 My Math Forum I don't understand the use of these trig functions for a force vector
 User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 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
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,033
Thanks: 1621

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 $\displaystyle cos(\theta+ \pi/2)= cos(\theta)(cos(\pi/2)- sin(\theta)sin(\pi/2)= cos(\theta)(0)- sin(\theta)(1)= -sin(\theta)$. $\displaystyle cos(\theta+ \pi/2)i + sin(\theta+ \pi/2)j= -sin(\theta)i + cos(\theta)j$. Tags force, functions, trig, understand, vector Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post williamwong0402 Geometry 3 December 11th, 2016 10:36 PM eddm87 Trigonometry 3 October 3rd, 2016 03:45 AM TaylorM0192 Algebra 1 September 13th, 2009 06:24 PM yoyosuper8 Calculus 6 April 26th, 2009 08:26 PM Creephun Real Analysis 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      