Asked by Jenn
Angle x is in the second quadrant and angle y is in the first quadrant such that sinx=5/13 and cosy=3/5, determine and exact value for cos (x+y).
I have no idea how to even start this question. Could someone please help me?
I have no idea how to even start this question. Could someone please help me?
Answers
Answered by
stephanie
x=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))
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))
Answered by
Jenn
Well, 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?
Answered by
Reiny
Make 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
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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