but Y is negative so we need to focus on the 30 because that's a reference angle so based on SOA TOA it's cosine 330 is going to be adjacent over hypotenuse adjacent is root3 hypotenuse is 2 so it's positive < tk3 over 2 so that's the same as cosine 11i / 6 now let's say if you have tangent - 150 how would you evaluate it what's tan50 so what you can do is convert the negative angle into a positive angle by looking for the positive co-terminal angle a coterminal angle is simply another number but it's an angle that lands in the same position so to find a c-terminal angle simply add 360 to a negative angle - 150 + 360 that's going to be 210 so - 150 and 210 are coterminal angles they occupy the same position on the number I mean on the unit circle so 210 is in quadrant 3 three and here's 180 so the reference angle is 30 so we have a 30 60 90 triangle so across the 30 is 1 across the 60 is root3 across the 90 is 2 and X and Y are both negative so you need to understand this positive 210 is 210° going this way a positive angle rotates counterclockwise in the opposite direction of a clock negative 150 you need to go the other way this is negative 150 notice that you end it in the same location 150 is a clockwise rotation it follows the direction of a clock but these two angles you can see that if you add- 150 + 360 get 210 so it looks like a full circle so now let's evaluate tangent to 10 based on sooa tangent is opposite over hypotenuse and a reference angle of 210 because he is 30 so opposite is1 I might have said opposite over hypotenuse but if I did it's opposite over adjacent SOA TOA so I meant to say opposite of adjacent so opposite is1 / adjacent is which is3 and so the negatives cancel so it's POS 1k3 and if we rationalize it it's going to be < tk3 over 3 so that's tangent of -50 now let's say if you have secant let's say 300 how would you evaluate secant 300 you need need to know that secant Theta is equal to 1/ cosine and you need to know the other functions cose Theta is 1/ sin Theta and Cent Theta is 1 over tangent which is also the same as cosine over s keep in mind tangent is s over cosine but coent is cosine over s so if you want to evaluate secant you need to evaluate cosine secant is 1/ cosine so what's cosine of 300 so let's draw our picture 300 is somewhere over here and so the difference between 300 and the x axis which is 360 we can see that we have a reference angle of 60 so across the 30 is 1 across the 60 is < tk3 across the hypotenuse is 2 so X is positive and in quadrant 1 but Y is negative because Y is going down in Quadrant 4 I'm meant to say Quadrant 4 not quadrant 1 so in Quadrant 4 x is positive but Y is negative so cosine 300 is associated with cosine 60 and cosine is adjacent over hypotenuse based on soaa so adjacent to 60 is one and the hypotenuse is two so cosine 300 is going to be positive 1 over 2 so keep in mind cosine is positive in quadrants 1 and four because X is positive on the right side but cosine is negative in quadrants 2 and three because X is negative on the left side so if cosine 300 is 12 secant 300 is reciprocal of cosine it's going to be 2 over one and so that's the answer for that so now let's try another one let's say if we wish to evaluate cosecant of 240 feel free to pause the video and see if you can figure this one out see if you can get the answer cosecant is 1 / s or 1 over s 240 so let's draw 240 240 is in quadrant three so 240s over here it differs from the xaxis 180 by 60 so the reference angle is 60 and this angle here is 30 so across the 30 is 1 across the 60 is root3 across the 90 is two so both X and Y are negative in quadrant 3 so if we wish to evaluate s 240 we need to focus on the reference angle to 60 based on to TOA sign is opposite ID hypotenuse so therefore sin 240 is < tk3 / 2 so if that's the case for S 240 then what is cose 240 equal to so it's going to be the reciprocal of this fraction so we just got to flip that fraction so it's going to be -2 over radical 3 now we have a radical on the bottom so we do need to rationalize it let's multiply the top and bottom by3 so cosecant 240 is -2 < tk3 / 3 so that's the answer for that so let's say if you want to evaluate coent of 600 how would you do it so if you get a very large angle subtract it by 360 get the C terminal angle 600 - 360 is 240 so we're looking for a cotangent of 40 so let's draw the triangle so 240 is somewhere over here and it differs by 60 from 180 so we have a 30 6090 triangle this is -1 < tk3 and 2 so if tangent is opposite over adjacent and coent is a reciprocal of tangent then Cent must be adjacent over opposite so looking at the 60 adjacent to the 60 is1 and opposite to the 60 is3 so Coan of 240 which is the same as Coan 600 it's 1 over < tk3 and if we rationalize it it's going to be < tk3 over 3 so keep in mind tangent 600 would be root3 1 but coent of 600 is 1k3 which is ro33 this reciprocal of tangent so now let's say if we have secant let's say -1 piun / 6 so what would you do so we have a NE angle and it's in radians and we have a secant function so how would you figure out this one so let's go ahead and convert radians into degrees let's not worry about the negative sign right now so let's multiply by 180 over Pi now I believe we did this one already 180 over 6 is 30 30 * 11 is 330 so this is- 330 now because it's negative let's add 360 to it to make it positive - 330 plus 360 that's positive3 which is a reference angle because it's an acute angle between 0 and 90 so we're looking for secant 30 so whenever you want to evaluate secant you want to find out cosine because secant is 1/ cosine so here we could just draw a generic 30 6090 triangle because we're looking for cosine 30 across the 30 is 1 across the 60 is root3 across the 90 is 2 so cosine is adjacent / hypotenuse so cosine 30 is adjacent which is < tk3 / the hypotenuse of 2 therefore secant 30 is reciprocal of cosine 30 so it's 2 over < tk3 and if we rationalize it we're going to get get the answer 2 < tk3 over 3 and so now you know how to evaluate secant cosecant and coent functions as well as s cosine and tangent functions now here's the question for you let's say if sin Theta is equal to 3 over 5 how can you find the other five trig functions cosine tent secant cose andent so if sin is 3 5 draw a triangle in Quadrant One sign is opposite over hypotenuse so the opposite is the three part the hypotenuse is five this is the 34 five triangle if you use the Pythagorean theorem you could come up with the same equation it's A2 + b 2 is equal to c^ 2 so it's 3^ 2 plus the missing side Square which we'll call b^ S and we have the hypotenuse which is five 32 is 9 and 5^ S that's 5 * 5 which is 25 25 - 9 is 16 and the square root of 16 is four so thus we have the 345 triangle once you find the third side of the triangle you could find the other trick functions so cosine Theta is adjacent / hypotenuse so that's 4 over 5 and tangent Theta is opposite over adjacent so that's 3 over 4 so now that we have that we could find the others cose Theta is the reciprocal of s and if sin Theta is 3 5 C secant is 5 over3 you just flip it and secant is the reciprocal of cosine so that's going to be 5 over 4 and coent is the reciprocal of tangent so if tangent is 3 over 4an is 4 over 3 so now let's say if cosine Theta is 7 over 25 and let's say the angle let's say Theta is less than 180 but greater than 90 what quadrant is the angle in between 90 and 180 the quadrant is number two so let's draw a triangle in Quad Quant 2 so here's there now cosine is -7 over 25 based on SOA TOA we know that cosine is adjacent ided hypotenuse so therefore this is -7 and this is 25 keep in mind X is negative in quadrant 2 anytime you go to the left the X values are negative so now what's the missing side you need to know your special triangles so we went over the 3 four five triangle the next one is the 5 12 13 triangle and there's the 8 15 17 triangle and the 7 24 25 triangle so the missing side is 24 there's some other special triangles too I've seen like the 9 40 41 and also there's the 11 606 one triangle so now that we have cosine we could find the other stuff so sin thet is opposite over hypotenuse opposite is 24 so it's 24 over 25 tangent Theta is opposite of adjacent that's 247 so secant Theta is the reciprocal of cosine that's going to be -25 / 7 and cosecant Theta is the reciprocal of so that's 25 24 and finally thet is the reciprocal of tangent so it's -7 over 24 so now it's your turn let's say if tangent Theta is equal to 8 over 15 and Theta is less than 270° but greater than 180 so the first thing you need to find is the quadrant so clearly this is quadrant 3 so let's draw a triangle in quadrant dream so here's our box and here's Theta so this is the 85 17 triangle opposite to Theta is the 8 jent is 15 the hypotenuse is 17 in quadrant 3 X and Y are negative so now let's find sin Theta sin Theta is opposite over hypotenuse opposite is8 adjacent is15 hypotenuse is 17 so it's going to be8 over 17 cosine Theta that's adjacent over hypotenuse so5 177 Cent Theta the reciprocal of tangent so that's 15 8 and cose that's going to be -7 8 and secant -17 over 15 now let's say that secant Theta is let's say it's 13 over 5 and let's say that sin Theta is less than zero go ahead and find the other five trig functions F free to pause the video now the first thing we needs to do is identify the quadrant so if secant is positive that means Co sign is positive because cosine is 1 / SEC and S is negative in which quadrant is cosine positive and S negative cosine is associated with x s is associated with Y so if x is positive we need to go to the right and if sign or Y is negative you need to go down so therefore this is in quadrant four so Theta is over here now secant is 13 over five which means that cosine is 5 over 13 so the adjacent side is five hypotenuse is 13 the Miss side must be 12 but notice that Y is going down so it's -2 so now we can find sin Theta that's -12 13 and cosecant Theta is 1 / s so that's going to be -13 /2 tangent Theta is opposite over adjacent so that's -12 over 5 and Coan Theta is-5 over2 now sometimes you may not have a special triangle to go by so let's say if sin Theta is 2 over 5 and let's say it's in quadrant one go ahead and find the other trig functions so we're going to follow the same pattern so opposite is two and the hypotenuse is five what's the Miss inside so if we use the Pythagorean theorem we could say that a is 2 2 we're looking for B and C is 5 so 2^ 2 is 4 5^ 2 is 25 25 - 4 is 21 so b^2 is equal to 21 which means that b is square otk of 21 so that's the missing side so now let's find cosine cosine Theta is going to be adjacent over hypotenuse that's < TK 21 over 5 tangent Theta that's opposite over Jason that's 2 over < tk21 and you got to rationalize it so this is going to be 2 < TK 21 over 21 now if you want to find coent Theta don't flip this uh fraction flip this one it's easier so Cent Theta is going to be justun 21 over two so you don't have to rationalize anymore secant Theta is one over cosine so it's going to be 5 over < tk21 and that you do have to rationalize so that's going to be 5 < TK 21 over 21 and finally uh the other one cosecant Theta is just 5/ 2 now let's say if cosine Theta is let's say say it's 3 over 5 and let's say the quadrant is one again so we know this is three this is five the missing side is four 3 four five triangle but using this if cosine is 3 over 5 how can you find the double angle sin 2 thet there's a formula for this sin 2 Theta is 2 sin Theta cosine Theta and we know that sign based on the triangle it's opposite over hypotenuse so it's 4 over5 and we already have the value for cosine that's 3 over5 2 is the same as 2 over 1 4 * 3 is 12 * 2 is 24 so it's 24 over 25 that's how you can evaluate the double angle using the triangle now if you want to find cosine to we can use the equation cosine 2us sin cosine s that's 3 over 5^ 2 sin squar s is 4 over 5 3 2 is 9 5 2 is 25 4 S is 16 9 - 16 is -7 so it's -7 over 25 now now if you want to find tangent 2 thet you can simply just divide sin 2 thet by cosine 2 thet so we have the value of sin 2 Theta in the last example we said it was 24 over 25 and we have the value of cosine 2 Theta it's -7 over 25 if you multiply the top fraction by 25 and the Bottom by 25 the 25s will cancel so tangent 2 Theta is simply -24 over 7 by the way if you want more examples I've created another video on I believe it's entitled trigonometry pre-calculus overview review test something like that it's somewhere on YouTube and it covers a lot of examples on trigonometry and questions like what we're dealing with right now and there's another one verifying uh trigonometric identity you can check out those videos because uh that topic most students have uh great difficulty with so if you ever come to a point where you have to know how to prove one side is equal to the other side if you have to prove or verify the identity uh check out the video I've created one might be of great help to you so now let's move move on to our next topic so let's say if sin Theta is equal to x / 3 What is the value of cosine Theta sometimes you make you may see a question like this and if you do H what do you do I mean how how do you solve it so let's draw a triangle and let's assume that it's in quadrant one because everything is positive if there's no negative sign assume it's in quadrant one so here's Theta now we know that sign is opposite over hypotenuse so we got to find a missing side so let's solve for b we'll call B as a missing side so A2 + b s is equal to c^2 according to the Pythagorean theorem so a is X we're looking for b c^ 2 that's 3^ 2 which is 9 so to isolate b^ s we need to subract subract both sides by x^2 so b^2 is 9 - x^2 and to solve for b we got to take the square root so B let me uh get rid of this B here so therefore B is the square root of 9 - x^2 so now we can figure out what cosine is so cosine is adjacent over hypotenuse so that's radical 9 - x^2 over 3 and that's how you do it so if I want to find tangent it's opposite over adjacent opposite to Theta is X adjacent isun 9 - x^2 so it's going to be x overun 9 - x^2 and you can easily find the other stuff coant is the reciprocal of s so cosecant is 3x secant is the reciprocal cosine that's 3 over radal 9 - x^2 and you could rationalize it if you want to and cotangent that's going to be radical 9 - x^2 / X now the last thing that we need to talk about is uh inverse functions so for example sin 30 if you use the unit circle if you look for the Y value sin 30 is a half so therefore the inverse sign of a half is 30 so s of an angle is equal to a value 30° is the angle2 is the value the inverse sign of a value is the angle whenever you're dealing with an inverse function basically you need to switch the variables you would switch X and Y in this case we switching the angle with the value but now you have to be careful because the inverse functions has limitations whereas the sign function its domain goes on forever from negative Infinity to Infinity so here's an example to illustrate that s of 150 is also a half however the inverse sign of a half and this is important it does not equal 150 question is why well as we mentioned before the inverse function has a restricted domain so let's go over the rules that you need to know regarding inverse functions and their restrictions so whenever you're dealing with the inverse function s can only be in quadrants 1 and four it can't be in quadrant 2 and 150 is in quadrant 2 and that's why the inverse sign of a half doesn't give you 150 it gives you 30 because 30 is in quadrant 1 so inverse sign only works in one and four and you need to know the angles and that it's between 90 and 90 so let's say if you want to find the inverse sign of negative a half well we know that sign is negative in Quadrant 4 and we know that sin 330 is negative a half however the inverse sign does not equal 330 even though 330 is in Quadrant 4 330 is not between 90 and 90 inverse sign is limited between 90 and 90 now s of -30 is also a half -30 and 330 are coterminal angles they're both right here you can go 330 in this direction or you can go -30 in this direction so they're co-terminal angles they differ by 360 so they're both in Quadrant 4 however -30 is in the range of 90 to 90 that's the range of the inverse sign function so -30 works so the inverse sign of negative a half is indeed -30 and not 330 even though both are in the right quadrant -30 is in the restricted domain of inverse sign which is between 90 and 90 so make sure you're aware of that the next trig function that we need to talk about is inverse cosine inverse cosine exists between quadrants one and two so the restricted domain for inverse cosine is from 0 to 180 and for inverse tangent it's like inverse sign it's quadrant 1 and four so none of them exists in quadrant three for the inverse functions so s and tangent is from 90 to 90 quadrants 1 and 4 inverse cosine is quadrant 1 and two from 0 to 180 so knowing that let's evaluate inverse cosine of 12 feel free to pause the video and figure out what inverse cosine of a half is so we know that cosine of 60 is2 and also cosine of 300 is a half now 300 is in Quadrant 4 and inverse cosine doesn't exist in Quadrant 4 so we can't use a 300 so it's 60 so inverse cosine of a half is 60 now what about inverse cosine of < tk3 over2 so cosine of if you use the unit circle and you look at the X values cosine of 150 is < tk3 over2 cosine is negative in quadrants 2 and 3 and cosine 210 is also3 2 however 210 is in quadrant 3 so we can't use that so therefore it's 150 150 is in quadrant 2 and it's within the restricted domain it's between 0 and 180 so now what about inverse tangent of one so this has to be in quadrant one because it's positive inverse 10 is positive in quadrant 1 and if you use the unit circle it's 45 now what about inverse tangent of < tk3 what about that one so tangent is negative in quadrants 2 and four tan 120 is < tk3 and tan of 3 100 is also < tk3 and also tan -60 is also < tk3 so inverse tan is negative in Quadrant 4 we can't use 120 now 300 and -60 is in Quadrant 4 but we can't use 300 because it's outside of the restricted domain it has to be between 90 and 90 so 60 fits the requirements and so that's the answer so whenever you're dealing with inverse sign and inverse tangent in Quadrant 4 the angle is going to be between 90 to Zer it's not going to be between 270 and 360 make sure you do not pick that answer so that is it for this video we've covered a lot of trig and hopefully you found it to be educational so thanks for watching and have a great day