SOLUTION: hoe to find the exact value of cos(arcsin(-1/2) + arctan(2/3))=

Algebra ->  Trigonometry-basics -> SOLUTION: hoe to find the exact value of cos(arcsin(-1/2) + arctan(2/3))=      Log On


   



Question 293364: hoe to find the exact value of cos(arcsin(-1/2) + arctan(2/3))=
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
how to find the exact value of cos(arcsin(-1/2) + arctan(2/3))
------------
Form: cos(a+b) = cos(a)*cos(b)-sin(a)*sin(b)
Your "a" is arcsin(-1/2) = -30 degrees
Your "b" is arctan(2/3) = 33.69 degrees
-------------------
Your Problem:
= cos(-30)cos(33.69)-sin(-30)sin(33.69)
---
= (sqrt(3)/2)(3/sqrt(13)) - (-1/2)(2/sqrt(13))
=========================================
Cheers,
Stan H.
=========================================

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
find the exact value of cos(arcsin(-1/2) + arctan(2/3))
--------------------
Plot the points and assign the angles A and B
A = arcsin(-1/2) = -30º
cos(A) = sqrt(3)/2
sin(A) = -1/2
----------------
B = arctan(2/3)
gives the point (2,3), a distance of sqrt(13) from the Origin.
cos(B) = 3sqrt(13)/13
sin(B) = 2sqrt(13)/13
---------------------
cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
= sqrt(3)/2 * 3sqrt(13)/13 - (-1/2)*2sqrt(13)/13
= 3sqrt%2839%29%2F26+%2B+sqrt%2813%29%2F13
---------------------------
or, = %28sqrt%2813%29%2F26%29%2A%283sqrt%283%29+%2B+2%29