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 June 4th, 2011, 12:37 AM #1 Newbie   Joined: May 2011 Posts: 17 Thanks: 0 Radians? I'm having trouble with this problem: Find the gradient of y = 3sin2t + 4cos2t when t = 2. It also tells me to use radians, although it was never explained to me so I really don't know how to apply them. Can someone please help me to understand?
 June 4th, 2011, 02:30 PM #2 Global Moderator   Joined: May 2007 Posts: 6,663 Thanks: 649 Re: Radians? Since you need to take derivatives, the derivative of sin x is cos x and the derivative of cos x is -sin x, WHEN x IS IN RADIANS.
 June 4th, 2011, 04:29 PM #3 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 $y\,=\,f(t)\,=\,3\sin{2t}\,+\,4\cos{2t}\,\Rightarro w\,\nabla f\,=\,\frac{\partial f}{\partial t}\,=\,6\cos{2t}\,-\,8\sin{2t}\,\Rightarrow\,\therefore\,\nabla f(2)\,=\,6\cos{4}\,-\,8\sin{4}$
 June 4th, 2011, 11:08 PM #4 Newbie   Joined: May 2011 Posts: 17 Thanks: 0 Re: Radians? I know that much, but if I do 6cos4 - 8sin4 on a calculator I get roughly 5.428 and the answer is 2.133. The only thing I'm not doing is using radians, because I don't know where to use them or how to apply them to the situation.
 June 4th, 2011, 11:32 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Definition of radian: ?/180 = 1°, e.g. ?/4 = (?/4)(180/?) = 180/4 deg = 45°. 4 = k? ? k*3.14, hence k = 4/3.14. We'll use 4 ? 4?/3.14. (4?/3.14)(180/?) = 720/3.14 deg ? 229.3°, hence 6 cos 4 - 8 sin 4 ? 6 cos 229.3° - 8 sin 229.3° ? 2.15. The given answer by you is 2.133, but 2.15 is close enough.
 June 5th, 2011, 01:28 AM #6 Newbie   Joined: May 2011 Posts: 17 Thanks: 0 Re: Radians? Thanks, I just tried this for myself and I've been getting the right answers. When trig functions are not involved is there any need/harm to use radians?
June 5th, 2011, 02:07 PM   #7
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Re: Radians?

Quote:
 Originally Posted by Phazon Thanks, I just tried this for myself and I've been getting the right answers. When trig functions are not involved is there any need/harm to use radians?
Any time you do any mathematics requiring functions of angles (trig functions are the most common), you need to use radians.

 June 5th, 2011, 05:05 PM #8 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 See this.
 June 6th, 2011, 03:49 AM #9 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Radians? Trig functions have many uses that have nothing to do with triangles and it is possible to define trig functions without using angles at all so that when you are asked to find sin(x) when x= 4 you don't have to worry about whether x is in degrees or radians just as you don't have to worry about units for x if you are asked to evaluate $x^2$ when x= 4. If you want to connect those with the trig functions you connect with right triangles, then you have to think "radians".
June 6th, 2011, 04:12 PM   #10
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Re: Radians?

Quote:
 Originally Posted by HallsofIvy Trig functions have many uses that have nothing to do with triangles and it is possible to define trig functions without using angles at all so that when you are asked to find sin(x) when x= 4 you don't have to worry about whether x is in degrees or radians just as you don't have to worry about units for x if you are asked to evaluate $x^2$ when x= 4. If you want to connect those with the trig functions you connect with right triangles, then you have to think "radians".
This is very misleading. Sin(4) is quite different when 4 is radians than when 4 is degrees. Also, and more important, there are many areas, such as calculus, where trig functions are used and it is very messy to use anything but radians.

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