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 November 21st, 2017, 06:03 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 = |V|cos(θ+90)i + |V|sin(θ+90)j I was hoping some one could explain how they got their answer ? Last edited by skipjack; November 22nd, 2017 at 01:36 PM.
November 21st, 2017, 06:25 PM   #2
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x component in the negative direction

y component in the positive direction

see diagram ...
Attached Images
 vector_quadII.jpg (16.6 KB, 25 views)

 November 21st, 2017, 06:32 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, 06:55 PM #4 Math Team     Joined: Jul 2011 From: Texas Posts: 3,033 Thanks: 1621 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 01:34 PM.
 November 21st, 2017, 06:58 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 01:33 PM.
 November 27th, 2017, 06:15 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Since in the xi+ yj form, the angle is typically measured from the positive x-axis, 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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