SOLUTION: tan(sin^-1 3/4-cox^-1 1/7)

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Question 799078: tan(sin^-1 3/4-cox^-1 1/7)
Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
My solution may not be elegant, but I think it's right.
Given expression
(1) tan(arcsin(3/4) - arccos(1/7))
Let
(2) a = arcsin(3/4), then
(3) sin(a) = 3/4 and using
(4) sin%5E2%28a%29+%2B+cos%5E2%28a%29+=+1 we get
(5) cos%5E2%28a%29+=+1+-+%283%2F4%29%5E2 or
(6) cos%28a%29+=+sqrt%287%29%2F4
Now let
(7) b = arccos(1/7), then
(8) cos(b) = 1/7 and using
(9) sin%5E2%28b%29+%2B+cos%5E2%28b%29+=+1 we get
(10)+sin%5E2%28b%29+%2B+%281%2F7%29%5E2+=+1 or
(11) sin%5E2%28b%29+=+1+-+%281%2F7%29%5E2 or
(12)+sin%28b%29+=+%284%2F7%29%2Asqrt%283%29
Now a biggy!
Use the following identities
(13) tan%28a-b%29+=+%28sin%28a-b%29%29%2F%28cos%28a-b%29%29
(14) sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
(15) cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
There may be a more direct way to get (13), but I dont have an extensive table of trig identities in my brain. These are the only ones I remember.
Anyway substitute (3), (6), (8), and (12) into (14) and (15) to get
(16) and
(17)
Now put (16) and (17) into (13) to get the solution to the problem. Note that all four terms have the common denominator, 28. Since it is in a ratio, it cancels out to give us
(18)