 # #10 5.04 Precalculus questions Anonymous
account_balance_wallet \$5

### Question Description denisjames
School: University of Virginia   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${8 + 4\sqrt{3} \over 4} = 2 + \sqrt{3}$

flag Report DMCA  Review Anonymous
Excellent job Brown University

1271 Tutors California Institute of Technology

2131 Tutors Carnegie Mellon University

982 Tutors Columbia University

1256 Tutors Dartmouth University

2113 Tutors Emory University

2279 Tutors Harvard University

599 Tutors Massachusetts Institute of Technology

2319 Tutors New York University

1645 Tutors Notre Dam University

1911 Tutors Oklahoma University

2122 Tutors Pennsylvania State University

932 Tutors Princeton University

1211 Tutors Stanford University

983 Tutors University of California

1282 Tutors Oxford University

123 Tutors Yale University

2325 Tutors