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November 27, 2014

November 27, 2014

Posted by **Jenn** on Sunday, November 8, 2009 at 10:24pm.

I have no idea how to even start this question. Could someone please help me?

- Math -
**stephanie**, Sunday, November 8, 2009 at 10:39pmx=arcsin(5/13)

y=arccos(3/5)

so find those calculations and plug it into cos(x+y) or in other words

cos(x+y)=(arcsin(5/13)+arccos(3/5))

- Math - I still don't understand -
**Jenn**, Sunday, November 8, 2009 at 11:19pmWell, I understand what you mean by the solution, but I am getting a really long decimal as the answer, so that's not an excat value. Please, can you help me some more?

- Math -
**Reiny**, Monday, November 9, 2009 at 8:42amMake diagrams of your triangles in the corresponding quadrants

if sinx = 5/13 and is in II

then cosx = -12/13

if cosy = 3/5 and is in I

then siny = 4/5

cos(x+y) = cosxcosy - sinxsiny

= (-12/13)(3/5) - (5/13)(4/5)

= -56/65

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