#10 5.04 Precalculus questions

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May 15th, 2015

sine, sin(x+y) = sin(x)cos(y) + cos(x)sin(y).

 With x=135 and y=30,

 we have \[\sin(165) = {\sqrt{2} \over 2}{\sqrt{3} \over 2} +{\sqrt{2} \over 2}{1 \over 2} = {\sqrt{6} + \sqrt{2} \over 4}\]

 Along the same lines,

 cos(x+y) = cos(x)cos(y) - sin(x)sin(y), so we have \[\cos(165) = {\sqrt{2} \over 2}{\sqrt{3} \over 2} - {\sqrt{2} \over 2}{1 \over 2} = {\sqrt{6} - \sqrt{2} \over 4}\]

 Now I guess you could use the identity for tangent, 

but it's easier at this point to say 

\[\tan(165) = {\sin(165) \over \cos(165)} = {\sqrt{6} + \sqrt{2} \over 4} {4 \over \sqrt{6} - \sqrt{2}} = {\sqrt{6} + \sqrt{2} \over \sqrt{6} - \sqrt{2}} \] 

Which, by multiplying both the top and bottom by sqrt(6)+sqrt(2),

 simplifies to\[{8 + 4\sqrt{3} \over 4} = 2 + \sqrt{3}\]

May 15th, 2015

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May 15th, 2015
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May 15th, 2015
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