document.write( "Question 786238: How do I prove cos(pi/6)=sqrt(3)/2 knowing that sin(pi/6)=1/2? \n" ); document.write( "

Algebra.Com's Answer #477851 by KMST(5328) $\"\"$ $\"About$

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If you were taught that
\n" ); document.write( " $\"%28sin%28x%29%29%5E2%2B%28cos%28x%29%29%5E2=1\"$ you would use that and the fact that $\"pi%2F6\"$ is in the first quadrant, where sine and cosine are positive.
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\n" ); document.write( "Otherwise, if you were only taught that sine and cosine are trigonmetric ratios that apply to right triangles, use a right triangle with a $\"pi%2F6\"$ angle and hypotenuse length 1. Then invoke the Pythagorean theorem. The measures of the legs of that right triangle are $\"cos%28pi%2F6%29\"$ and $\"sin%28pi%2F6%29\"$ .
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\n" ); document.write( "In fact, if you split an equilateral triangle in half using a median (connecting the midpoint of one side to the opposite vertex), you would get two congruent right triangles with one $\"pi%2F6\"$ angle. Fron that idea, you can deduce both values, $\"sin%28pi%2F6%29\"$ and $\"sin%28pi%2F6%29\"$ . The shorter leg of those traingles, opposite the $\"pi%2F6\"$ angle is half of the side of the equilateral triangle, that is now the right triangle's hypotenuse. Hence $\"sin%28pi%2F6%29=1%2F2\"$ \n" ); document.write( "

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