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A = arcsin(1/8)
B = arccos(1/8)
The goal is to find A+B
As a slight detour, let's find out what sin(A+B) is
Recall from the list of trig identities that
sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
Let's compute each piece
sin(A) = sin(arcsin(1/8))
sin(A) = 1/8
The sine and inverse sine functions cancel each other out.
cos(B) = cos(arccos(1/8))
cos(B) = 1/8
We have similar cancellation going on.
cos(A) = cos(arcsin(1/8))
cos(A) = 3*sqrt(7)/8
See the figure below
I used the pythagorean theorem to determine the 3*sqrt(7) portion
sin(B) = sin(arccos(1/8))
sin(B) = 3*sqrt(7)/8
The steps are similar to the previous section
Now we can put everything together
sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
sin(A+B) = (1/8)*(1/8) + ( 3*sqrt(7)/8 )( 3*sqrt(7)/8 )
sin(A+B) = 1/64 + 63/64
sin(A+B) = (1 + 63)/64
sin(A+B) = 64/64
sin(A+B) = 1
A+B = arcsin(1)
A+B = pi/2
arcsin(1/8) + arccos(1/8) = pi/2
Answer: pi/2 radians (which is equivalent to 90 degrees)
As pointed out by the other tutor, there's nothing particularly special about the 1/8. It turns out that arcsin(x) + arccos(x) = pi/2 for any x on the interval of
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