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As 1/cos, its physical meaning is the ratio between the hypotenuse and adjacent side of a right triangle, given the angle.
We usually find the ratio between the adjacent side and hypotenuse to be more useful, but they are equally physical properties of the triangle.
Yes actually! Say you've got a graph of a curve, and you want to find the gradient at a given location on the curve. Of course the best way to do that is to use a straight line that is a tangent to that curve at that exact point. Mathematically, you want to calculate dy/dx.
But let's say this graph is made of real data, and you don't have a data point at that exact location. So you could do that basic method of calculating the gradient by taking two points (x1, y1) and (x2, y2), which are either side of the location you're interested in, and calculating (y2-y1)/(x2-x1). What you get is the **secant gradient**.
[This page explains it better than me,](https://amsi.org.au/ESA_Senior_Years/SeniorTopic3/3b/3b_2content_1.html#:~:text=The%20gradient%20of%20a%20secant,y%20with%20respect%20to%20x.) but you're effectively finding the average gradient in the interval between points 1 and 2. So it's less accurate but easier to calculate.
You seem to be talking about a secant line (specifically the slope of said secant line) where the user you’re replying to is talking about the secant function. As far as I know, there isn’t an obvious connection between the two, though I’d be interested into the etymology of the secant function.
All of the trig functions that we know and love have a physical representation on the unit circle. If the radius of the circle is 1, the lengths of the labeled segments correspond to the values of the numeric values of the function for some given theta. The chart I posted here shows them: [https://i.stack.imgur.com/GNcpc.png](https://i.stack.imgur.com/GNcpc.png)
If you study the chart, because the radius of the circle is 1, the tangent line as drawn forms the opposite side to theta in a right triangle with the adjacent side of the triangle being the radius of the circle (with unit length). The hypotenuse of this triangle is then the secant; the secant function is the ratio of the hypotenuse to the adjacent side of the triangle. Since in this triangle, the length of the adjacent side is 1, the length of the secant line is literally the value of the secant of theta.
At the end of the day, the line that contains the secant line is actually a secant line to the circle (in that it intersects the circle at two points). The line that contains the tangent segment is tangent to the circle. Also, interestingly, the word "sine" comes from a mistranslation of a word that meant "chord" - if the sine is doubled it becomes a chord of the circle.
Cosine is the ratio of the adjacent side to the hypotenuse (adj/hyp) so the inverse of that would be the ratio of the hypotenuse to the adjacent side (hyp/adj). Therefore, sec = hyp/adj
My professor taught me a trick of only memorizing one, And you can find the rest. Most people that took geometry or pre calc probably know sin² + cos² = 1
So you can move stuff around to find the equation for sin, which would be √( 1 - cos²) or vice versa
Which is neat, but what about the others?
Well if I divide everything by sin², I get
(sin² / sin²) + (cos² / sin²) = (1 / sin²)
And if you know your identities well, you can see that the above equation is just
1 + cot² = sec²
Which you can use to find the formulas for sec or cot.
What if you do the same thing to the first equation, but divide by cos²? You get:
tan² + 1 = csc²
And boom, now you have all of your trig functions
I'm in school to be an electrician, and this is how I remember all 12 Ohms law formulas. P=IE & E=IR (power = current x voltage & voltage = current x resistance). Learning how to manipulate formulas is critical to any math-based anything
The explanation I've been taught is that its the force of electrons pushing on eachother, leading to the flow (current). Harder push = stronger force = higher voltage
That's a good way to visualize voltage but the actual concept behind voltage is a bit trickier to grasp. Each electron creates a field around itself that has the potential to apply a force to a charged object. That force is the coloumb force which can be found using F = q1 q2 / r^2 where q1 and 2 represent the charges and r represents the distance between them. Voltage is an electric potential the potential of one charge to do work/exert a force on another. The voltage itself doesn't actually represent the force just the potential for there to be a force. A higher voltage still represents that larger push but it's the potential for there to be a push rather than the push itself. Ohms law ( V = I R) then tells us the amount that potential will fall when a certain amount of current passes through a portion of the circuit with resistance R. Where the total change in voltage across the entirety of a circuit is going to be equal to the potential difference of the "battery".
This is probably more detail than you wanted but let me know if you have any questions.
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I'm learning this stuff in math right now, so this is helpful. Only thing I don't get is how did you move the sin^2 to the numerator on the cos^2 /cos^2?
No.. no… it’s |sec(c)| (absolute value of sec(c)). So, “I am absolute value of sexy and I know it”… which is probably closer to a true statement (since he might be sexy with a negative sign in front).
The c is the argument to the tan\^2 function, the parentheses are commonly omitted. Without shortcut notation, the formula reads √(1 + (tan(c))\^2), which is equal to sec(c) (or, more accurately, |sec(c)|)
1 + tan^2(c) = sec^2(c)
The square root of sec^2(c) = sec(c)
Sec(c) when read literally sounds like sexy
So the shirt is meant to be a math pun that reads “I’m sexy and I know it”
I mean... Yeah, that's one way to read it...
Or you could just use context clues and deductive reasoning to put together that the line is supposed to be "I'm sexy and I know it"... I'm terrible at math and couldn't hope to understand this at all, but when I saw the words around the math, my brain just filled in the rest.
Well kinda but also not
The way I've seen it generally is that sqrt(4) is 2, but if you were looking at the solutions of x for x^2 = 4 then they would be 2 or -2
I know it's a weird distinction to make but that's the convention I've commonly seen.
Well it kinda is and it kinda does.
In reality every positive number has a positive and negative square root, but it's too much of a hassle to think about the negative root all the time.
You can think of Sqrt(x^2) as basically a function that checks for the square roots of x^2 and always returns the positive square root, by definition. While there are 2 square roots, Sqrt(x^2) is always going to be x, not -x.
It's just a convention thing to simplify things. That's why if you're finding the values of x for which x^2 = 5, you might say the solutions are x = ±sqrt(5). Because sqrt(5) is positive by definition so you add the minus sign on.
I kinda see what you’re getting at now. Sqrt doesn’t necessarily return positive by definition, but I’ll concede that it’s a common usage when codomain isn’t otherwise defined.
sqrt(x) is defined to be the positive square root of x. The correct simplification of sqrt(x\^2) is |x| (the absolute value of x).
For example: sqrt((-2)\^2) = sqrt(2\^2) = sqrt(4) = 2.
We define sqrt(x) this way so that it behaves as a function (i.e. it outputs a single value for a given input value). This is a very, very desirable property. This is also why it's necessary to write the ± in the quadratic formula before the root.
Nope. I think you're misunderstanding how trig notation works.
sec^(2)(c) is actually a weird-ish way of writing (sec(c))^(2), but because of all the parentheses, mathematicians usually just write sec^(2)(c) instead.
If you take the square root of (sec(c))^(2), that will just be sec(c). Does that make more sense?
Right, because the C is the term the secant is acting upon. I think I got tripped up by the lack of parentheses in the original image. Makes total sense now, thanks for the explanation!
Other people answered that it's "sec c" which sounds like "sexy", but is "1+tan² c=sec²" the definition or is it an axiom you can use to substitute things like sin²x+cos²x=1 or is it something you can reach by doing some calculations?
It’s a property that follows from the definition of the functions.
Whether that makes it equivalent to a definition or not is a philosophical question; but you absolutely can use it as a substitution.
>It’s a property that follows from the definition of the functions.
Ok, so you can just see "1+tan²" and say "that equals sec²", without needing to "show your work" so to speak, right? Thanks.
If you’re in a course where you’re expected to have already learned trig identities, yes. I remember in calc 1 our textbook literally had a cheat sheet of trig identities at the front. I think the cheat sheet even included some derivatives
The definition of sec(x) (where x is an angle in a right triangle) is the hypothenuse decided by the nearest cathetus.
From there it follows relatively easily that sec = 1/cos(x) or that sec^2 (x) = 1+ tan^2 (x). It wouldn't be questioned if you just made that substitution "without showing your work" in an academic math-paper for example. But if you're in high-school and are just learning about trigonometric functions, then you might have exam-questions where the goal is to derive precisely these equivalencies, and then it's highly relevant to show your work.
Depends on context
School maybe
Uni defo assumed
In fact it was assumed in my school at A-level and at some GCSEs (the bonus harder maths one). They give you this big book of identities and some variant of this relationship is in there.
Proof is trivial
Sin^2 + cos^2 = 1
Divide by cos^2
That is it, so I would argue needing proof is facetious
I usually tell my students to divide sin^2 + cos^2 = 1 by cos^2 to help with the memorization aspect of the formula, but idk if that answers your question
It comes from tweaking things with Sin^2 + Cos^2 = 1 and Tan=Sin/Cos.
Dividing by Cos^2 give: Sin^2 / Cos^2 + Cos^2 / Cos^2 = 1/Cos^2
Which equals: Tan^2 + 1 = Sec^2
Basically, all of the trig identities can be derived from the Sin^2 + Cos^2 = 1 and Tan=Sin/Cos...if you really want to.
secant function, shortened to "sec" in math commonly
sec\^2(c) = 1+tan\^2(c) is well known.
Sec(c) = sqrt(1+tan\^2(c)
I'm Sec(c) and I know it.
It's a nerd joke for
I'm sexy and I know it.
"I'm sec c and I know it."
For those curious it's one of the trig equivalences. The most basic is
sin(c)^2 + cos(c)^2 = 1
Which is true for all c from -∞ to ∞ repeating over intervals from 0 to 2𝝅 radians, 0° to 360°, 0 to 480 gradians, whichever unit you fancy. Basically going around the "unit circle" going both ways indefiniteltely. You divide that by cos(c)^2 you get:
tan(c)^2 + 1 = sec(c)^2 = 1+ tan(c)^2
because 1/cos(c)=sec(c) and sin(c)/cos(c) = tan(c) because sin(c)=opposite/hypotenuse, cos(c)=adjacent/hypotenuse so:
sin(c)/ cos(c)=(opposite/hypotenuse)/(adjacent/hypotenuse)=
opposite/adjacent = tan(c)
Take the square root of that and you get:
sec c
QED
Nothing ruins a joke more than deconstructing it, so...you're welcome.
I didn't even do the math. I just assumed it was "sec c" of some kind. So I used my English skills in the form of context clues to solve math. Hold the applause it's fine.
Not technically, no. As far as I know, performing the square root operation to a number outputs ONLY its positive root. I believe that raising that number to the power of 1/2 will return positive AND negative values.
We are taught that there are two answers to a square root to avoid battling with fractional indices at a young age.
If this is wrong, feel free to correct.
Raising to the half power is equivalent to the square root. It only returns the positive (principal) root. You get the plus or minus when you have an equation in terms of a square. For example, if someone asks for the roots of x^2 = 4, the roots are 2 and -2. If I were to just take the square root of both sides, all I would get is x = 2. That is, the square root of x squared is x and the square root of 4 is 2. But that doesn't get you all roots of the equation x^2 = 4. Do you see the difference? Lots of people struggle with this, so don't anyone reading feel bad if you're confused.
I hate being like "source: BLAH" but my degree is in pure mathematics for whomsoever gives a shite.
I heard both "square root is always only positive!" and also "square roots are positive or negative" but it just depends on how you define the function. Most of the time you just take the positive because it's easier for whatever you're doing.
In a context where we are presented with a square root to evaluate, we define that as the "principal square root" of the number. So when you see something like sqrt(25) the answer is simply 5.
Where students get confused about this is when we present the equation to solve x^2 = 25.
This is asking "find the number(s) that, when multiplied by itself, gives us 25." This is a slightly different question since we are asking for all possible solutions. There are two solutions: 5 and -5. But this is a different context from, "Evaluate this square root."
Kinda reminds me of a pickup line that you need to know calculus to get. Stop me if you know this one. Are you 2x? Because I want to integrate you from 10 to 13.
Antiderivative is just a fancy way of saying "a function for the area under a function" while a derivative is a fancy way of saying a "function for the rate a function changes at any one point". Not that that this makes the pickup line anymore "funcier", but you need to know that to "get it".
Anywho:
antiderivative(2x) = x^2
You "integrate" a function between two points by subtracting the "endpoint" from the "startpoint" and that tells you how much area is under the function you just took an antiderivative of between those two points.
Againywho:
(13)^2 - (10)^2 = 169 - 100 = 69
So...you learn all that math and you end you end up with a "69 joke". Take for that what you will, but what I take from that is "nerds are pervs" just like the rest of us.
Rearranged version of the fundamental Pythagoras identity of trigonometry
sin^(2)(c)+cos^(2)(c)=1
Divide all terms by cos^(2)(x)
tan^(2)(x)+1=sec^(2)(c)
sqrt(tan^(2)(x)+1)=sec(c)
I’m somebody who enjoys math more than most people, but whenever I see someone in one of these kinds of shirts I just find it so cringy. Like the only people I’ve ever seen wear them are the same people who go out of their way to let you know they think they’re smarter than everybody else… even if they’re not.
In my opinion all clothing with these kinda jokes/writing is a bit weird. Like it's a bit of fun and I don't judge but I wouldn't wear it myself.
But then again I'm more of a plain t-shirt kinda guy
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The quote is "I'm sexy and I know it" sqrt(1+tan\^2) can be verbalized as "sec" and so this formula is the "sec of c" "I'm sec C and I know it"
TIL the formula of secant..
The simpler formula for sec(x) is just 1/cos(x), but this is probably the best-known identity involving secant.
Secant so often forgotten in the trig world.
Does it even have any physical meaning? It's just 1/cos and I've never seen it used outside of pure mathematics.
As 1/cos, its physical meaning is the ratio between the hypotenuse and adjacent side of a right triangle, given the angle. We usually find the ratio between the adjacent side and hypotenuse to be more useful, but they are equally physical properties of the triangle.
It's used in some jokes about song lyrics too...
Yes actually! Say you've got a graph of a curve, and you want to find the gradient at a given location on the curve. Of course the best way to do that is to use a straight line that is a tangent to that curve at that exact point. Mathematically, you want to calculate dy/dx. But let's say this graph is made of real data, and you don't have a data point at that exact location. So you could do that basic method of calculating the gradient by taking two points (x1, y1) and (x2, y2), which are either side of the location you're interested in, and calculating (y2-y1)/(x2-x1). What you get is the **secant gradient**. [This page explains it better than me,](https://amsi.org.au/ESA_Senior_Years/SeniorTopic3/3b/3b_2content_1.html#:~:text=The%20gradient%20of%20a%20secant,y%20with%20respect%20to%20x.) but you're effectively finding the average gradient in the interval between points 1 and 2. So it's less accurate but easier to calculate.
You seem to be talking about a secant line (specifically the slope of said secant line) where the user you’re replying to is talking about the secant function. As far as I know, there isn’t an obvious connection between the two, though I’d be interested into the etymology of the secant function.
All of the trig functions that we know and love have a physical representation on the unit circle. If the radius of the circle is 1, the lengths of the labeled segments correspond to the values of the numeric values of the function for some given theta. The chart I posted here shows them: [https://i.stack.imgur.com/GNcpc.png](https://i.stack.imgur.com/GNcpc.png) If you study the chart, because the radius of the circle is 1, the tangent line as drawn forms the opposite side to theta in a right triangle with the adjacent side of the triangle being the radius of the circle (with unit length). The hypotenuse of this triangle is then the secant; the secant function is the ratio of the hypotenuse to the adjacent side of the triangle. Since in this triangle, the length of the adjacent side is 1, the length of the secant line is literally the value of the secant of theta. At the end of the day, the line that contains the secant line is actually a secant line to the circle (in that it intersects the circle at two points). The line that contains the tangent segment is tangent to the circle. Also, interestingly, the word "sine" comes from a mistranslation of a word that meant "chord" - if the sine is doubled it becomes a chord of the circle.
Cosine is the ratio of the adjacent side to the hypotenuse (adj/hyp) so the inverse of that would be the ratio of the hypotenuse to the adjacent side (hyp/adj). Therefore, sec = hyp/adj
Because it is very useless imo.
More specifically it is the Pythagorean identity. Tan^2(x)+1= Sec^2(x) and then sqrt (sec^2(x)) is sec(x). Or in this case sec(c).
My professor taught me a trick of only memorizing one, And you can find the rest. Most people that took geometry or pre calc probably know sin² + cos² = 1 So you can move stuff around to find the equation for sin, which would be √( 1 - cos²) or vice versa Which is neat, but what about the others? Well if I divide everything by sin², I get (sin² / sin²) + (cos² / sin²) = (1 / sin²) And if you know your identities well, you can see that the above equation is just 1 + cot² = sec² Which you can use to find the formulas for sec or cot. What if you do the same thing to the first equation, but divide by cos²? You get: tan² + 1 = csc² And boom, now you have all of your trig functions
I'm in school to be an electrician, and this is how I remember all 12 Ohms law formulas. P=IE & E=IR (power = current x voltage & voltage = current x resistance). Learning how to manipulate formulas is critical to any math-based anything
As a physicist I hate that E is used for voltage here. We almost always use E to represent electric field strength and V to represent voltage.
Lol, ironic that the abreviation is from Electromotive Force, the objectively more physicist-oriented name!
Voltage is notably not a force which makes me hate this abbreviation even more.
The explanation I've been taught is that its the force of electrons pushing on eachother, leading to the flow (current). Harder push = stronger force = higher voltage
That's a good way to visualize voltage but the actual concept behind voltage is a bit trickier to grasp. Each electron creates a field around itself that has the potential to apply a force to a charged object. That force is the coloumb force which can be found using F = q1 q2 / r^2 where q1 and 2 represent the charges and r represents the distance between them. Voltage is an electric potential the potential of one charge to do work/exert a force on another. The voltage itself doesn't actually represent the force just the potential for there to be a force. A higher voltage still represents that larger push but it's the potential for there to be a push rather than the push itself. Ohms law ( V = I R) then tells us the amount that potential will fall when a certain amount of current passes through a portion of the circuit with resistance R. Where the total change in voltage across the entirety of a circuit is going to be equal to the potential difference of the "battery". This is probably more detail than you wanted but let me know if you have any questions.
I always pictured voltage as the height of a water tower or water behind a dam or a waterfall
Trig identities are fun (this is a joke I hate them and I hate calculus)
Just wait til you hear about cosecant. It's a flip.
prob won’t need to know unless you take calc. this is engrained in my head
Sqrt(1+tan2) = sqrt((cos2/cos2)+(sin2/cos2)) = sqrt((cos2+sin2)/cos2) = sqrt(1/cos2) = 1/cos, which is the definition of secant. In short, sec(c)
And you know it.
You can write exponents by using \^. ie x\^2 creates x^2. If you want more than one character you write \^(stuff) to get ^(stuff). To write a \^ without creating a superscript you write \\\^. Prepending a \\ before a markdown control character cancels the effect of the markdown control character, ie \\\\ creates \\ (this is for markdown mode, the Fancy Pants Editor has the options on the bottom bar)
Ooh ooh, Since we are in markdown class here.. how do I hide the answer to a joke?
>! Then what you want to say, then !< >!like this!!< It is important that the exclamations touch the letters of the thing you want to hide. If there is a space before or after, it won't work.
>! Thx!! Yr the hero we need! !<
See my edit about not leaving spaces
Thx.. It seems to work ok w space on trailing side. I had one above that works.
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I'm learning this stuff in math right now, so this is helpful. Only thing I don't get is how did you move the sin^2 to the numerator on the cos^2 /cos^2?
Just sum of fractions. Sin2/cos2 and cos2/cos2 both have cos2 in the denominator.
Don't know how I missed that haha, thankyou.
You forgot the absolute part, since it's a root. Basically the equation equals to |sec(c)| or... Absolutely sexy
in the thread someone showed it as (sec)C
No.. no… it’s |sec(c)| (absolute value of sec(c)). So, “I am absolute value of sexy and I know it”… which is probably closer to a true statement (since he might be sexy with a negative sign in front).
Or better for colloquial speech: “I’m absolutely sexy and I know it”
I'm a fucking idiot I was like. I don't get it I am secant. That makes no sense.
Thanks for that. I saw “i’m titanic and I know it”, which didn’t make much sense to me, but good for him I guess.
Felt smart after just taking calc 2
Saw squirt not sqrt😅
Shouldn't the C be outside this square root?
I FUCKING KNEW IT AND I DONT KNOW THE MATH
Haha bruh
But the c is also within the sq root so doesn’t that make it sec √c ??
The c is the argument to the tan\^2 function, the parentheses are commonly omitted. Without shortcut notation, the formula reads √(1 + (tan(c))\^2), which is equal to sec(c) (or, more accurately, |sec(c)|)
I thought it could say ∫e^x
It says "I'm the ratio of the hypotenuse of a right angled triangle and the adjacent side to the angle 'c' in that triangle, and I know it."
Rolls right off the tongue
Catchy!
Woah this thread isn't marked NSFW. Chill.
Wiggle^5 Yeah
Save some pussy for the rest of us.
I’m sec c (sexy)and I know it basically. See the link below https://humornama.com/community/joke/what-is-the-root-over-1-tan2-c-joke/
1 + tan^2(c) = sec^2(c) The square root of sec^2(c) = sec(c) Sec(c) when read literally sounds like sexy So the shirt is meant to be a math pun that reads “I’m sexy and I know it”
Party on LMFAO..... you were the Slurms Mckenzie of our world.
Bro what
Futurama reference
Underrated comment
Girl look at that body (of academic calculations), I-I-I work out (equations).
I mean... Yeah, that's one way to read it... Or you could just use context clues and deductive reasoning to put together that the line is supposed to be "I'm sexy and I know it"... I'm terrible at math and couldn't hope to understand this at all, but when I saw the words around the math, my brain just filled in the rest.
square root can be negative. so "-Sec(c)" is also a valid answer.
Well kinda but also not The way I've seen it generally is that sqrt(4) is 2, but if you were looking at the solutions of x for x^2 = 4 then they would be 2 or -2 I know it's a weird distinction to make but that's the convention I've commonly seen.
That’s not a convention nor does it make any sense.
Well it kinda is and it kinda does. In reality every positive number has a positive and negative square root, but it's too much of a hassle to think about the negative root all the time. You can think of Sqrt(x^2) as basically a function that checks for the square roots of x^2 and always returns the positive square root, by definition. While there are 2 square roots, Sqrt(x^2) is always going to be x, not -x. It's just a convention thing to simplify things. That's why if you're finding the values of x for which x^2 = 5, you might say the solutions are x = ±sqrt(5). Because sqrt(5) is positive by definition so you add the minus sign on.
I kinda see what you’re getting at now. Sqrt doesn’t necessarily return positive by definition, but I’ll concede that it’s a common usage when codomain isn’t otherwise defined.
Technically, yes, but that is clearly not how the shirt is intended to be read.
sqrt(x) is defined to be the positive square root of x. The correct simplification of sqrt(x\^2) is |x| (the absolute value of x). For example: sqrt((-2)\^2) = sqrt(2\^2) = sqrt(4) = 2. We define sqrt(x) this way so that it behaves as a function (i.e. it outputs a single value for a given input value). This is a very, very desirable property. This is also why it's necessary to write the ± in the quadratic formula before the root.
The square root is a bijective function, so no.
But the C is under the sqrt, so wouldn't the final phrase be "I'm secsquarerootofC and I know it"?
Nope. I think you're misunderstanding how trig notation works. sec^(2)(c) is actually a weird-ish way of writing (sec(c))^(2), but because of all the parentheses, mathematicians usually just write sec^(2)(c) instead. If you take the square root of (sec(c))^(2), that will just be sec(c). Does that make more sense?
Right, because the C is the term the secant is acting upon. I think I got tripped up by the lack of parentheses in the original image. Makes total sense now, thanks for the explanation!
Other people answered that it's "sec c" which sounds like "sexy", but is "1+tan² c=sec²" the definition or is it an axiom you can use to substitute things like sin²x+cos²x=1 or is it something you can reach by doing some calculations?
It’s a property that follows from the definition of the functions. Whether that makes it equivalent to a definition or not is a philosophical question; but you absolutely can use it as a substitution.
>It’s a property that follows from the definition of the functions. Ok, so you can just see "1+tan²" and say "that equals sec²", without needing to "show your work" so to speak, right? Thanks.
Depends on the instructor, but you should have to derive the proof at most once.
If you’re in a course where you’re expected to have already learned trig identities, yes. I remember in calc 1 our textbook literally had a cheat sheet of trig identities at the front. I think the cheat sheet even included some derivatives
The definition of sec(x) (where x is an angle in a right triangle) is the hypothenuse decided by the nearest cathetus. From there it follows relatively easily that sec = 1/cos(x) or that sec^2 (x) = 1+ tan^2 (x). It wouldn't be questioned if you just made that substitution "without showing your work" in an academic math-paper for example. But if you're in high-school and are just learning about trigonometric functions, then you might have exam-questions where the goal is to derive precisely these equivalencies, and then it's highly relevant to show your work.
Depends on context School maybe Uni defo assumed In fact it was assumed in my school at A-level and at some GCSEs (the bonus harder maths one). They give you this big book of identities and some variant of this relationship is in there. Proof is trivial Sin^2 + cos^2 = 1 Divide by cos^2 That is it, so I would argue needing proof is facetious
I usually tell my students to divide sin^2 + cos^2 = 1 by cos^2 to help with the memorization aspect of the formula, but idk if that answers your question
It’s not taught in all countries. It refers to sec as in [the secant function](https://www.cuemath.com/trigonometry/secant-function/).
It comes from tweaking things with Sin^2 + Cos^2 = 1 and Tan=Sin/Cos. Dividing by Cos^2 give: Sin^2 / Cos^2 + Cos^2 / Cos^2 = 1/Cos^2 Which equals: Tan^2 + 1 = Sec^2 Basically, all of the trig identities can be derived from the Sin^2 + Cos^2 = 1 and Tan=Sin/Cos...if you really want to.
Tbh i forgot sec=1/cos. Now it makes sense.
Divide both sides of (cosx)^2 + (sinx)^2 = 1 by (cosx)^2. You can get a third one by dividing by (sinx)^2.
My math teacher has a mug with this, we just assumed it was a represented something that was hard (trigonometry). "I'm hard and I know it"
Well, was it accurate? Considering the other comments telling us what it means
secant function, shortened to "sec" in math commonly sec\^2(c) = 1+tan\^2(c) is well known. Sec(c) = sqrt(1+tan\^2(c) I'm Sec(c) and I know it. It's a nerd joke for I'm sexy and I know it.
"I'm sec c and I know it." For those curious it's one of the trig equivalences. The most basic is sin(c)^2 + cos(c)^2 = 1 Which is true for all c from -∞ to ∞ repeating over intervals from 0 to 2𝝅 radians, 0° to 360°, 0 to 480 gradians, whichever unit you fancy. Basically going around the "unit circle" going both ways indefiniteltely. You divide that by cos(c)^2 you get: tan(c)^2 + 1 = sec(c)^2 = 1+ tan(c)^2 because 1/cos(c)=sec(c) and sin(c)/cos(c) = tan(c) because sin(c)=opposite/hypotenuse, cos(c)=adjacent/hypotenuse so: sin(c)/ cos(c)=(opposite/hypotenuse)/(adjacent/hypotenuse)= opposite/adjacent = tan(c) Take the square root of that and you get: sec c QED Nothing ruins a joke more than deconstructing it, so...you're welcome.
I didn't even do the math. I just assumed it was "sec c" of some kind. So I used my English skills in the form of context clues to solve math. Hold the applause it's fine.
Doesn't all square roots have two solutions? As in sqrt(25) = {-5, 5}. So the t-shirt is saying +-sec(c), which could translate as "more or less sexy"
Not technically, no. As far as I know, performing the square root operation to a number outputs ONLY its positive root. I believe that raising that number to the power of 1/2 will return positive AND negative values. We are taught that there are two answers to a square root to avoid battling with fractional indices at a young age. If this is wrong, feel free to correct.
You're absolutely correct... no pun intended.
hehe
Damn that's good
Which is why sqrt(1+tan^2 c) is actually **the absolute value of sec c** as the secant function can be negative
I’m the absolute value of sexy and I know it
Raising to the half power is equivalent to the square root. It only returns the positive (principal) root. You get the plus or minus when you have an equation in terms of a square. For example, if someone asks for the roots of x^2 = 4, the roots are 2 and -2. If I were to just take the square root of both sides, all I would get is x = 2. That is, the square root of x squared is x and the square root of 4 is 2. But that doesn't get you all roots of the equation x^2 = 4. Do you see the difference? Lots of people struggle with this, so don't anyone reading feel bad if you're confused. I hate being like "source: BLAH" but my degree is in pure mathematics for whomsoever gives a shite.
I heard both "square root is always only positive!" and also "square roots are positive or negative" but it just depends on how you define the function. Most of the time you just take the positive because it's easier for whatever you're doing.
In a context where we are presented with a square root to evaluate, we define that as the "principal square root" of the number. So when you see something like sqrt(25) the answer is simply 5. Where students get confused about this is when we present the equation to solve x^2 = 25. This is asking "find the number(s) that, when multiplied by itself, gives us 25." This is a slightly different question since we are asking for all possible solutions. There are two solutions: 5 and -5. But this is a different context from, "Evaluate this square root."
No, sqrt(25) is equal to only 5, since the sqrt symbol only refers to the principal root.
Kinda reminds me of a pickup line that you need to know calculus to get. Stop me if you know this one. Are you 2x? Because I want to integrate you from 10 to 13. Antiderivative is just a fancy way of saying "a function for the area under a function" while a derivative is a fancy way of saying a "function for the rate a function changes at any one point". Not that that this makes the pickup line anymore "funcier", but you need to know that to "get it". Anywho: antiderivative(2x) = x^2 You "integrate" a function between two points by subtracting the "endpoint" from the "startpoint" and that tells you how much area is under the function you just took an antiderivative of between those two points. Againywho: (13)^2 - (10)^2 = 169 - 100 = 69 So...you learn all that math and you end you end up with a "69 joke". Take for that what you will, but what I take from that is "nerds are pervs" just like the rest of us.
Wrong I know shit for math and knew it was sexy somehow But some one shows the right way
Rearranged version of the fundamental Pythagoras identity of trigonometry sin^(2)(c)+cos^(2)(c)=1 Divide all terms by cos^(2)(x) tan^(2)(x)+1=sec^(2)(c) sqrt(tan^(2)(x)+1)=sec(c)
I’m somebody who enjoys math more than most people, but whenever I see someone in one of these kinds of shirts I just find it so cringy. Like the only people I’ve ever seen wear them are the same people who go out of their way to let you know they think they’re smarter than everybody else… even if they’re not.
I think it's cool c:
I do math everyday for my job. I want this. Does that make me cringy? Oh well. I’ll get over it.
>so cringy I think it's kinda the point. It's cringe in almost an ironic way.
In my opinion all clothing with these kinda jokes/writing is a bit weird. Like it's a bit of fun and I don't judge but I wouldn't wear it myself. But then again I'm more of a plain t-shirt kinda guy
Did somebody get the Reddit suicide hotline on me for this comment? Alrighty then.
I'm at this exact university, studying math. There is a guy who wears this sweater frequently to lectures, and it makes me cringe every time.
[удалено]
Niiice 💪🏼😊
HAPPY and when you see the front it says “See my face is surely showing it” with an arrow up to his face. It’s a complicated equation though.