Trigonometry

[RNtoMD87]

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Hey, I'm currently taking trig, and while I am able to memorize equations, I am trying to conceptualize it, especially the different wave forms. 3b1b was a great resource for conceptualizing some calc concepts, does anyone have any resources to really teach trig in this manner? I would like to learn how to apply this to daily life. I've always know the similar triangles and using Pythagorean theorem but the waves I'm basically using from pure memorization and the Pythagorean identity I am only able to derive, but not truly UNDERSTAND.

 

[nobaddays]

These types of visualizations always helped me make sense of what the trig functions actually represent and why the wave graphs all look the way they do.

 

[RNtoMD87]

Thanks baddays

Wow yeah that makes a whole lot more sense. I wonder why they don't teach this in school. Its straight memorization.

 

[deleted938090]

Pythagorean identity:
Consider a circle equation centered at the origin (0, 0) with radius r. From the distance formula, we have (distance from a point lying on the circle) = r. We rewrite this in algebra terms, letting (x, y) be the point on the circle.
sqrt( (x-0)^2 + (y-0)^2 ) = r.
We square both sides and then we get:
x^2 + y^2 = r^2.

Angles are measured from the +x axis counterclockwise. Imagine you have an angle with one ray at +x and another intersecting the circle, forming an angle of theta from the +x to the intersection to the circle. Now, draw lines to the X and Y axes from that point on the circle. Using sin(theta) = opposite/hypotenuse, you should notice that it's always true that sin(theta) = y / r, where y is the y coordinate of that point. Similarly, cos(theta) = adjacent/hypotenuse = x/r.

Now, we have:
Sin(theta)^2 + Cos(theta)^2 = (y/r)^2 + (x/r)^2 = (x^2+y^2)/(r^2). However, we already know from the distance formula that x^2 + y^2 = r^2. Therefore, (x^2+y^2)/(r^2) = 1 by dividing both sides by r^2. Thus, for any angle, it is always true that:
1. sin theta = y / r
2. cos theta = x / r
3. sin^2 + cos^2 = 1


**Real life application!**
Imagine that the ray on the x axis is any direction you're looking at. It's always true that sin(theta)'s gives the y component, which is perpendicular to your direction (hence is the component perpendicular) and cos(theta) gives the x component, which is parallel to your direction.

 

[RNtoMD87]

Pythagorean identity:
Consider a circle equation centered at the origin (0, 0) with radius r. From the distance formula, we have (distance from a point lying on the circle) = r. We rewrite this in algebra terms, letting (x, y) be the point on the circle.
sqrt( (x-0)^2 + (y-0)^2 ) = r.
We square both sides and then we get:
x^2 + y^2 = r^2.

Angles are measured from the +x axis counterclockwise. Imagine you have an angle with one ray at +x and another intersecting the circle, forming an angle of theta from the +x to the intersection to the circle. Now, draw lines to the X and Y axes from that point on the circle. Using sin(theta) = opposite/hypotenuse, you should notice that it's always true that sin(theta) = y / r, where y is the y coordinate of that point. Similarly, cos(theta) = adjacent/hypotenuse = x/r.

Now, we have:
Sin(theta)^2 + Cos(theta)^2 = (y/r)^2 + (x/r)^2 = (x^2+y^2)/(r^2). However, we already know from the distance formula that x^2 + y^2 = r^2. Therefore, (x^2+y^2)/(r^2) = 1 by dividing both sides by r^2. Thus, for any angle, it is always true that:
1. sin theta = y / r
2. cos theta = x / r
3. sin^2 + cos^2 = 1


**Real life application!**
Imagine that the ray on the x axis is any direction you're looking at. It's always true that sin(theta)'s gives the y component, which is perpendicular to your direction (hence is the component perpendicular) and cos(theta) gives the x component, which is parallel to your direction.

I knew everything until the real life application, and that lost me.

I think my problem is unless in real life I’m looking at a triangle (height of a tower based off its shadow kind of thing), it’s hard for me to conceptualize.

Like it’s hard for me to mentally see relationships as a sine wave. Say like blood pressure. Or cost vs sales, etc.

 

[longhaul3]

Are you familiar with torque? A force applied at some distance from a fixed point produces rotation. The amount of torque increases with the distance from the center (the principle behind long wrenches and other levers) but also with the angle at which the force is applied.

Think about struggling to loosen a tight lug nut on the side of the road. You have your wrench out and are trying everything you can to no avail. What do you do? Put the wrench on horizontally and step/jump on the end of it. The reason this works is that the torque is determined by the sine of the angle at which the force is applied. If you jumped on the wrench parallel to its axis (sin0 = 0), you could push as hard as you wanted and get exactly nowhere. The closer to perpendicular (sin90 = 1, or 100% of the force) you can apply the force, the more torque you'll produce. For angles in between, you'll get a smaller but non-zero percentage of the force translated into torque. The closer to 90 degrees, the greater the sine of the angle, so the greater the torque.

If you think of sine/cosine as percentages of the whole, or as orthogonal components of a single vector with some 2-d direction, this turns out to be T = F sin(theta), ignoring the radius.

The values of sine for various thetas are given by the Pythagorean theorem as above, but in decimal form they all work out to be percentages that increase as theta gets closer to 90 degrees.

 

[RNtoMD87]

Hmm that actually helps out somewhat. I've always known that the further from the axis produced more torque and at 90 degrees more torque was produced from experience working on HMMWVs or my personal vehicle, but never thought of it in terms of trigonometry. I think I'm going to focus on this angle for a bit. Thanks a lot for that. (Hah just noticed the pun I made)


Ive always just rationalized that lets say you're applying the force perpendicularly (like from the 90 degree position) that the object is absorbing the force, rather than being moved by it, and the further you get to 180 or 0, the force isn't being absorbed, but acting upon the object.