December 31st, 2014, 06:44 PM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  Differentiation
I'm given $\displaystyle y=\frac{cos2x}{1sinx}$ to differentiate. My answer is $\displaystyle \frac{cosxcos2x2(sin2x)(1sinx)}{(1sinx)^2}$ But the given answer is $\displaystyle \frac{4sin(2x)3cos(x)+cos(3x)}{2(sin(x)1)^2}$. 
January 1st, 2015, 12:04 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond 
Post your work and we'll help you identify and correct any errors.

January 1st, 2015, 01:10 AM  #3 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
$\displaystyle y=\frac{cos2x}{1sinx}$ $\displaystyle \frac{dy}{dx}=\frac{(1sinx)(2sin2x)(cos2x)(cosx)}{(1sinx)^2}$ $\displaystyle =\frac{2sin2x+2sin^22x+cosx(cos2x)}{(1sinx)^2}$ $\displaystyle =\frac{(2)(sin2x)\left [ 1sinx \right ]+cosx(cos2x)}{(1sinx)^2}$ $\displaystyle =\frac{cosxcos2x2(sin2x)(1sinx)}{(1sinx)^2}$ The given answer is $\displaystyle \frac{4sin(2x)3cos(x)+cos(3x)}{2(sin(x)1)^2}$ 
January 1st, 2015, 10:43 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond 
The given answer is correct. Your mistake is in assuming $\displaystyle (\sin(x))(2\sin(2x))=2\sin^2(2x)$ Correct that, then use the identities $\displaystyle \cos(\theta)\cos(\varphi)= 2\sin\left(\frac{\theta+\varphi}{2}\right)\sin \left(\frac{\theta\varphi}{2}\right)$ and $\displaystyle \cos(\theta)+\cos(\varphi)=2\cos\left(\frac{\theta +\varphi}{2}\right)\cos\left(\frac{\theta\varphi}{2}\right)$ to simplify further. 
January 3rd, 2015, 09:31 AM  #5 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
What this symbol $\displaystyle \varphi$ means?

January 3rd, 2015, 10:31 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 
It's a variable, phi, like theta is a variable.


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