document.write( "Question 1159222: Use half-angle formulas to find the exact values.
\n" ); document.write( "a. cos 15(degrees)
\n" ); document.write( "b. sin 20(degress)30'\r
\n" ); document.write( "\n" ); document.write( "I got \"sqrt%283%29\"/2 for a and -\"sqrt%282%29\"/2 for b which are wrong. What did I do wrong?
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Algebra.Com's Answer #782211 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "I'm not sure what part b is trying to say because the notation is a bit strange. \r
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\n" ); document.write( "\n" ); document.write( "I'll show you how to do part a\r
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\n" ); document.write( "\n" ); document.write( "\"cos%28x%2F2%29+=+sqrt%28+%281%2Bcos%28x%29%29%2F2%29\" Half angle identity for cosine, when cosine is positive. Keep in mind that cos(15) is a positive value as the angle 15 degrees is in Q1 where cosine is positive.\r
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\n" ); document.write( "\n" ); document.write( "\"cos%2830%2F2%29+=+sqrt%28+%281%2Bcos%2830%29%29%2F2%29\" Plug in x = 30\r
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\n" ); document.write( "\n" ); document.write( "\"cos%2815%29+=+sqrt%28+%281%2Bsqrt%283%29%2F2%29%2F2%29\" Replace cos(30) with sqrt(3)/2\r
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\n" ); document.write( "\n" ); document.write( "\"cos%2815%29+=+sqrt%28+%282%2F2%2Bsqrt%283%29%2F2%29%2F2%29\" Rewrite 1 as 2/2\r
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\n" ); document.write( "\n" ); document.write( "\"cos%2815%29+=+sqrt%28+%28%282%2Bsqrt%283%29%29%2F2%29%2F2%29\" Combine the upper fractions\r
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\n" ); document.write( "\n" ); document.write( "\"cos%2815%29+=+sqrt%28+%28%282%2Bsqrt%283%29%29%2F2%29%2A%281%2F2%29%29\" \r
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\n" ); document.write( "\n" ); document.write( "\"cos%2815%29+=+sqrt%28+%282%2Bsqrt%283%29%29%2F4+%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"cos%2815%29+=+sqrt%28+2%2Bsqrt%283%29+%29%2Fsqrt%284%29\" Break up the square root\r
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\n" ); document.write( "\n" ); document.write( "\"cos%2815%29+=+sqrt%28+2%2Bsqrt%283%29+%29%2F2\" Simplify sqrt(4) into 2. This is as simplified as it gets.\r
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\n" ); document.write( "\n" ); document.write( "-----------------------------------\r
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\n" ); document.write( "\n" ); document.write( "Side note: if you use a calculator, you should find
\n" ); document.write( "\"cos%2815%29+=+0.96592582628907\" (make sure your calculator is in degree mode)\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%28+2%2Bsqrt%283%29+%29%2F2+=+0.96592582628907\"\r
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\n" ); document.write( "\n" ); document.write( "both are approximations. Since we get the same approximation, this helps confirm the answer.\r
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\n" ); document.write( "\n" ); document.write( "Another way to confirm the answer is to compute \"%28+sqrt%282%2Bsqrt%283%29%29+%29%2F%282%29-cos%2815%29\" and you should get zero as a result (or some very small number due to rounding error). I'm using the idea that if \"x+=+y\", then \"x-y+=+0\"
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