now let's talk about evaluating inverse sine functions so let's say if we want to find the value of the inverse sine of one half what is that equal to what do you think the answer is now sine of what angle is equal to one-half if we could figure this out then we could find out what the value of arc sine one-half is now we know sine pi over six which is the same as 30 degrees is equal to one half but also sine five pi over six is equal to one half as well so then which one do we use does the inverse sine of one half is it equal to pi over six five pi over six or is it both it turns out that the arc sine of one half is equal to pi over six but it does not equal five pi over six now why is that the answer is due to the range of the arc sine function the range is from negative pi over two to pi over two which we mentioned that before and five pi over six is not in this range that's why five pi over six is not the answer pi over 6 is between negative 90 and 90. that is the answer so going to back to the unit circle you need to know that the arc sine function exists in quadrants 1 and four so if this value is positive it's going to be in quadrant one because sine is positive in quadrant one if we had a negative value we need to use the answer in quadrant four and we have to pick an answer that's within the range from negative pi over two to pi over two so even though quadrant four is between 270 and 360 that is not in this range you need to use negative 90 to 90. so keep that in mind also you can confirm this with your calculator if you type arc sine of one-half and if it's in degrees you should get 30 degrees you won't get 150 which is 5 pi over 6. if you put in radians then you should get pi over six or maybe the decimal form of pi over six so arc sine does not exist in quadrants two and three if you have an answer in these quadrants you will get it wrong now let's work on some other examples what is arc sine of the square root of three over two so go ahead and find the answer now we know that sine of let's use degrees sine of 60 degrees is equal to the square root of three over two also sine of 300 is also root 3 over 2. and at the same time i take that back sine of 300 is negative square root 3 over 2. sine of 120 is positive root 3 over 2. so we have two possible answers 60 and 120 so which one should we use now 60 is in quadrant one which is okay 120 is in quadrant two and the arc sine function does not exist in quadrant two so the answer is 60 degrees which we can write as pi over three in mind pi is 180 180 divided by 3 is 60. so that's the answer for this example arc sine of root 3 over 2 is 60. now what about arc sine of negative one-half what's the answer for that so sine of what angles is equal to negative one-half let's make a list and then let's eliminate the wrong answers so we know sine 30 is one half and sine 150 is one half sine 210 is negative one half 210 150 they all have a reference angle of 30 and 210 is in quadrant three sine of 330 which is in quadrant four that's also negative one half and also sine of negative 30 that's in quadrant four as well that's negative one half so which one is the correct answer so we know that the inverse sine function exists in quadrants one and four it does not exist in two and three so 210 is in quadrant three so that answer is eliminated now 330 is in quadrant four and negative 30 is also in quadrant four so this is 330 and this is negative 30. so which one is the answer because they both exist in the appropriate quadrant now don't forget about the range of the arc sine function it's from negative pi over two to pi over two or negative ninety to positive ninety three thirty is not in that range three thirty is the same as eleven pi over six so your answer has to be between negative 90 degrees and 90 degrees so this answer is within the appropriate range and it's in the right quadrant so therefore sine arc sine of negative one-half is negative 30 or negative pi over six that's the answer that we want now let's work on another example go ahead and evaluate arc sine of negative square root 2 over 2. so sine of what angles is equal to negative root 2 over 2. we know that sine 45 is equal to the square root of 2 over 2 but it's a positive answer not a negative answer sine of 225 is equal to negative root 2 over 2 and sine of 315 is also equal to that and the coterminal angle of 315 which is negative 45 sine negative 45 is negative root 2 over 2. now we're going to eliminate this answer because that's in quadrant 3 and the inverse sine function does not exist there and we're going to eliminate this answer because it's not between negative 90 and 90. so this is the only answer that is within the range of negative 90 to 90 degrees and it's in quadrant 4. so arc sine of negative square root two over two is therefore equal to negative 45 degrees which is the same as negative pi over four now what about these two what is arc sine of zero arc sine of one and the inverse sine of negative one find the values of these three things now we know that sine of zero is equal to zero so this answer is zero degrees zero degrees it's within the range of negative pi over two to pi over two now arc sine of one is pi over two so that's also in the range of the arc sine function arc sine of negative one is negative pi over two and so those are some other values that you want to be familiar with now let's move on to evaluating inverse cosine functions so let's evaluate inverse cosine one half so what's the answer for this problem find the exact value of this function well we know that cosine of pi over three that's cosine of sixty degrees is equal to one-half cosine of five pi over three which is in quadrant four is also equal to one-half cosine is positive in quadrants one and four but negative in quadrants two and three so which one is the answer is it pi over three or five pi over three now recall that the range of a cosine function i mean of a r cosine function is from zero to pi so therefore our cosine only exists in quadrants 1 and 2. so if you have a positive value like what we do in this case like what we have here you want to make sure your angle is between 0 and 90. if it's negative it has to be between 90 which is pi over 2 and 180 which is pi so cosine or our cosine does not exist in quadrants three and four so just keep that in mind therefore we can't use five pi over three that's in quadrant four so our cosine one half is equal to pi over 3 which is the same as 60 degrees and so that's the answer now what about this one what is the inverse cosine of negative square root 3 divided by 2 so take a minute and try that problem cosine of 30 degrees is positive square root 3 over 2 but we need a negative value cosine is negative in quadrants two and three cosine one fifty is negative root three over two and cosine two ten is also negative root three over two we can eliminate 30 because it's not negative it's positive and we can eliminate 210 because it's not in the range of zero to pi 210 is in quadrant three and the arc cosine does not exist in quadrant three 150 on the other hand is in quadrant two and the r cosine function exists in quadrants one and two and also 150 is in the range it's between 0 and 180 so 150 is the answer which is the same as 5 pi over 6 so arc cosine negative root 3 over 2 is equal to 5 pi over 6. go ahead and try this one arc cosine negative square root two over two so if you have a negative value you need to choose the angle in quadrant two now we know that cosine of 45 degrees is positive square root two over two but it's not negative so but this is the reference angle that we need in quadrant two the angle is 135 which has the same reference angle as 45 and this is negative square root 2 over 2. so therefore our cosine negative root 2 over 2 is 135 and this answer is in quadrant two it's between zero and pi and this is the same as three pi over four so that's the answer for this problem now go ahead and try these arc cosine of zero arc cosine one and arc cosine negative one and remember the range is from zero to pi now we know that cosine of pi is equal to negative one cosine of pi over two is zero cosine of three pi over two is zero and cosine of zero is one so our cosine of zero must be pi over two based on this one here if cosine pi over two is zero arc cosine of zero is pi over two and pi over two is within the range so that's the answer now what about arc cosine of one well cosine zero is one so therefore arc cosine one must be equal to zero and zero is included in the range so that answer is acceptable three pi over two is not between zero and pi three pi over two is 270 degrees it's not between zero and one so we can get rid of that one now pi is in the range if cosine pi is equal to negative one arc cosine of negative one is pi so these are the answers now let's talk about how to find the exact value of our tangent functions so let's start with the arc tangent of zero tangent of what angle is equal to zero how can we figure this out it turns out that tangent of zero degrees is equal to zero so here's how we know that we know that tangent of an angle is equal to y over x and at zero degrees we have the point one comma zero so y is zero x is one zero divided by one is zero so tangent of zero degrees has a value of zero so this is the angle and this is the value when dealing with inverse trig functions you have the value on the inside and you're looking for the angle so the angle is zero degrees now what about this one what is the inverse tangent of one what's the answer for that so tangent of what angle is equal to one this occurs when y and x have the same value at an angle of 45 degrees we have the point square root 2 over 2 comma square root 2 over 2. so tangent of 45 is equal to y over x and whenever you divide two numbers that are the same they will equal to one so our tangent of one is 45 degrees which is the same as pi over four that's the answer now if arc tangent of 1 is equal to pi over 4 what is the value of arc tangent of negative 1 now you need to know that the inverse tangent function has a range that's similar to the inverse sine function and that is from negative pi over 2 to pi over 2. so if we were to draw the unit circle the inverse tangent function like the inverse sine function exists only in quadrants one and four between negative pi over two and pi over two the inverse cosine function exists in quadrants one and two that is between zero and pi so the only answer that's going to give us negative one is at 45 degrees that is negative 45 in quadrant four so this has to be negative pi over 4 which is in this region right here we can't use 7 pi over 4 because that's not in this range so we have to use negative pi over 4. now what is the value of the inverse tangent of the square root of three so because we have a positive value the answer has to be in quadrant one so let's use the 30 16 90 reference triangle for this example across the 30 is 1 across the 60 is root 3 across the 90 is 2. we know tangent is opposite over adjacent if we choose the 30 degree angle opposite whoa is one adjacent to it is root three one over root three is not the same as the square root of three so therefore the angle that we have to use is 60. opposite to 60 is the square root of 3 adjacent to 60 is 1. so tangent of 60 degrees is equal to the square root of 3. so arctan of square root 3 is 60 degrees or pi over 3 which is in quadrant one if you get an answer in quadrant one it's okay because arc tan arc sine are cosine they can all exist in quadrant one now what about the arc tangent of negative square root three divided by three what is that equal to so tangent of what angle is equal to negative root 3 over 3. now if tangent 60 is the square root of 3 we know tangent 30 has to be root 3 over 3. so 30 is the reference angle but we need it to be in quadrant 4. so it's going to be negative 30. we can confirm it tangent of negative 30 which is negative pi over 6 is equal to sine of negative pi over 6 or negative 30 divided by cosine of negative 30. sine negative 30 is negative one-half sine is negative in quadrant four but cosine is positive cosine negative 30 is root three over two if we multiply the top and bottom by two we can get rid of these and so we're left with negative one over the square root of three now we have to rationalize it so this becomes a negative square root of three over three therefore tangent of negative pi over six is equal to negative square root three over three so the arctan of negative root three over three has to be negative pi over six this exists in quadrant four and it's within a range of negative pi over two to pi over two so that's the answer so now you know how to evaluate inverse trigonometric functions so to review remember this arc cosine exists in quadrants one and two so that's with a range of 0 to pi so as long as your answer is between 0 and pi or within quadrant 1 and 2 you should be okay when evaluating our cosine functions the arc cosine function does not exist in quadrants three or four so i'm just going to put an x in blue now for arc sine which i'm going to use red arc sine exists in quadrants one and quadrant four so that's within a range of negative pi over two to pi over two so arc sine does not exist in quadrants two and three and then finally our tangent is the same for arc sine so arctan exists in quadrants four and one and that is between negative pi over two and pi over two it does not exist in two and three none of them exist in quadrant three so you should never use an angle there so remember arc cosine has a range of zero to pi so your answer has to be in that range for arc sine and arc tangent when getting an answer it has to be between negative pi over two and pi over two now let's say if we have a composition of an inverse trig function what should we do so for example let's say if we want to evaluate arc sine of sine pi over three what's the answer for this particular example it turns out that arc sine and sine will cancel giving us an answer of pi over three and we can do that because pi over three is in quadrant one and arc sine exists in quadrant one but let's take it one step at a time so first let's evaluate sine pi over 3 or sine 60 degrees so using the unit circle or using the 30 60 90 reference triangle you know that sine of 60 is equal to the opposite side which is square root 3 divided by the hypotenuse 2. so sine 60 or sine pi over 3 is the square root of 3 over 2. now we need to evaluate arc sine arc sine exists in quadrants one and four now because this is positive we don't want the answer in quadrant four because arc sine is negative in quadrant four so therefore the angle has to be 60. it's going to be pi over 3 because that is in quadrant 1 and arcsin exists there so in this case the final answer is simply pi over 3. we were able to cancel these two you