i presume this should be Sin^-1[sin(11pi/14)]Sin^-1(sin11(pi)/14)
short answer, if you sign something, and then take the sin^-1 of the answer, you just get back the original something you started with. obvious huh?
so Sin^-1[sin(11pi/14)] = 11pi/14 ........without working anything out! trick question, boy they are sly.
if you don't what i said before, i'll prove it
let's say we had sin^-1(sinx) = y
=> sinx = siny
=> x = y
so the original thing we applied sine to is the answer
same thing: Cos^-1[Cos(7(pi)/5)] = 7pi/5Cos^-1[Cos(7(pi)/5)]
this question is a little more interesting. let's start by following the hint.Sin[2csc^-1(5/3)] Hint: Let csc^-1(5/3) = theta, then use the appropriate double angle formula.
i don't know how to type theta, so let's just call it x
let csc^-1(5/3) = x
so Sin[2csc^-1(5/3)] = sin(2x) = sinxcosx + sinxcosx = 2sinxcosx
and they expected this to make it easier?! oh, but it is easier. let's examine csc^-1(5/3) a bit more
let csc^-1(5/3) = y
=> 5/3 = csc(y)
=> 5/3 = 1/sin(y) ..........this is a trig identity
=> 3/5 = sin(y) .............i took the inverse of both sides
=> sin^-1(3/5) = y
so we see that csc^-1(5/3) = sin^-1(3/5)
Aha. we have a sin^-1 translation of our guy, so we know what sinx is right off the bat (see the trick in the previous questions). now all that's left is to find cosx.
we had x = csc^-1(5/3) = sin^-1(3/5)
so cosx = cos[sin^-1(3/5)], now let's examine sin^-1(3/5)
let sin^-1(3/5) = z
=> 3/5 = sin(z)
now recall that sin^2(z) + cos^2(z) = 1
=> (3/5)^2 + cos^2(z) = 1
=> cos^2(z) = 1 - (3/5)^2 = 16/25
=> cos(z) = sqrt(16/25) = 4/5
so we have cos(z) = 4/5
=> cos^-1(4/5) = z ..............but wait! sin^-1(3/5) = z
so sin^-1(3/5) = cos^-1(4/5)
so cos[sin^-1(3/5)] = cos(cos^-1(4/5)) = 4/5 ....remember the trick
(note, we could also find out what cosx is by using a right-triangle, see diagram below)
sin(2x) = 2sinxcosx
this problem was very involved, if you didnt get it the first time around, go over it, it's really simple when you see how to use the trick.