
Trigonometry Trigonometry Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 21st, 2017, 05:18 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 07:13 AM. 
October 21st, 2017, 05:46 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 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, 06:40 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 07:26 AM.  
October 21st, 2017, 07:43 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,040 Thanks: 1063  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 07:28 AM.  
October 22nd, 2017, 01:14 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, 04:34 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 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, 05:20 AM  #7  
Senior Member Joined: Jun 2015 From: England Posts: 844 Thanks: 252  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 05:23 AM.  
October 22nd, 2017, 08:34 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,293 Thanks: 1684 
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, 12:31 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, 12:45 PM  #10 
Senior Member Joined: Sep 2015 From: USA Posts: 2,040 Thanks: 1063  

Tags 
equation, identity, prove 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Formal name for the "0 identity" of a function  Adam Ledger  Number Theory  11  May 6th, 2016 04:40 PM 
A "simple" application of dirac delta "shift theorem"...help  SedaKhold  Calculus  0  February 13th, 2012 11:45 AM 
"separate and integrate" or "Orangutang method"  The Chaz  Calculus  1  August 5th, 2011 09:03 PM 
sample exerimentneed help finding "statistic" and "result"  katie0127  Advanced Statistics  0  December 3rd, 2008 01:54 PM 