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November 21st, 2017, 07:03 PM  #1 
Newbie Joined: Nov 2017 From: uk Posts: 3 Thanks: 0  I don't understand the use of these trig functions for a force vector
Hi I am trying to learn mechanics with static particles. I have a vector that points up and left with an angle theta with respect to the vertical axis. The answer sheet says that the direction is therefore: V = −V sinθi + V cosθj But I don't see how it's a negative sin trig for i component and positive cos trig for j component. I can understand it more like: V = Vcos(θ+90)i + Vsin(θ+90)j I was hoping some one could explain how they got their answer ? Last edited by skipjack; November 22nd, 2017 at 02:36 PM. 
November 21st, 2017, 07:25 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 
x component in the negative direction y component in the positive direction see diagram ... 
November 21st, 2017, 07:32 PM  #3 
Newbie Joined: Nov 2017 From: uk Posts: 3 Thanks: 0 
Thank you for the reply. I've not seen it done like that before. We have only been taught doing it like this so far: Are there other ways to visualise it without thinking about changing the i and j axis signs? It's still not 100% obvious to me why the vertical j in your picture is using "cos" and the horizontal negative i is "sin" ? I would naturally had used "cos" for i and "sin" for j in that image. 
November 21st, 2017, 07:55 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 
Look at the position of $\theta$ in the right triangle ... ... the horizontal component is opposite to $\theta$ and the vertical component adjacent. Remember SOHCAHTOA? Last edited by skipjack; November 22nd, 2017 at 02:34 PM. 
November 21st, 2017, 07:58 PM  #5 
Newbie Joined: Nov 2017 From: uk Posts: 3 Thanks: 0 
Ohhhh, I see it now! It's so obvious too! Thank you. Last edited by skipjack; November 22nd, 2017 at 02:33 PM. 
November 27th, 2017, 07:15 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
Since in the xi+ yj form, the angle is typically measured from the positive xaxis, you can also write that as $\displaystyle cos(\theta+ \pi/2)i+ sin(\theta+ \pi/2)j$. But $\displaystyle sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)$ and $\displaystyle cos(\theta+ \phi)= cos(\theta)cos(\phi) sin(\theta)cos(\phi)$ So $\displaystyle sin(\theta+ \pi/2)= sin(\theta)cos(\pi/2)+ cos(\theta)sin(\pi/2)= sin(\theta)(0)+ cos(\theta)(1)= cos(\theta)$ and $\displaystyle cos(\theta+ \pi/2)= cos(\theta)(cos(\pi/2) sin(\theta)sin(\pi/2)= cos(\theta)(0) sin(\theta)(1)= sin(\theta)$. $\displaystyle cos(\theta+ \pi/2)i + sin(\theta+ \pi/2)j= sin(\theta)i + cos(\theta)j$. 

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force, functions, trig, understand, vector 
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