 We're going to continue talking about inverse trig functions in particular a Trig function and its inverse cancel each other out However the following two conditions must hold before you can do this first You have to be sure that the innermost function whatever it may be is Define that the given value Next you must be sure the given value is in the range of the outer function These are the two conditions we will check Before cancelling out a trig function and its inverse first example Find the value of each of the following inverse cosine of cosine of negative 2 pi over 3 before I go any further Remember cosine has a special property I'm focusing on the cosine on the inside Cosine is an even function therefore it eats negative signs So my expression is actually equal to What's on the screen here? Now let's check those two conditions before we can cancel out. We have to check the two conditions First and foremost is cosine of 2 pi over 3 defined Does it give you a real number answer and the answer is yes You can take the cosine of any angle and it'll always be defined second Look at cosine inverse on the outside is 2 pi over 3 in the range of cosine inverse And the answer to that is yes 2 pi over 3 is in the range It's in the range of Of inverse cosine because remember the range of inverse cosine is Zero to pi so as cool as it may seem The inverse An original function cancel each other out leaving you with an answer of 2 pi over 3 Now you might be wondering well I could have just cancelled out the inverse and the actual cosine function to begin with and The answer is no you have to check these good two conditions always Part B. You may want to go ahead and try canceling out the two functions, but no don't do that You must check your first condition condition number one Is tangent of 3 pi over 2 defined that would be 270 degrees that would be 270 degrees That's the bottom portion of the unit circle which has ordered pair Zero negative one Tangent of 3 pi over 2 That would be Sign over cosine negative 1 over 0 And guess what that is not defined So I can't pursue this question any further Since the inner trig function is not defined at 3 pi over 2 The answer is undefined so condition one was not met Now Time for a brain teaser first Sign of 5 pi over 8 is it defined? The answer is yes. Of course. It is sign is defined for any at any angle So sign of 5 pi over 8 is defined however The range so now I have to look at the range of the outer function of inverse sign the range of inverse sign is Negative pi over 2 the pi over 2 and my question for you is does 5 pi over 8 fall in that range and it does not so the range of sign inverse is negative pi over 2 to pi over 2 but 5 pi over 8 is in Quadrant 2 and this doesn't mean we stop and we give up because furthermore the sign of An angle in quadrant 2 sign is Positive and quadrant 2 the sign of an angle The sign function of an angle in quadrant 2 has the same value as an angle Quadrant 1 Because sign is positive and quadrant 1 sign is positive and quadrant 2 so it doesn't matter Where the angle is you're gonna get a positive answer for sign either way That's because sign is positive in both quadrants 1 and 2 so what I have to do to make this Inverse trig expression defined is I have to move 5 pi over 8 to quadrant 1 I Have to move 5 pi over 8 the quadrant 1 In other words, I'm literally taking a quadrant 2 angle and finding the reference angle in quadrant 1 So I have pi minus 5 pi over 8 Which is 8 pi over 8 minus 5 pi over 8 Which gives me a quadrant 1 angle of 3 pi over 8 So 5 pi over 8 3 pi over 8 when you apply a sign to these functions You get the same answer either way So we're going to take 5 pi over 8 and Exchange it for 3 pi over 8 replace 5 pi over 8 with 3 pi over 8 Now Yes, sign is defined at 3 pi over 8. Yes 3 pi over 8 is in the range of inverse sign now We can cancel out The inverse sign function and sign function and the answer here is 3 pi over 8 So that one had a little bit of a trick to it. We had to take that inside angle measure and move it to a Angle measure which was defined in the range of the inverse sign function Now just for fun, let's find the inverse of the trig function f of x equals 5 sine x minus 6 We'll state the domain and the range of the inverse So first things first I have the function f of x equals 5 sine x minus 6 I Need to find the domain of the original function and Then the range of the original function Remember I'm finding an inverse. So it's important that I find a domain that is 1 to 1 So my domain has to be negative pi over 2 to pi over 2 my range Normally my range is negative 1 to 1 for the sine function, but we have some transformations happening here We're taking our answer of the sine function Which is normally between negative 1 and 1 and we're multiplying it by 5 then we're taking the way 6 by order of operations So I'm going to multiply negative 1 and 1 both by 5 That gives you negative 5 comma 5 and then I'm going to take away 6 from each Which gives me negative 11 to negative 1 That would be the domain in the range of this original function now time to find the inverse Well to find the inverse I'm first going to write my function in terms of x and y Because the key to finding an inverse is to switch x and y x and y switch places X and y switch places Then I want to solve for y So add 6 to both sides x plus 6 equals 5 sine y Divide both sides by 5 and you get x plus 6 Over 5 is sine y Then all I have to do is apply the inverse sine function to both sides So you get sine inverse of x plus 6 over 5 is equal to y Sine inverse and sine will cancel each other out on the right-hand side So my inverse function f inverse of x is Sine inverse of x plus 6 over 5 That is the inverse trig function Now time for The domain and time for the range of the inverse And there is really no math you have to do here because literally you can take the domain of the original and It's equal to the range of the inverse So the range of the inverse is negative pi over 2 to pi over 2 and The range of the original is equal to the domain of the inverse So negative 11 to negative 1 and This answers the question we have the inverse trig function We have the domain and range of the original function then we have the domain and the range of the inverse and That concludes our discussion on inverse trig functions for now More to come soon. 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