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December 25th, 2011, 11:18 PM   #1
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trigonometric value

Without using a trigonometric table, calculate the value of:

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December 26th, 2011, 01:23 AM   #2
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Using product-to-sum identities, cos(60°) = 1/2 and cos(36°) - sin(18°) = 1/2 (which I proved here),

sin(12°)sin(48°)sin(54°) = (1/2)(cos(36°) - 1/2)sin(54°)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ = (1/4)(2cos(36°)sin(54°) - sin(54°))
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ = (1/4)(1 + sin(18°) - sin(54°))
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ = (1/4)(1 - (cos(36°) - sin(18°)))
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ = (1/4)(1 - 1/2)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ = 1/8.

Last edited by skipjack; October 18th, 2015 at 03:16 AM.
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December 26th, 2011, 06:50 PM   #3
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Re: trigonometric value

First, a bit of advice for your ...

For the degree symbol, use ^{\circ} and for the dot multiplication symbol use \cdot. Then use \( and \) to enclose expressions that are taller than one character, such as variables with exponents, or fractions to automatically get the right sized bracketing symbols. This works for any bracketing symbols.

I edited your post to include these. Now to the problem, and much less elegantly than [color=#00BF00]skipjack[/color]:

Using a product-to-sum identity and cos(-x) = cos(x) we have:





Using sin(90 - x) = cos(x), we have , thus:



Using a triple-angle identity for sine on the left and a double-angle identity for cosine on the right, we have:





Factoring yields:



Knowing we find:



Using a double-angle identity for cosine, we find:





Putting it all together, we find:







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December 27th, 2011, 11:29 AM   #4
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Re: trigonometric value





Using product-to-sum:









Using triple angle identities in sin(54)=sin(3×18):



Factorize:





We have:



:mrgreen:

Last edited by skipjack; October 18th, 2015 at 03:07 AM.
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December 27th, 2011, 11:43 AM   #5
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Re: trigonometric value

How did you know:

?
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December 28th, 2011, 01:13 AM   #6
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Re: trigonometric value

Yes. Let's see:

We have:













Thus we have :


Last edited by skipjack; February 25th, 2016 at 04:28 AM.
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December 28th, 2011, 10:57 AM   #7
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That proof is equivalent to one that appears in the post I linked to above.
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