SOLUTION: Evaluate exactly: sin[arctan(5/sqrt15) - arcsin(3/sqrt10)]

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Question 884510: Evaluate exactly:
sin[arctan(5/sqrt15) - arcsin(3/sqrt10)]

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
sin%28arctan%285%2Fsqrt%2815%29%29+-+arcsin%283%2Fsqrt%2810%29%29%29%29+

Let alpha%22%22=%22%22arctan%285%2Fsqrt%2815%29%29

and beta%22%22=%22%22arcsin%283%2Fsqrt%2810%29%29

arctan%285%2Fsqrt%2815%29%29 means the angle whose tangent is 5%2Fsqrt%2815%29.
The tangent is opposite%2F%28adjacent%29 so let's draw a right triangle
containing angle alpha by using the numerator of 5%2Fsqrt%2815%29, which
is 5, for the opposite side, and using the denominator of 5%2Fsqrt%2815%29,
which is sqrt%2815%29 for the opposite side.

arcsin%283%2Fsqrt%2810%29%29 means the angle whose sine is 3%2Fsqrt%2810%29.
The sine is opposite%2F%28hypotenuse%29 so let's draw another right triangle
containing angle beta by using the numerator of 3%2Fsqrt%2810%29, which is
3, for the opposite side, and using the denominator of 3%2Fsqrt%2810%29, which
is sqrt%2810%29 for the hypotenuse.





We calculate the hypotenuse of the first right triangle and
the adjacent side to beta in the second one:

c%5E2=a%5E2%2Bb%5E2
c%5E2=%28sqrt%2815%29%29%5E2%2B5%5E2
c%5E2=15%2B25
c%5E2=40
c=sqrt%2840%29
c=sqrt%284%2A10%29
c=2sqrt%2810%29

c%5E2=a%5E2%2Bb%5E2
%28sqrt%2810%29%29%5E2=a%5E2%2B3%5E2
10=a%5E2%2B9
1=a%5E2
sqrt%281%29=a
1=a

Put those values on the triangles:



Now

sin%28arctan%285%2Fsqrt%2815%29%29+-+arcsin%283%2Fsqrt%2810%29%29%29%29+%22%22=%22%22sin%28alpha-beta%29%22%22=%22%22sin%28alpha%29cos%28beta%29-cos%28alpha%29sin%28beta%29%22%22=%22%22

 %22%22=%22%22 5%2F%282%2A10%29%22%22-%22%223sqrt%2815%29%2F%282%2A10%29 %22%22=%22%22 5%2F20%22%22-%22%223sqrt%2815%29%2F20 %22%22=%22%22 %285-3sqrt%2815%29%29%2F20

Edwin