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 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
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Quote:
 Originally Posted by greg1313 $\displaystyle \sin^2\! x=4\sin^2\! \frac x2 \cos^2\! \frac x2 \implies \sin^2\! \frac x2 = \frac{\sin^2\! x}{4\cos^2\! \frac x2}$
I'm very confused, how did sin^2x equal that?

The only identity I see that matches it is a power-reducing formula which gives me 1-cos x over 2.

Last edited by skipjack; October 22nd, 2017 at 07:26 AM.

October 21st, 2017, 07:43 PM   #4
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Quote:
Originally Posted by materialartist09
Quote:
 Originally Posted by greg1313 $\displaystyle \sin^2\! x=4\sin^2\! \frac x2 \cos^2\! \frac x2 \implies \sin^2\! \frac x2 = \frac{\sin^2\! x}{4\cos^2\! \frac x2}$
I'm very confused, how did sin^2x equal that?

The only identity I see that matches it is a power-reducing formula which gives me 1-cos x over 2.
$\sin^2(x) = \sin^2\left(2 \frac x 2\right) =$

$\left(\sin\left(2 \frac x 2\right)\right)^2 =$

$\left(2 \sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\right)^2 =$

$4 \sin^2\left(\frac x 2\right)\cos^2\left(\frac x 2\right)$

Last edited by skipjack; October 22nd, 2017 at 07:28 AM.

October 22nd, 2017, 01:14 AM   #5
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after writing the R.H.S. in terms of sin(x) and cos(x)

using the formula cos(2x) = 1 - 2(sinx)^2
Attached Images
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 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 $\displaystyle \sin^2 \frac x2 = \frac{1 - \cos x}{2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x - \cot^2 x}{2\csc^2 x - 2\csc x\cot x}$ Thanks from materialartist09
October 22nd, 2017, 05:20 AM   #7
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Quote:
 Originally Posted by materialartist09 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.
The thing to understand is the difference between an identity and an equality .

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 right-hand side of the original equation and seeking to simplify it. The right-hand side of the original equation has removable discontinuities where $\cos x$ = -1, so the original equation isn't, strictly speaking, an identity. Thanks from greg1313
October 22nd, 2017, 12:31 PM   #9
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Quote:
 Originally Posted by greg1313 $\displaystyle \sin^2 \frac x2 = \frac{1 - \cos x}{2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x - \cot^2 x}{2\csc^2 x - 2\csc x\cot x}$
okay i understand, thank you so much!

I do have 1 more question though, looking at my sheet, I can't really figure out how 1-cos^2x over 2+2cosx equals the right side of the equation.

October 22nd, 2017, 12:45 PM   #10
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Quote:
 Originally Posted by materialartist09 okay i understand, thank you so much! I do have 1 more question though, looking at my sheet, I can't really figure out how 1-cos^2x over 2+2cosx equals the right side of the equation.
divide everything by $\sin^2(x)$

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