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1/24/2017 10:24:45 PM
13
Please Help Me With Math
Can someone explain the trigonometric ratios? (I think that's what they're called; they're sine, cosine, and tangent) And the inverses?
A math site says that sin^-1 (Opposite / Hypotenuse) = θ but isn't that regular sine, because the "sin^-1" part has no value?
See, I don't understand. Please help me.
English
#Offtopic
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2 RepliesIt all depends on what you are given and what you are looking for. If your are given an angle (let's say 20°)and the length of the side opposite the angle (let's say 5) and are asked to find the hypotenuse, you would have the equation sin(20)=5/x, x=5/sin(20) = 14.6 If you are looking for the angle and are given the opposite length (5) and the hypotenuse (15), the equation is sin(theta)=5/15, theta=sin^-1(1/3) = 19.5°
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1 ReplyYou know, I feel pretty retarded right now.
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2 RepliesUnless you plan on becoming a math major or get into physics, then you will never use that crap again. I went from 11th grade high school math to finishing a Masters degree without ever using sine, cosine, or tangents. I've also been in a highly technical career for 8 years now without using it. There are two things to infer from this: 1. I'm getting too old to be so involved in games that I post on their forums 2. Don't fret this portion of your life. You don't have to learn things you'll never use for eternity, unless you dig that kinda thing.
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Edited by Inspectorfreely: 1/25/2017 3:53:31 PM[quote]Can someone explain the trigonometric ratios? (I think that's what they're called; they're sine, cosine, and tangent) And the inverses? A math site says that sin^-1 (Opposite / Hypotenuse) = θ but isn't that regular sine, because the "sin^-1" part has no value? See, I don't understand. Please help me.[/quote] [b]SOHCAHTOA[/b] [b]S[/b]in = [b]O[/b]pposite/[b]H[/b]ypotenuse [b]C[/b]os = [b]A[/b]djacent/[b]H[/b]ypotenuse [b]T[/b]an = [b]O[/b]pposite/[b]A[/b]djacent There are also Sec, Cos but flipped around, Csc, Sin but flipped around, and Cot, Tan but flipped around. The inverses are used to find the angle (θ) adjacent or across from it. This number can be in degrees or radians, but I'm guessing your teacher just wants you to find it degrees. For example, if you want to find the angle of a right triangle that has a height of 5, a hypotenuse of 7, and the angle is opposite of the height, then you would want to use sin^-1. Work: θ = sin^-1 (5/7) θ = 45.6°
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https://youtu.be/U06jlgpMtQs
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Instructions unclear, dick stuck in calculator
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4 RepliesOkay, so in trig, I know you use triangles a lot, but you must think about everything on the circle. I'm going to explain the unit circle, if you already know about this or have any questions, just ask me. I have to deal with trig in almost everything I do in college, so I'm sure I can help somehow. Imagine a circle, centered at the origin of the traditional 4 quadrant graph. With a radius of 1. Say you drew a line from the center of the circle to the edge of it, somewhere in the first quadrant. How can we describe the angle that that line makes from the x-axis? Well, lets think about it, in the first quadrant, what happens to the line when the angle gets bigger? Rather, recall when you described a line as the ratio of the y to the x components, or rather, rise/run. Think about the line in that way, instead of just being a line. So, when the angle gets bigger, what happens to the rise and run? The answer is that, from 0° to 90° , the x-component (the run) starts at 1, since the radius is 1, and as the angle grows, the x-component decreases. The y-component (the rise), starts at 0, and increases as the angle grows. What trig functions allow us to do, is describe that angle by using the components of the line. They use the ratio of the given components to determine this angle. Sin uses the y component, and Cos uses the x component. If r is the radius, Sin(x)=y/r, Cos(x) = x/r. When you raise a number to a negative exponent, its the same thing as saying its reciprocal. x^-1 is the same thing as 1/x. So, sin^-1 would be r/y, or hypotenuse/opposite.
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Edited by whiteLI0N: 1/25/2017 2:33:15 AMSin(theta) = opposite/hypotenuse Sin^-1(opposite/hypotenuse) = theta Think of it like an algebra equation. You eliminate sin() from one side by using sin^-1() on both sides.
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Well you certainly live up to your name...
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Read title as "Please help me with meth." 😂
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An inverse undoes the action of whatever it is the inverse of. Sin^-1 takes a value (opposite/hyp) and spits out an angle, while sin does the opposite and eats the angle. Is that what you are asking?
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Inb4CellarDoor
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2 RepliesI'm just gonna say 42... Works in most of these"ihaveno-blam!-ingidea" situations...
