"Prove the following equation is an identity"

Oct 2017
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0
cali


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.
 
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greg1313

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Oct 2008
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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 $\sin^2 x$:

$$\sin^2 \frac x2 = \frac{\sin^2 x}{4\cos^2 \frac x2} = \frac{1 - \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x - \cot^2 x}{2\csc^2 x - 2\csc x\cot x}$$
 
Oct 2017
8
0
cali
\(\displaystyle \sin^2\! x=4\sin^2\! \frac x2 \cos^2\! \frac x2 \implies \sin^2\! \frac x2 = \frac{\sin^2\! x}{4\cos^2\! \frac x2}\)
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.
 
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romsek

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Sep 2015
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\(\displaystyle \sin^2\! x=4\sin^2\! \frac x2 \cos^2\! \frac x2 \implies \sin^2\! \frac x2 = \frac{\sin^2\! x}{4\cos^2\! \frac x2}\)
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 = $

$4 \sin^2\left(\frac x 2\right)\cos^2\left(\frac x 2\right)$
 
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greg1313

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\(\displaystyle \sin^2 \frac x2 = \frac{1 - \cos x}{2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x - \cot^2 x}{2\csc^2 x - 2\csc x\cot x}\)
 
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Jun 2015
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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?
 
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skipjack

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Dec 2006
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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.
 
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Oct 2017
8
0
cali
\(\displaystyle \sin^2 \frac x2 = \frac{1 - \cos x}{2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x - \cot^2 x}{2\csc^2 x - 2\csc x\cot x}\)
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.
 

romsek

Math Team
Sep 2015
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USA
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 $\sin^2(x)$