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October 21st, 2017, 06:18 PM  #1 
Newbie Joined: Oct 2017 From: cali Posts: 8 Thanks: 0  "Prove the following equation is an identity" I've always struggled with these kinds of problems, since there's so many kinds of ways to do it. Often, whenever I try to solve and prove identities, I end up nowhere. :/ So some tips and advice would be awesome. Last edited by skipjack; October 22nd, 2017 at 08:13 AM. 
October 21st, 2017, 06:46 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
Using $\sin2x = 2\sin x\cos x$: $$\sin^2 x=4\sin^2 \frac x2 \cos^2 \frac x2 \implies \sin^2 \frac x2 = \frac{\sin^2 x}{4\cos^2 \frac x2}$$ Using $\cos \frac x2 = \pm\sqrt{\frac{1 + \cos x}{2}}$ and a Pythagorean identity: $$\frac{\sin^2 x}{4\cos^2 \frac x2} = \frac{1  \cos^2 x}{2 + 2\cos x}$$ Divide top and bottom of $\frac{1  \cos^2 x}{2 + 2\cos x}$ by $\sin^2 x$: $$\sin^2 \frac x2 = \frac{\sin^2 x}{4\cos^2 \frac x2} = \frac{1  \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x  \cot^2 x}{2\csc^2 x  2\csc x\cot x}$$ 
October 21st, 2017, 07:40 PM  #3  
Newbie Joined: Oct 2017 From: cali Posts: 8 Thanks: 0  Quote:
The only identity I see that matches it is a powerreducing formula which gives me 1cos x over 2. Last edited by skipjack; October 22nd, 2017 at 08:26 AM.  
October 21st, 2017, 08:43 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,171 Thanks: 1141  Quote:
$\left(\sin\left(2 \frac x 2\right)\right)^2 = $ $\left(2 \sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\right)^2 = $ $4 \sin^2\left(\frac x 2\right)\cos^2\left(\frac x 2\right)$ Last edited by skipjack; October 22nd, 2017 at 08:28 AM.  
October 22nd, 2017, 02:14 AM  #5 
Newbie Joined: Apr 2017 From: Bhadohi, U.P., India Posts: 26 Thanks: 1  after writing the R.H.S. in terms of sin(x) and cos(x)
using the formula cos(2x) = 1  2(sinx)^2

October 22nd, 2017, 05:34 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
$\displaystyle \sin^2 \frac x2 = \frac{1  \cos x}{2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1  \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x  \cot^2 x}{2\csc^2 x  2\csc x\cot x}$

October 22nd, 2017, 06:20 AM  #7  
Senior Member Joined: Jun 2015 From: England Posts: 891 Thanks: 269  Quote:
For an equation in x, like you have An identity has the same value or result, or is true for all x or for every x. An equality has the same value or result, or is true for some x but not for every x. So to prove that an equation is an identity you aim is to prove that it holds good for every value of x. So $\displaystyle {x^2}  4 = 0$ is true for some values of x and is an equation but $\displaystyle {\left( {x  4} \right)^2} = {x^2}  8x + 16$ is true for all values of x so is an identity Does this help? Last edited by studiot; October 22nd, 2017 at 06:23 AM.  
October 22nd, 2017, 09:34 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
In both of greg1313's solutions, there's a sign error in the final step. Note that these solutions are easier to find by starting with the rather complicated righthand side of the original equation and seeking to simplify it. The righthand side of the original equation has removable discontinuities where $\cos x$ = 1, so the original equation isn't, strictly speaking, an identity. 
October 22nd, 2017, 01:31 PM  #9  
Newbie Joined: Oct 2017 From: cali Posts: 8 Thanks: 0  Quote:
I do have 1 more question though, looking at my sheet, I can't really figure out how 1cos^2x over 2+2cosx equals the right side of the equation.  
October 22nd, 2017, 01:45 PM  #10 
Senior Member Joined: Sep 2015 From: USA Posts: 2,171 Thanks: 1141  

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