in this video we're going to focus on the unit circle so in a new circle you need to know that this is the first quadrant this is quadrant two quadrant three and quadrant four perhaps you heard of the expression all students take calculus in quadrant one all sine cosine tangent functions are positive so that's the all part and all students take calculus in quadrant two sine is positive in Quadrant three tangent is positive in Quadrant four cosine is positive those are some things to know cosine is associated with the x value sign is associated with the Y value so this is zero degrees 90 180 270 this is 30 45 and 60 and then there's 120-150 well actually before 1:50 we have 135 and then 150 after 150 the next angle in degrees is to 10 then to 25 and to 40 after which we have 300 315 and 330 now you need to know the angles in radians that correspond to these values so 30 is the same as PI over 6 150 is 5 PI over 6 to 10 is 7 PI over 6 and 330 is 11 PI over 6 all of these share the same reference angle which is 30 degrees next you have the PI over 4 so 45 is 1 PI over 4 135 is 3 PI over 4 to 25 is 5 PI over 4 and 315 that's 7 PI over 4 and 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 over 2 180 is PI and 270 is 3 PI over 2 0 and 360 are coterminal angles 360 is also equal to 2 pi by the way so now 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 xxx its X is going to be root 3 over 2 and Y is 1/2 at 45 its root 2 over 2 and that's 60 it's going to be 1/2 comma root 3 over 2 and at 90 its 0 comma 1 now if you know the values for the first quadrant it's going to be the same for the other quadrants input the signs may change so in Quadrant 2 X is negative but Y is positive so notice that you have 1 0 on the right side 180 is going to be negative 1 0 now I 30 we have root 3 over 2 comma one half at 150 it's going to be very similar only X is going to be negative so it's going to be negative root 3 over 2 comma positive 1/2 135 shares the same reference angle as 45 so therefore they're going to have the same values so it's going to be negative root 2 over 2 comma root 2 over 2 and 120 and 60 they share the same reference angle so it's going to be negative 1/2 and positive root 3 over 2 so as you can see the values on the left is the same as the values on the right don't difference is the x value is a 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 negative root 3 over 2 comma negative 1/2 225 is similar to 45 but negative so negative root 2 over 2 comma negative root 2 over 2 and 240 similar to 60 so we're going to have negative 1/2 and negative 3 over 2 now 0 1 2 7 is opposite to 90 so instead of being 0 1 we're going to have 0 negative 1 okay now for the quadrant 4 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 positive 3 positive root 3 over 2 comma negative 1/2 and 315 is similar to 45 but X I mean but Y is still negative so it's root 2 over 2 comma negative root 2 over 2 and 300 has a reference angle of 60 so it's going to be 1/2 comma negative root 3 over 2 so x and y are positive in Quadrant 1 in Quadrant 2 X is negative Y is positive in Quadrant 3 x and y are both negative or in Quadrant 4 X is positive Y is negative so that's the 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 sine cosine or tangent function but before we get into that let's talk about how to use the unit circle to evaluate sine and cosine so let's say if we want to figure out what's sine of 60 is sine is associated with the Y value sine of 60 is root 3 over 2 cosine is associated with the x value cosine 60 is 1/2 now if you want to find tangent 60 tangent 60 is y over X is going to be root 3 over 2 divided by 1/2 the twos are going to cancel so tan 60 is simply root 3 so here's another example let's say if we want to evaluate sine 180 sine is going to be the Y value sine 180 is zero cosine 180 is the x value it's negative one tangent 180 is y over X it's 0 over negative 1 which is 0 now let's say if we want to evaluate 300 sine I mean not 300 but 330 sine 330 is negative 1/2 is the Y value cosine 330 that's positive root 3 over 2 tangent 330 is negative 1/2 divided by root 3 over 2 it's Y over X the twos will cancel so it's going to be negative 1 over root 3 and if you rationalize it it's going to be negative root 3 over 3 so that's how you can use the unit circle to evaluate sine cosine and tangent 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 sine of 30 degrees how can we do so without the help of the unit circle so you need to know the 30-60-90 triangle and this would commit it to memory is going to be very helpful across the 30 here's one across the 60 is root 3 across the 90 or the hypotenuse it's 2 now there's something called sylco Toa and we're going to focus on the cell part because S stands for sine sine sine 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 sine theta is opposite divided by hypotenuse so we're looking for sine 30 opposite to 30 is 1 so 1 is opposite to it 2 is the hypotenuse root 3 is adjacent to 30 so sine 30 is opposite which is 1 divided by the hypotenuse which is across the box so that's 2 so sine 30 is 1/2 and if you go to the unit circle if you locate 30 and if you look at the y-coordinate it should be 1/2 so now let's say if we wish to evaluate cosine of 30 degrees so now we need to you look at the car part of silver Toller so cosine is equal to adjacent divided by hypotenuse adjacent is root 3 so it's going to be root 3 divided by the hypotenuse which is 2 so cosine 30 is root 3 over 2 now the next thing we're going to look at is tangent of 30 so if we focus on a total part tan 30 its opposite over adjacent opposite is 1 adjacent is root 3 so it's going to be 1 divided by root 3 now you don't want to have a radical on the 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 root 3 so 1 times root 3 well that's just 3 3 root 3 times root 3 is the square root of 9 and the square root of 9 is 3 so tan 30 is root 3 over 3 now the next thing needs to be familiar with is the 45 45 90 degree triangle across the 45 is a 1 and across the 90 degree angle or die potenuse is root 2 so if you want to evaluate let's say sine of 45 sine of 45 you can use any of the 45 angles its opposite which is 1 divided by the hypotenuse of root 2 and you do have to rationalize it but when you do you're going to get root 2 over 2 sine 45 that's what it is root 2 over 2 cosine 45 is going to be the same thing it's going to be adjacent and divided by hypotenuse which is 1 over root 2 which rationalize 2 root 2 over 2 root 2 over 2 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 1 and adjacent is 1 so it's 1 over 1 tan 45 is simply 1 well that's great but what if we have an angle that's not in Quadrant 1 so for example let's say if we wish to evaluate sine of 120 120 it's 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 1 is the same as the angle in quadrant 1 so you don't need it if your angles in quadrant one in quadrant two the reference angle is 180 minus the angle in quadrant 2 in quadrant three 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 degrees 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 30-60-90 triangle so across the 30 we said is a 1 across the 60 is root 3 and across the 90 is the 2 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 positive root 3 the hypotenuse is always going to be positive so sine of 120 is equal to sine of 60 the only thing you have to worry about is if it's going to be positive or negative now sine 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 sine 60 is positive now sine 120 sine is associated with the Y value so assigned positive or negative in Quadrant 2 notice that in Quadrant 2 the Y value goes up so therefore Y is positive which means that sine 120 is positive so sine is positive in Quadrant 2 remember all students take calculus the S part is for quadrant 2 sine is positive so sine 120 is equal to positive sine 60 so now using sohcahtoa we can evaluate sine 60 so it's going to be opposite opposite to 60 is the positive root 3 divided by the hypotenuse which is 2 so sine 60 and therefore sine 120 is root 3 over 2 now it's your turn evaluate cosine of 240 so 240 is in Quadrant three so that's 240 degrees and this is 180 so keep in mind a reference angle in Quadrant three it's equal to the quadrant three angle minus 180 so it's 240 minus 180 which is 60 so this angle inside the triangle is 60 degrees 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 30-60-90 triangle again now across the 30 we have a one across the 60 is root three across the 90s two now 240 is in Quadrant three 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 our reference angle but cosine is negative in Quadrant three so let's put a negative sign so now let's evaluate it so cosine is adjacent divided by hypotenuse adjacent to the 60 is negative one and hypotenuse is 2 so it's going to be negative one over two keep in mind cosine 60 is positive one over two but we need to add the negative sign so overall it should be negative one over two because cosine 242 40s in Quadrant three and cosine is negative over all in Quadrant dream so that's cosine 240 negative 1/2 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 over X or it's also opposite over hypotenuse but let's draw so 330 is going to be somewhere over here so this entire angle is 330 degrees now the reference angle in Quadrant 4 is 360 minus the angle in quadrant four then the angle is 330 so therefore we have a reference angle of 30 degrees so now we can turn this into a triangle the 30-60-90 triangle across the 30 is one across the 60 is root 3 and hypotenuse is always 2 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 two and four but tangent is positive in one and three so it's going to negative 1030 so now let's focus on tan 30 tangent is opposite opposite to 30 is negative 1 divided by adjacent adjacent or next to 30 is root 3 so it's going to be negative 1 over root 3 overall tangent is negative in Quadrant 4 now let's rationalize it let's multiply top and bottom by root 3 so therefore tan 330 is negative root 3 over 3 try this one cosine 225 so 225 is in Quadrant three and sine and cosine are negative in Quadrant three so notice the difference between 225 in the x axis 225 - 180 that's a 45 difference so what we have is the 45-45-90 triangle so across the 45 is one and across the 90 is root 2 because this is in Quadrant three 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 negative 1 divided by the hypotenuse root 2 and don't forget to rationalize it so cosine 225 is negative root 2/2 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 45-45-90 time 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 degrees the XY coordinate on a unit circle is 0 comma 1 and tangent is y divided by X tangent is sine divided by cosine so it's sine 90 over cosine 90 now it turns out that sine 90 is the y-coordinate it's 1 cosine 90 is 0 whenever you get a 0 in the denominator of a fraction is undefined so tangent 90 is undefined if you were to type in tan 90 in your calculator is going to tell you what are you doing it's going to be going to get an error so tan 90 is undefined tan 270 is also undefined cotangent 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 sine of 5 PI over 6 how would you do it so here we have the angle in radians and if you're going to use the 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 30-60-90 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 PI's will cancel so 180 is basically 18 times 10 and 18 is 6 times 3 so we can cancel a 6 so we have leftover is 5 times 3 which is 15 and 15 times 10 that's 150 so therefore 5 PI over 6 is equal to 150 degrees so we're looking for a sign of 150 so let's draw it 150 is in Quadrant two and it's 30 degrees 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 one across the 60 is root three across the 90s two now in Quadrant two we know that X is negative but Y is positive so now sine 150 is associated with sine 30 so let's focus on the 30 degree angle the reference angle so sine is opposite over hypotenuse based on sohcahtoa so opposite to the 30 is 1 and the hypotenuse is 2 so therefore sine of 150 which is the same as sine 5 PI over 6 it's 1/2 and you could check that with the unit circle so now it's your turn try this example what is cosine of 11 PI over 6 so let's convert radians into degrees so let's multiply by 180 over PI so we can cancel the PI's now 18 divided by 6 is 3 so 180 divided by 6 is 30 so we have is 11 times 13 11 is 10 plus 1 so let's distribute the 30 30 times 10 is 330 times 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 degree reference angle so it's a 30-60-90 triangle again across the 30 is a 1 across the 60 is root 3 across the 90s - when watching 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 sohcahtoa it's cosine 330 is going to be adjacent over hypotenuse adjacent is root 3 hypotenuse is 2 so it's positive root 3 over 2 so that's the same as cosine 11 PI over 6 now let's say if you have tangent negative 150 how would you evaluate it what's tan negative 150 so what you can do is convert the negative angle into a positive angle by looking for the positive coterminal angle a coterminal angle is simply another number but it's an angle that lands in the same position so the find the coterminal angle simply add 360 to a negative angle negative 150 plus 360 that's going to be 210 so negative 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 three and here's 180 so the reference angle is 30 so we have a 30-60-90 triangle so across the 30 is one across the 60 is root 3 across the 90s - and x and y are both negative so he needs to understand this positive 210 is 210 degrees 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 the same location negative 150 is a clockwise rotation it follows the direction of a clock but these two angles you can see that if you add negative 150 plus 360 to get the 210 so it looks like a full circle so now let's evaluate tangent 210 based on sohcahtoa tangent is opposite over hypotenuse and the reference angle of 210 because he is 30 so opposite is negative 1 I might have said opposite over hypotenuse but if I did its opposite over adjacent so cut to 1 so I meant to say opposite of adjacent so opposite is negative 1 divided by Jason is which is negative root 3 and so the negatives cancel so it's positive 1 over root 3 and if we rationalize it it's going to be root 3 over 3 so that's tangent of negative 150 now let's say if you have secant let's say 300 how would you evaluate secant 300 you need to know that secant theta is equal to 1 over cosine and you need to know the other functions cosecant theta is 1 over sine theta and cotangent theta is 1 over tangent which is also the same as cosine over sine keep my tangent is sine over cosine but cotangent is cosine of a sine so if you want to evaluate secant you need to evaluate cosine secant is 1 over 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 sixteen so across the 30 is one across the 60 is root three across hypotenuse is 2 so X is positive in quadrant one but y is negative because y is going down in Quadrant four I'm going to stay quadrant 4 not quadrant one so in quadrant four X is positive but Y is negative so cosine 300 is associated with cosine 60 and cosine is adjacent over hypotenuse based on sohcahtoa so adjacent to 60 is 1 and the hypotenuse is 2 so cosine 300 is going to be positive 1 over 2 so keep in mind cosine is positive in Quadrant 1 and 4 because X is positive on the right side the cosine is negative in quadrants two and three because X is negative on the left side so if cosine 300 is 1/2 secant 300 is reciprocal of cosine it's going to be 2 over 1 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 over sine or 1 over sine 240 so let's draw 240 240 is in Quadrant 3 so two 40s over here it differs from the x-axis 180 by 60 so the reference angle 60 and this angle here is 30 so across the 30 is 1 across the 60 is root 3 across the 90s too so both x and y are negative in Quadrant 3 so if you wish to evaluate sine 240 we need to focus on the reference angle the 60 based on sohcahtoa sine is opposite divided by hypotenuse so therefore sine 240 is negative root 3 divided by 2 so that's the case for sine 240 then what is cosecant 2 for equal 2 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 negative 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 by 3/3 so cosecant 240 is negative 2 root 3 divided by 3 so that's the answer for that so let's say if you want to evaluate cotangent of 600 how would you do it so if you get a very large angle subtract it by 360 get the coterminal angle 600 minus 360 is 240 so we're looking for a cotangent of 240 so let's draw the triangle so 240 is somewhere over here and it differs by 60 from 180 so we have a 30-60-90 triangle this is negative one negative root three and two so if tangent is opposite over adjacent and cotangent is the reciprocal of tangent then cotangent must be adjacent over opposite so looking at the 16 adjacent to the 60 is negative one and opposite to the 60 is negative root three so co'tin of 240 which is the same as kill 10 600 it's 1 over root 3 and if we rationalize it it's going to be root 3 over 3 so keep in mind tangent 600 would be root 3 over 1 but cotangent of 600 is 1 over root 3 which is root 3 over 3 this reciprocal of tangent so now let's say if we have secant let's say negative 11 PI over 6 so what would you do so we have a negative 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 so 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 writing 180 over 6 is 30 30 times 11 is 330 so this is negative 330 now because it's negative let's add 360 to it to make it positive negative 330 plus 360 that's positive 30 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 divided by cosine so here we could just draw a generic 30-60-90 triangle because we're looking for cosine 30 across the 30 is 1 across the six days root 3 across the 90s 2 so cosine is adjacent divided by hypotenuse so cosine 30 is adjacent which is root 3 divided by the hypotenuse of 2 therefore secant 30 is the reciprocal of cosine 30 so it's 2 over root 3 and if we rationalize it we're going to get the answer to root 3 over 3 and so now you know how to evaluate secant cosecant and cotangent functions as well as sine cosine and tangent functions now here's the question for you let's say if sine theta is equal to 3 over 5 how can you find the other five trig functions cosine tangent secant cosecant cotangent so if sine is 3 over 5 draw a triangle in Quadrant 1 sine is opposite over hypotenuse so the opposite is the 3 part the hypotenuse is 5 this is the 3 4 5 triangle if you use the Pythagorean theorem you could come up with the same equation it's a squared plus B squared is equal to C squared so it's 3 squared plus the missing side squared which we'll call B squared and we have the hypotenuse which is five three squared is nine and five squared that's five times five which is twenty five 25 minus 9 is 16 and the square root of 16 is 4 so this we have the 3 4 5 triangle once you find the third side of the triangle you could find the other trig functions so cosine theta is adjacent divided by hypotenuse so that's four over five and tangent theta is opposite over adjacent so that's three over four so now that we have that we can find the others cosecant theta is the reciprocal of sine and if sine theta is 3 over five cosecant is five over 3 just flip it and secant is the reciprocal of cosine so that's going to be five over four and cotangent is the reciprocal of tangent so if tangent is three over four co'tin is four over three so now let's say if cosine theta is negative seven 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 Quadrant two so here's theta now cosine is negative 7 over 25 based on sohcahtoa we know that cosine is adjacent divided by hypotenuse so therefore this is negative 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 4 5 triangle the next one is the 5 12 13 triangle and there's the 8 15 17 triangle and a 7 24:25 triangle so the missing side is 24 there's some other special triangles - I've seen like the 940 41 and also there's the 11 60 61 prime so now that we have cosine we can find the other stuff so sine theta is opposite over hypotenuse opposite is 24 so it's 24 over 25 tangent theta is opposite over adjacent that's 24 over negative 7 so secant theta is the reciprocal of cosine so that's going to be negative 25 over 7 and cosecant theta is the reciprocal of sine so that's 25 over 24 and finally cotangent theta is the reciprocal of tangent so it's negative 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 degrees but greater than 180 so the first thing you need to find is the quadrant so clearly this is quadrant dream so let's draw a triangle in quadrant dream so here's our box and here's theta so this is the 8 15 17 triangle opposite to theta is the 8 adjacent is 15 the hypotenuse is 17 in Quadrant 3 x and y are negative so now let's find sine theta sine theta is opposite over hypotenuse opposite is negative 8 adjacent 19:15 hypotenuse is 17 so it's going to be negative 8 over 17 cosine theta that's the jacent over hypotenuse so negative 15 over 17 cotangent theta the reciprocal of tangent so that's 15 over 8 and cosecant that's going to be negative 17 over 8 and Sica negative 17 over 15 now let's say that secant theta is let's say it's 13 over 5 and let's say that sine theta is less than zero go ahead and find the other five trig functions feel free to pause the video now the first thing we need to do is identify the quadrant so if secant is positive that means cosine is positive because cosine is 1 over secant and sine is negative in which quadrant is cosine positive and sine negative cosine is associated with X sine is associated of Y so if X is positive we need to go to the right and if sine or Y is negative you need to go down so therefore this is in Quadrant four so theta is over here now C CAP it is 13 over five which means that cosine is 5 over 13 so the adjacent side is five hypotenuse is 13 the missing side must be 12 but notice that Y is going down so it's negative 12 so now we can find sine theta that's a negative 12 over 13 and cosecant theta is 1 over sine so that's going to be negative 13 over 12 tangent theta is opposite over adjacent so that's negative 12 over 5 and co'tin theta is negative 5 over 12 now sometimes may not have us but your triangle to go by so let's say if sine theta is two over five 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 5 what's the missing side so if we use the Pythagorean theorem we could say that a is 2 we're looking for B and C is 5 so 2 squared is 4 5 squared is 25 25 minus 4 is 21 so B squared is equal to 21 which means that B is the square root of 21 so that's the missing side so now let's find cosine cosine beta is going to be adjacent over hypotenuse so that's root 21 over 5 tangent theta that's opposite over adjacent it's 2 over root 21 and you got to rationalize it so this is going to be 2 root 21 over 21 now if you want to find cotangent theta don't flip this fraction flip this one it's easier so cotangent theta it's going to be just root 21 over 2 so you don't have to rationalize anymore secant theta is 1 of a cosine so it's going to be 5 over root 21 and that you do have to rationalize so that's going to be 5 root 21 over 21 and finally the other one cosecant theta it's just 5 over 2 now let's say it's cosine theta is let's say it's three over five and let's say the quadrant is one again so we know this is three this is five the missing side is 4 3 4 5 triangle but using this if cosine is 3 over 5 how can you find the double angle sine 2 theta there's a formula for this sine 2 theta is 2 sine theta cosine theta and we know that sine based on the triangle it's opposite over hypotenuse so it's 4 over 5 and we already have the value for cosine that's 3 over 5 2 is the same as 2 over 1 4 times 3 is 12 times 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 2 theta we can use the equation cosine squared minus sine squared cosine squared that's 3 over 5 squared sine squared sine is 4 over 5 3 squared is 9 5 squared is 25 4 squared is 16 9 minus 16 is negative 7 so it's negative 7 over 25 now if you want to find tangent to theta you can simply just divide sine 2 theta by cosine 2 theta so we have the value of sine two theta in the last example we said it was 24 over 25 and we have the value of cosine 2 theta it's a negative 7 over 25 if you multiply the top fraction by 25 and the bottom by 25 the 20 fives will cancel so tangent 2 theta is simply negative 24 / 7 by the way if you want more examples I've created another video on if it believes entitled trigonometry precalculus overview review tests 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 trigonometric identity you can check out those videos because that topic most students have 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 check out the video I've created one might be of great help to you so now let's move on to our next topic so let's say if sine theta is equal to X divided by 3 what is the value of cosine theta sometimes you make you may see a question like this and if you do huh what do you do how do how do you solve 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 and now we know that sine is opposite over hypotenuse so you got to find a missing side so let's solve for B we'll call B is massage so a squared plus B squared is equal to C squared according to the Pythagorean theorem so a is X we're looking for B C squared that's 3 squared which 9 so to isolate B squared we need to subtract both sides by x squared so B squared is 9 minus x squared and to solve for B we got to take the square root so B let me get rid of this B here so therefore B is the square root of 9 minus x squared so now we can figure out what cosine is so cosine is adjacent over hypotenuse so that's radical 9 minus x squared over 3 and that's how you do it so if I want to find tangent its opposite over adjacent opposite so theta is X adjacent is root 9 minus x squared so it's going to be x over root 9 minus x squared and you can use the find the other stuff cosecant is the reciprocal of sine so cosecant is 3 over X secant is the reciprocal of cosine that's 3 over radical 9 minus x squared and you can rationalize it if you want to and cotangent that's going to be radical 9 minus x squared divided by X now the last thing that we need to talk about is inverse functions so for example sine 30 if you use the unit circle if you look for the Y value sine 30 is 1/2 so therefore the inverse sine of 1/2 is 30 so sine of an angle is equal to a value 30 degrees is the angle 1/2 is the value the inverse sine of a value is the angle whenever you're dealing with an inverse function basically you need to switch two variables you would switch X and Y in this case we're switching the angle the value but now you have to be careful because the inverse functions has limitations was a sine function its domain goes on forever from negative infinity to infinity so here's an example to illustrate that sine of 150 is also 1/2 however the inverse sine of 1/2 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 sine can only be in quadrants one and four in camp in quadrant two and 150 is in quadrant two and that's why the inverse sine of 1/2 doesn't give you 150 it gives you 30 because 30 is in quadrant one so inverse sine only works in one and four and you need to know the angles and that it's between negative 90 and 90 so let's say if you want to find the inverse sine of negative 1/2 well we know that sine is negative in Quadrant four and we know that sine 330 is negative 1/2 however the inverse sine does not equal 330 even though 330 is in Quadrant four 330 is not between negative 90 and 90 inverse sine is limited between negative 90 and 90 now sine of negative 30 is also negative 1/2 negative 30 and 330 are coterminal angles they're both right here you can go 330 in this direction or you can go negative 30 in this direction so their co-terminal almost they differ by 360 so they're both in Quadrant 4 however negative 30 is in the range of negative 90 to 90 that's the range of the inverse sine function so negative 30 works so the inverse sine of negative 1/2 is indeed negative 30 and not 330 even though both are in the right quadrant negative 30 is in the restricted domain of inverse sine which is between negative 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 exist between quadrant one and two so the restricted domain for inverse cosine is from zero to 180 and for inverse tangent it's like inverse sine its quadrant one and four so none of them exist in quadrant three for the inverse functions so sine and tangent is from negative 90 to 90 quadrant two running for inverse cosine is quadrant one and two from zero to 180 so knowing that let's evaluate inverse cosine of 1/2 feel free to pause the video and figure out what inverse cosine of 1/2 is so we know that cosine of 60 is 1/2 and also cosine of 300 is 1/2 now 300 is in Quadrant four and inverse cosine it doesn't exist in Quadrant four so we can't use a 300 so it's 60 so inverse cosine of 1/2 is 60 now what about inverse cosine of negative root 3 over 2 so cosine of if you use the unit circle and you look at the x-values cosine of 150 is negative root 3 over 2 cosine is negative in quadrants two and three and cosine two 10 is also negative root three over two however 210 is in Quadrant three so we can't use that so therefore it's 150 150 is in Quadrant two and it's within the restricted domain this between 0 and 180 so now what about inverse tangent of 1 so this has to be in quadrant one because it's positive in verse 10 is positive in Quadrant 1 and if you use the unit circle it's 45 now what about inverse tangent of negative root 3 what about that one so tangent is negative in quadrants two and four tan 120 is negative root three and tan of 300 is also negative root three and also tan negative 60 is also negative root three so in verse 10 is negative in Quadrant four we can't use 120 now 300 and negative 60 is in Quadrant four but we can't use 300 because it's outside of the restricted domain it has to be between negative 90 and 90 so negative 16 fits the requirements and so that's the answer so whenever you're dealing with inverse sine and inverse tangent in quadrant 4 the angle is going to be between negative 90 to zero 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