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# Trig challenge problem: maximum value

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the maximum value of the expression one over sine squared of theta plus three sine theta cosine theta plus five cosine squared of theta is let's rewrite this, so this is one over one over, so I have a sine squared of theta I have a sine squared of theta and then at least my brain whenever I see a sine squared of theta I always look for cosine squared of theta because I know when I take the sum of them it equals one I don't have just one cosine squared of theta here at five cosine squared of thetas let me just take one of them so I have a plus cosine the squared theta because I took one of them I only have four of these left so plus four cosine squared theta and then I have this stuff plus three sine of theta of cosine of theta so what this first step allowed me to do is just turn these two characters right over here sine squared theta plus cosine squared theta that is equal to one so we've simplified it to one over one over one plus now let's think about how we can write cosine squared of theta I'll write our identities here cosine squared of theta we've proved these in the trigonometry playlist This is equal to one plus cosine one plus cosine of two theta all of that over two and my goal here is that I really just wanna get everything well I just wanna simplify and maybe we will do calculus to find the minimum value of the denominator which will give us the maximum value of the numerator so let's say cosine squared theta is equal to this so four times this is just going to be so four times this before divide by two would be two times this numerator so it's going to be two plus two cosine of two theta that's this term right here and then this term this term right over here we could use the trig identity that sine of two theta is equal to two sine of theta cosine of theta or you divide both sides by two you get one half sine of two theta is equal to sine of theta cosine of theta so this is this right over here this part right over here is going to be one half sine of two theta multiplied by three so it's going to be plus three halves sine of two theta let's see this part over here clearly simplifies this is the three I'm gonna rewrite here this is one over three plus two cosine of two theta plus three halves sine of two theta now we really just looking for the minimum we're looking for the minimum value of the denominator which would give us the same as the maximum value of the numerator or just be one over at this minimum values let's see how how low we can get assuming were above zero let's see how low we can get for this denominator right here it's gonna look for its minimum value so one that we can do just to simplify things the minimum value of this is going to be the same the min The min of this thing I don't want to write it here because it can confuse the problem the minimum value of three plus two cosine two theta plus three halves sine of two theta is going to be the same thing as the minimum value of three plus I'm just gonna do the substitution that two theta is equal to X it simplify things a little bit you don't have to do that so three plus two cosine of X plus three halves sine of X so this is a pretty simple expression to see how we can figure out its minimum value and my temptation is to take the derivative find out where the derivative is equal to zero and then that will either be a maximum or a minimum point so let's take the derivatives the derivative of this expression right over here with respect to X would derivative the derivative of 3 is zero the derivative of two cosine of X is negative negative two sine of x derivative of three halves sine of X gonna be plus three halves cosine of x and that is going to be equal to zero we want to find where the slope of the zero because it's where the maximum or minimum point and let's see we can add we can add two sine of X of both sides so we get three halves cosine of X is equal to two sine of x and then we can divide both sides both sides of this equation by will by two thirds because I don't want to get too many steps so three fourth cosine of x is equal to sine of X and we can divide both sides by cosine of x so we get three over four is equal to sine of x over a cosine of x which is the same thing as the tangent of X so uh... the X value that gives us three or the tangent of the X sorry it's the three fourths is going to give us either a maximum or a minimum point so let's think about this let's think about this a little bit let me draw my unit circle so let's think about that two that two X value that will give us a tangent of three fourths so let we draw my unit circle that's a unit circle Let me draw a unit done one this is always the hardest part so let me draw this... ah! right there is my unit circle so how can I get a triangle or or well let's think about that way how can I get a triangle who were an angle is tangent of three fourth and remember tangent is opposite over adjacent right tangent is opposite over adjacent so if this is my triangle right over here if this is X opposite over adjacent is equal to three fourths so opposite could be three an adjacent could be four and we hopefully immediately recognize this this is the this is a three four five triangle because it's the right triangle three square plus four square is twenty-five which is five squares as this is three four five triangle now there's two tangent values so x could be like this and this obviously isn't isn't a unit hypotenuse over here but we can divide everything by five it would be so we could have this situation we could have this situation over here where this is X this is the unit circle the hypotenuse is one this is three over five and this is four over five this word tangent of X here tangent of X would give us a three fourth but it was going to give us a maximum value or a minimum value well over here both cosine both cosine of X and sine of x are going to be positive so both of these are going to be positive values so it's going probably maximize maximize the expression over here by the other x that gives us the same tangent remember the tangent is really just the slope of the radius of the unit circle would be the angle would be this angle this is the same this has the same tangent value so this X this x over here this X and this case the tangent is still the tangent is still going to be three fourths but over here the sine and cosine are negative so over here over here the X coordinate or the cosine is going to be negative four fifths and the sine value or the Y values is going to be negative three fifths and this will give us this will give us a minimum point because here everything is every...well, the sine and the cosine are negative so let's use let's use this x right over here and notice we don't even have to figure out what the X is because we know that if tangent over here is three fourths the either of both the sine is going to be the sine of the three fifths and the cosine is going to be four fifth or which will give some actual point or the tangent could be three fourths and then the sine will be negative three fifths and cosine will be negative four fifths so let's use these over here so the minimum is going to be equal to three plus when two times cosine of the X we're using this one over here so the two times cosine of X cosine of x here is negative four fifths negative four fifths and then plus three halves plus three halves times the sine of X sine of x here is negative three fifths negative three fifths and what is this going to be equal to this is going to be equal to three plus this is negative eight fifths three you should write three minus eight fifth three minus eight fifths minus nine tenths minus nine tenths and so this is going to be equal to we could put everything over ten thirty over ten minus sixteen over ten right that's eight fifths minus nine over ten and this gives us what this gives us five over ten five over ten or or one-half so the minimum value the minimum value for our denominator everything we've been dealing so far has been our denominator the minimum value of all of this business over here the minimum value is one half so the maximum value that this whole expression takes is when the minimum values one half so we get one over one half which is equal to two and we're done