Diving airplane as a projectile

[jbriggs444]

I have a doubt. When finding resultant vectors from a set of given vectors then what I have learned is take the triangles one by one and find the components by taking theta positive and after computing the magnitude of components put signs according to which direction they lie. After that when you find all the $\hat i$ and $\hat j$ add them and get the resultant.
Can I find it like what I did here in this thread by taking signs of the theta angle. Will I get the same components with proper signs.
What is the correct technical way to do it?

If you want a "hat" on a unit vector, you use \hat, e.g. \hat i which renders as $\hat i$

Yes, you can add a list of vectors by converting them all to components, adding the x and y components separately and then reporting the result either as $(\sum x, \sum y)$ or as $\hat i \sum x + \hat j \sum y$. [You'd use actual numbers and not sigmas, of course]

Yes, you can find the components of a vector in any manner that makes sense to you.

One method is:

"I know the component is negative since the angle is below the horizontal. I know I want the sine since I am looking for the length of the side opposite my angle. So I multiply the [positive] length by the [positive] sine and then negate because I already figured out that I want a negative result".

 

[rudransh verma]

So I multiply the [positive] length by the [positive] sine and then negate because I already figured out that I want a negative result".

How about doing it in a more textbook sense. By taking $\sin of -\theta$ if it is below the horizontal. Will I get the same result like -vertical component?

 

[jbriggs444]

How about doing it in a more textbook sense. By taking $\sin of -\theta$ if it is below the horizontal. Will I get the same result like -vertical component?

Yes. $-(290 \sin 30 ^\circ) = 290 \sin -30 ^\circ$

 

[rudransh verma]

Yes. $-(290 \sin \theta) = 290 \sin -\theta$

but what about if my vector is in second or third quadrant?

I know we usually take + magnitude multiple with positive sin/cos of angle and then put signs as - and + to x and y components in second quadrant because we know that’s how they exist in that quadrant in Cartesian coordinate system.

But how about not following this method and somehow take something like magnitude multiplied with sin of original angle measured from +x axis?

 

[jbriggs444]

but what about if my vector is in second or third quadrant?

So your vector is in the second quadrant. You can draw your triangle and measure the angle as 30 degrees.

You eyeball the situation and decide that the result will be positive. Using the triangle method, you find $+(290 \sin 30 ^\circ)$. Using your polar coordinates, you find $290 \sin 150 ^\circ$. Same result either way.

Or you are in the third quadrant. You draw your triangle and measure the angle as 30 degrees. You eyeball the situation and decide that the result will be negative. Using the triangle method you find $-(290 \sin 30 ^\circ)$. Using polar coordinates, you find $290 \sin 210 ^\circ$. Same result either way.

 

[kuruman]

Here is another method that you might find more to your liking.
1. Draw your vector $\vec A$ in a standard Cartesian axes frame.
2. Define angle $\theta$ starting from the positive x-axis to your vector counterclockwise. Note that this angle can be between 0° and 360°.
3. Figure out what $\theta$ is for your particular situation. For example, if your given angle is 30° below the positive x-axis in quadrant IV, then $\theta = 330^{\circ}$; if your given angle is 30° to the left of the positive y-axis in quadrant II, then $\theta = 120^{\circ}$; if your given angle is 40° to the left of the negative y-axis in quadrant III, then $\theta = 200^{\circ}$, and so on.
4. With this (unit circle) convention, you can always write the components as $A_x=A\cos\theta$ and $A_y=A\sin\theta$ which takes care of everything automatically. You don't have to figure out where the sine or cosine go or where to put negative signs.

 

[rudransh verma]

gative y-axis in quadrant III, then θ=200∘, and so on.

You mean 230. I get it. Thanks @jbriggs444 @kuruman . I will get back to you if I get further doubts. I got to check it.

 

[kuruman]

You mean 230. I get it. Thanks @jbriggs444 @kuruman . I will get back to you if I get further doubts. I got to check it.

Yes, I meant 230°.

 

[rudransh verma]

With this (unit circle) convention, you can always write the components as Ax=Acos⁡θ and everything is taken care of automatically. You don't have to figure out where the sine or cosine go or where to put negative signs.

I was right. I have a Big doubt. What right triangle is this in which we take cos of angle? Cos of obtuse angle(330 or 230)?

And what is this unit circle convention?

 

[kuruman]

I was right. I have a Big doubt. What right triangle is this in which we take cos of angle? Cos of obtuse angle(330 or 230)?

And what is this unit circle convention?

You are right with your "Big" doubt: there is no such thing as a right triangle in which one of the angles is greater than 90°. However, if you use the unit circle convention, there is no right triangle to use for finding the components, just the equation $$\vec A=A\cos\theta~\hat i+A\sin\theta~\hat j$$ with angle $\theta$ as defined in post #36 and in the video linked below.

Study this video carefully, especially the last 2½ minutes.

Edit: The equation in this post was edited to its current form. Initially, it was incorrectly written as $\vec A=A_x\cos\theta~\hat i+A_y\sin\theta~\hat j.$

 

[rudransh verma]

You are right with your "Big" doubt: there is no such thing as a right triangle in which one of the angles is greater than 90°. However, if you use the unit circle convention, there is no right triangle to use for finding the components, just the equation $$\vec A=A_x\cos\theta~\hat i+A_y\sin\theta~\hat j$$ with angle $\theta$ as defined in post #36 and in the video linked below.

Study this video carefully, especially the last 2½ minutes.

I watched the video and I again have a doubt. How can you be so sure that the new definition of trig function will hold for all the quadrants. Is it always true that the cos of angle is x coordinate and sin of angle is y coordinate even with 90 and above angles? Can you prove for bigger angles?
Also in post 36 there is A instead of 1. So does it follow the unit circle convention? Its A circle convention. I guess now there is a bigger circle with radius A but everything works the same.

 

[jbriggs444]

Is it always true that the cos of angle is x coordinate and sin of angle is y coordinate even with 90 and above angles? Can you prove for bigger angles?

It works for all angles from minus infinity to plus infinity.

The sine and cosine functions are periodic. $\sin ( x + 360^\circ) = \sin x = \sin ( x - 360^\circ)$. This makes sense if you think about it. It does not matter how many times the minute hand circles the clock face. Its y coordinate at 3 o'clock is always the same.

 

[kuruman]

Also in post 36 there is A instead of 1. So does it follow the unit circle convention? Its A circle convention. I guess now there is a bigger circle with radius A but everything works the same.

You can always write any vector $\vec A$ as the magnitude of the vector $A$ times a unit vector $\hat a$ in its direction, namely $\vec A=A~\hat a$. Now the unit circle convention says that I can always write $$\hat a=\cos\theta~\hat i+\sin\theta~\hat j$$ Put that in the first equation to get $$\vec A= A(\cos\theta~\hat i+\sin\theta~\hat j).$$ If you identify $A_x \equiv A\cos\theta$ and $A_y \equiv A\sin\theta$, you get $$\vec A=A_x~\hat i+A_y~\hat j.$$

 

[rudransh verma]

The sine and cosine functions are periodic. $\sin ( x + 360^\circ) = \sin x = \sin ( x - 360^\circ)$. This makes sense if you think about it. It does not matter how many times the minute hand circles the clock face. Its y coordinate at 3 o'clock is always the same.

Obvious. How about we talk about $\theta < 360^\circ$. $120^\circ, 135^\circ, 150^\circ$. $\sin$ value goes from 0 to +1 in 1st and then back to 0 in 2nd quadrant. $\sin(90+ \theta)= \cos\theta$ So for every $0<\theta<90$ we have $\cos\theta$. That said sin of bigger angles from 90 to 180 repeats it’s value as cos. $\sin$ forms Mirror images in unit circle. Eg $\sin 120^\circ= \sin 60^\circ= \cos 30^\circ$
$\sin 135^\circ= \sin 45^\circ= \cos 45^\circ$

 

[kuruman]

Perhaps it might be useful for you to memorize a couple of trigonometric identities for sums of angles. They are handy for translating sines and cosines with arguments outside the first quadrant to sines and cosines with arguments in the first quadrant. These are:$$\begin{align} & \sin(a \pm b)= \sin a \cos b \pm \cos a \sin b \nonumber \\& \cos(a \pm b)=\cos a\cos b \mp \sin a\sin b \nonumber \end{align}$$The idea is to write the given angle as the sum of two parts, one which is a multiple of $90^{\circ}$ and one which is in the 1st quadrant. For example, given $\theta=235^{\circ}$, you write, $235^{\circ}=180^{\circ}+55^{\circ}.$Then $$\begin{align}\sin(235^{\circ}) & =\sin(180^{\circ}+55^{\circ}) \nonumber \\ & = \sin(180^{\circ})\cos(55^{\circ}) +\cos(180^{\circ})\sin(55^{\circ}) \nonumber \\ & =0\times \cos(55^{\circ})+(-1)\times \sin(55^{\circ}) \nonumber \\ & =-\sin(55^{\circ}).\nonumber\end{align}$$My calculator says $\sin(235^{\circ})=-0.8191$ and $\sin(55^{\circ})=0.8191.$ Had I started with $\sin(270^{\circ}-35^{\circ})$ instead, I would have ended up with $-\cos(35^{\circ})$ which is also $-0.8191.$ Do you see how it works?

 

[rudransh verma]

My calculator says sin⁡(235∘)=−0.8191 and sin⁡(55∘)=0.8191. Had I started with sin⁡(270∘−35∘) instead, I would have ended up with −cos⁡(35∘) which is also −0.8191. Do you see how it works?

So basically all the trigonometric functions with arguments not in first quadrant are trig functions with arguments in first quadrant ?

I guess that’s why we are made to memorise values of all the trig functions from $0^\circ$ to $90^\circ$.

 

[kuruman]

So basically all the trigonometric functions with arguments not in first quadrant are trig functions with arguments in first quadrant ?

More precisely, all trig functions with arguments outside the first quadrant can be mapped into trig functions with arguments in the first quadrant.